(Sec:QuadraticForm)= # Quadratic forms (Subsec:Terminology)= ## Introduction and terminology The simplest functions from $\R^n$ to $\R$ are linear functions $$ f(x_1,\ldots,x_n) = \sum_{i=1}^n a_ix_i +b \, =\, a_1x_1 + a_2x_2 + \cdots + a_nx_n + b. $$ In short $f(\mathbf{x}) = \mathbf{a}^T\mathbf{x} + b$, for some vector $\mathbf{a}$ in $\R^n$ and some number $b$ in $\R$. This is the common notion of linearity in calculus. To be linear in the linear algebra sense the constant term $b$ must be zero. After that come the *quadratic functions* ::::{math} :label: Eq:QuadForms:GeneralQuadForm q(x_1,\ldots,x_n) = \sum_{i,j=1}^{n} a_{ij}x_ix_j + \sum_{i=1}^{n} b_ix_i + c, :::: where all parameters $a_{ij}$, $b_i$ and $c$ are real numbers. A quadratic function in the two variables $x_1$, $x_2$ thus becomes $$ q(x_1,x_2) = a_{11}x_1^2 + a_{12}x_1x_2 + a_{21}x_2x_1 + a_{22}x_2^2 + b_1x_1 + b_2x_2 + c, $$ Note that this can be written as $$ q(x_1,x_2) = \begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} b_1 & b_2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + c. $$ In general, a shorthand representation of Equation {eq}`Eq:QuadForms:GeneralQuadForm` becomes $$ q(\mathbf{x}) = \mathbf{x}^TA\mathbf{x} + \mathbf{b}^T\mathbf{x} + c, $$ for an $n\times n$-matrix $A$, a vector $\vect{b}$ in $\R^n$, and a number $c$ in $\R$. The part $\mathbf{x}^TA\mathbf{x}$ is called a **quadratic form**. ::::{prf:example} :label: Ex:QuadForms:Diagonalize For the matrix $A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}$ the corresponding quadratic form is $$ \begin{array}{rcl} q(\vect{x}) = \begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 4& 5 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} &=& \begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} x_1 + 2x_2 \\ 4x_1+ 5x_2 \end{pmatrix} \\ &=& x_1^2 + (2+4)x_1x_2 + 5x_2^2 \\ &=& x_1^2 + 6x_1x_2 + 5x_2^2. \end{array} $$ Note that the last expression does not uniquely determine the matrix. We can split the coefficient $6$ of the term $x_1x_2$ in a different way and will end up with a different matrix. If we distribute it evenly over $x_1x_2$ and $x_2x_1$ we get a _symmetric_ matrix. $$ \begin{array}{rcl} x_1^2 + 6x_1x_2 + 5x_2^2 &=& x_1^2 + (3+3)x_1x_2 + 5x_2^2 \\ &=& x_1^2 + 3x_1x_2 + 3x_2x_1 + 5x_2^2 \\ &=& x_1(x_1 + 3x_2) + x_2(3x_1 + 5x_2) \\ &=& \begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} x_1 + 3x_2 \\ 3x_1+ 5x_2 \end{pmatrix}\\ &=&\begin{pmatrix} x_1 & x_2 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 3 & 5 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. \end{array} $$ :::: The above example leads to the first proposition about quadratic forms. ::::{prf:proposition} :label: Prop:QuadForms:SymmMatrixA Every quadratic form $q(\mathbf{x})$ can be written uniquely as $$ q(\mathbf{x}) = \mathbf{x}^TA\mathbf{x} $$ for a _symmetric_ matrix $A$. This symmetric matrix $A$ is then called **the matrix of the quadratic form**. :::: ::::{prf:example} :label: Ex:QuadForms:SymmetricMatrix We will find the symmetric matrix $A$ for the symmetric form $$ q(x_1,x_2,x_3) = x_1^2 + 2x_2^2 + 5 x_3^2 - 4 x_1x_2 + 6 x_2x_3. $$ So we need a symmetric matrix $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix}$. From $$ \mathbf{x}^TA\mathbf{x} = a_{11}x_1^2 + a_{22}x_2^2 + a_{33}x_3^2 + 2a_{12}x_1x_2 + 2a_{13}x_1x_3 + 2a_{23}x_2x_3 $$ we read off $$ a_{11} = 1, \, a_{22} = 2, \, a_{33} = 5, \,\,\text{and}\,\, a_{12} = -2, \, a_{13} = 0, \, a_{23} = 3. $$ So $A = \begin{pmatrix} 1 & -2 & 0 \\ -2 & 2 & 3 \\ 0 & 3 & 5 \end{pmatrix}$. :::: If we restrict ourselves to two variables, we see that the graph of a linear function $z = a_1x_1 + a_2x_2 + b$ is a plane. :::{figure} Images/Fig-QuadForms-Plane1.svg :name: Fig:QuadForms:Plane1 :class: dark-light The plane $z = \frac13x_1 - x_2 +2$. ::: The graph of a quadratic function is a curved surface. {numref}`Figure %s ` and {numref}`Figure %s ` show two of these quadratic surfaces. :::{figure} Images/Fig-QuadForms-QuadSurface1.png :name: Fig:QuadForms:QuadSurface1 :class: dark-light :width: 50% The surface $z = -\frac13x_1^2 + \frac13x_2^2 + 2 $. ::: :::{figure} Images/Fig-QuadForms-QuadSurface2.png :name: Fig:QuadForms:QuadSurface2 :class: dark-light :width: 50% The surface $z = -\frac12x_1^2 - \frac14x_2^2 + x_1 - x_2 + 2$. ::: The shape of the surfaces is in most cases determined by the quadratic part $\vect{x}^TA\vect{x}$. The linear part is then only relevant for the position. ::::{prf:example} :label: Es:QuadForms:Shift Consider the quadratic surface described by $$ z = x_1^2 + 2x_1x_2 + 3 x_2^2. $$ We will apply the shift $$ \tilde{x}_1 = x_1-3, \,\, \tilde{x}_2 = x_2 + 2, $$ so $$ x_1 = \tilde{x}_1+3, \,\,x_2 = \tilde{x}_2- 2. $$ In the new variables $(\tilde{x}_1,\tilde{x}_2)$ we get $$ \begin{array}[t]{rcl} z &=&(\tilde{x}_1+3)^2 + 2(\tilde{x}_1+3)(\tilde{x}_2- 2) + 3 (\tilde{x}_2- 2)^2 \\ &=& \tilde{x}_1^2 + 2\tilde{x}_1\tilde{x}_2 + 3 \tilde{x}_2^2 +2\tilde{x}_1-6\tilde{x}_2 + 9. \end{array} $$ Note that the quadratic parts are the same, $$ x_1^2 + 2x_1x_2 - 3 x_2^2 \quad \text{versus} \quad \tilde{x}_1^2 + 2\tilde{x}_1\tilde{x}_2 - 3 \tilde{x}_2^2. $$ :::: ::::{prf:example} :label: Ex:QuadForms:Shift2 The surfaces defined by $$ \mathcal{S}_1: z = 2x_1^2 - 2x_1x_2 + x_2^2 \quad \text{and} \quad \mathcal{S}_2: z = 2x_1^2 - 2x_1x_2 + x_2^2 + 8x - 6y + 4 $$ are shifted versions of the same surface. Namely, $$ 2x_1^2 - 2x_1x_2 + x_2^2 + 8x - 6y + 4 = 2(x_1+1)^2 - 2(x_1+1)(x_2-2) + (x_2-2)^2 - 6. $$ Thus $\mathcal{S}_2$ is also described by $$ z + 6 = 2(x_1+1)^2 - 2(x_1+1)(x_2-2) + (x_2-2)^2. $$ This means that if $\mathcal{S}_1$ is translated over the vector $\left(\begin{array}{c} -1 \\ 2 \\ -6 \end{array}\right)$ it becomes the surface $\mathcal{S}_2$. :::: For the rest of the section we will therefore only look at the quadratic part $\vect{x}^TA\vect{x}$. One of the simplest quadratic forms results when we take $A = I = I_n$, the identity matrix. Then we have $$ q(\vect{x}) = \vect{x}^TI_n\vect{x} = \vect{x}^T\vect{x} =x_1^2 + x_2^2 + \cdots + x_n^2 = \vect{x}\ip\vect{x}. $$ For this quadratic form, it is clear that it will only take on non-negative values. And that $$ q(\vect{x}) = 0 \quad \iff \quad \vect{x}=\vect{0}. $$ Such a quadratic form is called _positive definite_. In the next subsection we will learn how to find out whether an arbitrary quadratic form has this property. (Subsec:SignOfQuadForm)= ## Diagonalisation of quadratic forms Let us first consider an example, to get some feeling for what is going on. ::::{prf:example} :label: Ex:QuadForms:CompleteSquares Consider the quadratic form $$ q(x_1,x_2) = x_1^2 + 4x_1x_2 + 3x_2^2 = \begin{pmatrix}x_1 & x_2 \end{pmatrix} \begin{pmatrix}1 & 2 \\ 2 & 3 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix} = \vect{x}^TA\vect{x} . $$ At first sight you might think that this quadratic form only takes on non-negative values. One way to show that this is not actually true is by _completing squares_. $$ x_1^2 + 4x_1x_2 + 3x_2^2 = (x_1 + 2x_2)^2 - 4x_2^2 + 3x_2^2 = (x_1 + 2x_2)^2 - x_2^2. $$ For the last expression, that does not contain a _cross term_, we can see how we can get a negative outcome. We can make the first term equal to $0$ by taking $x_2 = 1$ and $x_1 = -2$, and then have $$ q(x_1,x_2) = q(-2,1) = (-2+2\cdot1)^2 - 1^2 = -1 < 0. $$ :::: One way to describe in a more abstract/general way what we did in {prf:ref}`Ex:QuadForms:CompleteSquares` is the following. We can introduce new variables $y_1, y_2$ via the _substitution_ $$ \left\{ \begin{array}{rr} y_1 =& x_1 + 2x_2 \\ y_2 =& x_2. \end{array} \right. $$ For short, $$ \vect{y} = \left(\begin{array}{c} y_1 \\ y_2 \end{array}\right) = \left(\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}\right)\left(\begin{array}{c} x_1 \\ x_2 \end{array}\right) = M\vect{x}. $$ Then in terms of the new variables the quadratic form becomes $$ q(\vect{x}) = (x_1 + 2x_2)^2 - x_2^2 = y_1^2 - y_2^2 = \begin{pmatrix}y_1 & y_2 \end{pmatrix} \begin{pmatrix}1 & 0 \\ 0 &-1 \end{pmatrix}\begin{pmatrix}y_1 \\ y_2 \end{pmatrix} = \vect{y}^TD\vect{y}. $$ Actually, it proves slightly advantageous to express the substitution as $\vect{x} = P\vect{y}$ for an invertible matrix $P$. We then have the following proposition. ::::{prf:proposition} Quadratic form under a substitution :label: Prop:QuadForms:Substitution The substitution $\vect{x} = P\vect{y}$ brings the quadratic form $$ q(\vect{x}) = \vect{x}^TA\vect{x} $$ over to $$ \tilde{q}(\vect{y}) = \vect{y}^TP^TAP\vect{y}. $$ :::: ::::{admonition} Proof of {prf:ref}`Prop:QuadForms:Substitution` :class: tudproof If we put $\vect{x} = P\vect{y}$ we get $$ q(\vect{x}) = \vect{x}^TA\vect{x} = (P\vect{y})^TA(P\vect{y}) = \vect{y}^TP^T A P \vect{y}. $$ So in terms of $\vect{y}$ we have the quadratic form $$ \tilde{q}(\vect{y}) = \vect{y}^T \tilde{A} \vect{y} $$ where $$ \tilde{A} = P^T A P. $$ :::: ::::{prf:example} :label: Ex:QuadForms:CompleteSquaresCtd In {prf:ref}`Ex:QuadForms:CompleteSquares` we considered the substitution $$ \vect{y} = \left(\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}\right)\vect{x} $$ or, equivalently $$ \vect{x} = \left(\begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array}\right)^{-1}\vect{y} = \left(\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right)\vect{y} = P \vect{y} $$ to the quadratic form $$ q(\vect{x}) = \vect{x}^T\left(\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right)\vect{x}. $$ We then have $$ \begin{array}{rcl} P^TAP &=& \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}^T\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \\ &=& \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \end{array} $$ According to {prf:ref}`Prop:QuadForms:Substitution` we then get the quadratic form $$ \tilde{q}(y) = \vect{y}^TP^TAP\vect{y} = \vect{y}^T\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\vect{y} = y_1^2 - y_2^2. $$ This agrees with what we derived in {prf:ref}`Ex:QuadForms:CompleteSquares`. :::: The technique of completing the squares is one way to 'diagonalise' a quadratic form. It may be turned into an algorithm that also works for quadratic forms in $n$ variables, but we will not pursue that track. There is a route that is more in line with the properties of symmetric matrices. Suppose $A$ is a symmetric matrix. We have seen (cf. {prf:ref}`Thm:SymmetricMat:OrthogDiag`) that it can be written as $$ A = QDQ^{-1} $$ for an orthogonal matrix $Q$. The diagonal matrix $D$ has the eigenvalues of $A$ on its diagonal. Since $Q$ is an orthogonal matrix, we have $$ A = QDQ^{-1} = QDQ^T. $$ If we compare this to {prf:ref}`Prop:QuadForms:Substitution` the following proposition results. ::::{prf:proposition} :label: Prop:QuadForms:Diagonalize Suppose $q(\vect{x})$ is a quadratic form with matrix $A$, i.e., $$ q(\vect{x}) = \vect{x}^TA\vect{x}. $$ Let $Q$ be an orthogonal matrix diagonalising $A$. That is, $A = QDQ^{-1}$. Applying the substitution $\vect{x} = Q\vect{y}$ then yields the quadratic form $$ \vect{y}^TD\vect{y} = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2, $$ where $\lambda_1, \ldots, \lambda_n$ are the _eigenvalues_ of the matrix $A$. :::: ::::{admonition} Proof of {prf:ref}`Prop:QuadForms:Diagonalize` :class: tudproof If we make the substitution $\vect{x} = Q\vect{y}$ we find that $$ \vect{x}^TA\vect{x} = (Q\vect{y})^TQDQ^{-1}(Q\vect{y}) = \vect{y}^TQ^TQD Q^{-1}Q\vect{y} = \vect{y}^TD\vect{y}. $$ The last expression is indeed of the form $$ \vect{y}^T\begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{pmatrix}\vect{y} = \lambda_1y_1^2 + \lambda_2y_2^2 + \cdots + \lambda_ny_n^2, $$ where $\lambda_1,\lambda_2, \ldots, \lambda_n$ are the eigenvalues of $A$. :::: Let us see how the construction of {prf:ref}`Prop:QuadForms:Diagonalize` works out in an earlier example. ::::{prf:example} :label: Ex:QuadForms:CompleteSquaresCtd2 Consider again the matrix $A = \left(\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right)$ of {prf:ref}`Ex:QuadForms:CompleteSquaresCtd`. Its characteristic polynomial is given by $$ p_A(\lambda) = (1-\lambda)(3-\lambda)-4 = \lambda^2 - 4\lambda - 1. $$ The eigenvalues are $$ \lambda_1 = 2 + \sqrt{5}, \quad \lambda_2 = 2 - \sqrt{5}. $$ So if we take $Q = \begin{pmatrix} \vect{q}_1 & \vect{q}_2 \end{pmatrix}$, where $\vect{q}_1$ and $\vect{q}_2$ are corresponding eigenvectors of unit length, we find that the substitution $\vect{x} = Q\vect{y}$ leads to $$ \vect{x}^TA\vect{x} \,\stackrel{\scriptsize \vect{x} = Q\vect{y}}{\longrightarrow}\, \vect{y}^TD\vect{y} = (2 + \sqrt{5})y_1^2 - (2 - \sqrt{5})y_2^2. $$ Since $(2 + \sqrt{5})> 0$ and $(2 - \sqrt{5})<2-2=0$ we may again conclude that the quadratic form takes on both positive and negative values. :::: ::::{prf:remark} :label: Rmk:QuadForms:SylvesterLaw In {prf:ref}`Ex:QuadForms:CompleteSquaresCtd` and {prf:ref}`Ex:QuadForms:CompleteSquaresCtd2` we applied two different substitutions to the same quadratic form with the matrix $A = \left(\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right)$. They led to the two different quadratic forms $$ \vect{y}^TD_1\vect{y} = \vect{y}^T\begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}\vect{y} \quad \text{and} \quad \vect{y}^TD_2\vect{y} = \vect{y}^T\begin{pmatrix}2 + \sqrt{5} & 0\\ 0 & 2 - \sqrt{5}\end{pmatrix}\vect{y}. $$ The diagonal matrices do not seem to have much in common. However, they do. It can be shown that if for a symmetric $n\times n$-matrix $A$ it holds that $$ P_1^TAP_1 = D_1 \quad \text{and} \quad P_2^TAP_2 = D_2, $$ for two invertible matrices $P_1$, $P_2$, then the _signs_ of the values on the diagonals of $D_1$ and $D_2$ match in the following sense: if $p_1$, $p_2$ denote the numbers of positive diagonal elements of $D_1, D_2$, and $n_i$ are the numbers of negative diagonal elements, then $$ p_1 = p_2 \quad \text{and} \quad n_1 = n_2. $$ It follows that the numbers of zeros on the diagonals, $n - p_i - n_i$, $i = 1,2$, must also be equal for the two matrices. In the two examples we see that $p_1 = p_2 = 1$ and also $n_1 = n_2 = 1$, in accordance with the statement. The property is known as _[Sylvester's Law of Inertia](https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia){target="_blank"}_. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/5a3d937b-6ecb-4fe8-b805-424af7e7ac55?id=90077 :label: grasple_exercise_8_2_T1 :dropdown: :description: To garner some evidence for Sylvester's Law of Inertia. :::: The following proposition is a direct consequence of the diagonalisation ({prf:ref}`Prop:QuadForms:Diagonalize`). ::::{prf:proposition} :label: Prop:QuadForms:MaximumxTAx Let $q(x) = \vect{x}^TA\vect{x}$ be the quadratic form with matrix $A$ and suppose that $A$ has the (ordered) eigenvalues $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$. Then the maximum and the minimum value attained by $q(\vect{x})$ under the constraint $\norm{\vect{x}} = 1$ are $\lambda_1$ and $\lambda_n$. :::: The proof contains the same type of reasoning as the proof of {prf:ref}`Prop:SymmetricMat:Max||Ax||`. ::::{admonition} Proof of {prf:ref}`Prop:QuadForms:MaximumxTAx` :class: tudproof, dropdown Suppose that $\vect{u}_1, \vect{u}_2,\ldots,\vect{u}_n$ is an orthonormal basis of eigenvectors for $A$ for the eigenvalues $\lambda_1 \geq \cdots \geq \lambda_n$. First of all $$ q(\vect{u}_1) = \vect{u}_1^TA\vect{u}_1 = \vect{u}_1^T(A\vect{u}_1) = \vect{u}_1^T(\lambda_1\vect{u}_1) = \lambda_1\vect{u}_1^T\vect{u}_1 = \lambda_1. $$ Likewise $q(\vect{u}_n) = \lambda_n$, so $q(\vect{x})$ does take on the values $\lambda_1$ and $\lambda_n$. Next, for an arbitrary unit vector $\vect{x}$, which can always be written as $$ c_1\vect{u}_1 + c_2\vect{u}_2 + \cdots + c_n\vect{u}_n, \quad \text{with} \quad c_1^2 + c_2^2 + \cdots + c_n^2 = 1 $$ (cf. proof of {prf:ref}`Prop:SymmetricMat:Max||Ax||`), we deduce that $$ \begin{array}{rcl} \vect{x}^TA\vect{x} &=& (c_1\vect{u}_1 + \cdots + c_n\vect{u}_n)^T(c_1\lambda_1\vect{u}_1 + \cdots + c_n\lambda_n\vect{u}_n) \\ &=& c_1^2\lambda_1\vect{u}_1^T\vect{u}_1 + \cdots + c_n^2\lambda_n\vect{u}_n^T\vect{u}_n \\ &=& c_1^2\lambda_1 + \cdots + c_n^2\lambda_n. \end{array} $$ All cross terms $\vect{u}_i^T\vect{u}_j$ with $i\neq j$ drop out since $\vect{u}_i\ip\vect{u}_j=0$, for $i\neq j$, and $\vect{u}_i^T\vect{u}_i = 1$ by the assumption that the vectors $\mathbf{u}_i$ are unit vectors. Now, invoking $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ and $c_1^2 + c_2^2 + \cdots +c_n^2 = 1$, we see that $$ \begin{array}{rcl} \lambda_n = \left(c_1^2 +c_2^2 +\cdots + c_n^2\right)\lambda_n &=& c_1^2\lambda_n +c_2^2\lambda_n +\cdots + c_n^2\lambda_n \\ &\leq& c_1^2\lambda_1 +c_2^2\lambda_2 +\cdots + c_n^2\lambda_n, \end{array} $$ and also $$ \lambda_1 \geq c_1^2\lambda_1 +c_2^2\lambda_2 +\cdots + c_n^2\lambda_n, $$ so we may conclude that indeed $$ \lambda_n \,\leq\, c_1^2\lambda_1 +c_2^2\lambda_2 +\cdots + c_n^2\lambda_n \,\leq\, \lambda_1, $$ where the expression in the middle is equal to our $\vect{x}^T A \vect{x} = q(\vect{x})$. :::: (Subsec:PosDefMatrices)= ## Positive definite matrices Let's start with a list of definitions. ::::{prf:definition} Classification of Quadratic Forms :label: Dfn:QuadForms:DefiniteMatrix Let $A$ be a symmetric matrix and $q_A(\vect{x}) = \vect{x}^TA\vect{x}$ the corresponding quadratic form. If none of the above applies, then $q_A$ is called an **indefinite quadratic form**. The same classification is used for symmetric matrices. E.g., $A$ is a **positive definite matrix** if the corresponding quadratic form is positive definite. :::: Note that every quadratic form $\vect{x}^TA\vect{x}$ gets the value $0$ when $\vect{x}$ is the zero vector. That is the reason we exclude the zero vector in the definition of positive/negative definite. The classification of a quadratic form follows immediately from the eigenvalues of its matrix. ::::{prf:theorem} :label: Thm:QuadForms:Classification Suppose $q_A(\vect{x}) = \vect{x}^TA\vect{x}$ for the symmetric $n \times n$-matrix $A$. Let $\lambda_1, \ldots, \lambda_n$ be the complete set of (real) eigenvalues of $A$ Then And lastly :::: ::::{admonition} Proof of {prf:ref}`Thm:QuadForms:Classification` :class: tudproof This immediately follows from {prf:ref}`Prop:QuadForms:Diagonalize`. If we make the substitution $\vect{x} = Q\vect{y}$ with the matrix $Q$ of the orthogonal diagonalisation, i.e., $$ A = QDQ^{-1} = QDQ^T, \quad D = \left(\begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & \cdots & \lambda_n \end{array}\right), $$ the quadratic form transforms to $$ \tilde{q}(\vect{y}) = \vect{y}^TD\vect{y} = \lambda_1y_1^2 + \cdots + \lambda_ny_n^2. $$ Let us consider the case where all eigenvalues $\lambda_i$ are _positive_. Then the expression for $\tilde{q}(\vect{y})$ is positive for all $\vect{y} \neq \vect{0}$. It remains to show that then also $q(\vect{x}) > 0$ for all vectors $\vect{x}\neq \vect{0}$. Since $Q$ is an orthogonal matrix it is also an invertible matrix. So any _non-zero_ vector $\vect{x}$ can be written as $$ \vect{x} = Q\vect{y} $$ for a (unique) _non-zero_ vector $\vect{y}$. As a consequence, for any non-zero vector $\vect{x}$ we have $$ \vect{x}^TA\vect{x} = (Q\vect{y})^TA(Q\vect{y}) = \vect{y}^TQ^TAQ\vect{y} = \vect{y}^TD\vect{y} > 0. $$ Likewise the other possibilities of the signs of the eigenvalues may be checked. :::: ::::{exercise} :label: Exc:QuadForms:CheckTheorem Verify the validity of the second statement made in {prf:ref}`Thm:QuadForms:Classification`. :::: :::{admonition} Solution to {numref}`Exc:QuadForms:CheckTheorem` :class: solution, dropdown The transformed quadratic form is still $$ \tilde{q}(\vect{y}) = \vect{y}^TD\vect{y} = \lambda_1y_1^2 + \cdots + \lambda_ny_n^2, $$ and $q_A(\vect{x})=\tilde{q}(\vect{y})$ with $\vect{x} = Q\vect{y}$. Now we can chain the following equivalent statements: $$ \begin{align*} \text{$q_A$ is positive definite} &\iff \text{$q_A(\vect{x})\geq0$ for all $\vect{x}$} \\ &\iff \text{$\tilde{q}(\vect{y})\geq0$ for all $\vect{y}$} \\ &\iff \text{$\lambda_1y_1^2 + \cdots + \lambda_ny_n^2\geq0$ for all $\vect{y}$} \\ &\iff \text{$\lambda_j\geq0$ for $j=1,\ldots,n$}. \end{align*} $$ ::: ::::{prf:example} :label: Ex:QuadForms:CompleteSquares2 Consider the quadratic form $$ q(x_1,x_2,x_3) = 2x_1^2 + x_2^2 +x_3^2 - 2x_1x_2 - 2x_1x_3. $$ The matrix of this quadratic form is $$ A = \left(\begin{array}{cc} 2 & -1 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right). $$ The eigenvalues of $A$ are computed as $$ \lambda_1 = 3, \quad \lambda_2 = 1, \quad \lambda_3 = 0. $$ With {prf:ref}`Thm:QuadForms:Classification` in mind, we can conclude that the quadratic form is positive semi-definite but not positive definite. :::: ::::{exercise} :label: Exc:QuadForms:CompleteSquares2 This exercise nicely recapitulates the ideas of the section. There is a cameo of the concept of completing the square, but that is of minor importance.
  1. Show that the quadratic from in {prf:ref}`Ex:QuadForms:CompleteSquares2` can be rewritten as follows :::{math} :label: Eq:QuadForms:CompleteSquares2 q(x_1,x_2,x_3) = 2(x_1 - \tfrac12x_2 - \tfrac12x_3)^2 + \tfrac12(x_2 - x_3)^2. :::
  2. What is the corresponding transformation $\vect{y} = P\vect{x}$ that brings the quadratic form in diagonal form $\vect{y}^TD_2\vect{y}$, and what is the diagonal matrix $D_2$?
  3. By inspection of $D_2$ find the classification of $q$.
  4. By inspection of Equation {eq}`Eq:QuadForms:CompleteSquares2`, find a non-zero vector $\vect{x}$ for which $q(\vect{x}) = 0$.
  5. Check that the vector you found in iii. is an eigenvector of the matrix of the quadratic form, i.e., $A = \left(\begin{array}{cc} 2 & -1 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)$.
:::: :::{admonition} Solution to {numref}`Exc:QuadForms:CompleteSquares2` :class: solution, dropdown
  1. We perform direct computations: $$ \begin{align*} q(x_1,x_2,x_3) &= 2x_1^2 + x_2^2 +x_3^2 - 2x_1x_2 - 2x_1x_3 \\ &= 2\left(x_1^2 - x_1x_2 - x_1x_3\right)+ x_2^2 +x_3^2 \\ &= 2\left(\left(x_1 - \frac12x_2 - \frac12x_3\right)^2-\left(\frac12x_2+\frac12x_3\right)^2\right)+ x_2^2 +x_3^2 \\ &= 2\left(\left(x_1 - \frac12x_2 - \frac12x_3\right)^2-\frac14x_2^2-\frac14x_2x_3-\frac14x_3^2\right)+ x_2^2 +x_3^2 \\ &= 2\left(x_1 - \frac12x_2 - \frac12x_3\right)^2-\frac12x_2^2-\frac12x_2x_3-\frac12x_3^2+ x_2^2 +x_3^2 \\ &= 2\left(x_1 - \frac12x_2 - \frac12x_3\right)^2+\frac12x_2^2-\frac12x_2x_3+\frac12x_3^2 \\ &= 2\left(x_1 - \frac12x_2 - \frac12x_3\right)^2+\frac12\left(x_2^2-x_2x_3+x_3^2\right) \\ &= 2(x_1 - \tfrac12x_2 - \tfrac12x_3)^2 + \tfrac12(x_2 - x_3)^2. \end{align*} $$
  2. We only present the answer here: $$ P=\begin{pmatrix} -\frac{2}{\sqrt6} & 0 & \frac{1}{\sqrt2} \\ \frac{1}{\sqrt6} & -\frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\ \frac{1}{\sqrt6}& \frac{1}{\sqrt2} & \frac{1}{\sqrt2} \end{pmatrix}\text{ and }D_2=\begin{pmatrix}3&0&0\\0&1&0\\0&0&0\end{pmatrix} $$
  3. $q$ is positive semi-definite.
  4. The last term of Equation {eq}`Eq:QuadForms:CompleteSquares2` gives rise to chosing $x_2=x_3$. The first term then indicates $x_1=\frac12x_2+\frac12x_3=x_3$. Chosing $x_3=1$ gives $x_2=1$ and $x_1=1$. In vector form: $$ \vect{x}=\begin{pmatrix}1\\1\\1\end{pmatrix}. $$
  5. The vector $\vect{x}=\begin{pmatrix}1\\1\\1\end{pmatrix}$ is a non-zero scalar multiple of the third column of $P$, so it is an eigenvector.
::: (Subsec:ConicSections)= ## Conic sections A _conic section_ or _conic_ is a curve that results when a circular cone is intersected with a plane. {numref}`Figure %s ` shows the different shapes when the plane is *not* going through the apex. :::{figure} Images/Fig-QuadForms-ConicSections.png :name: Fig:QuadForms:ConeWithPlanes :class: dark-light Intersections of a cone with several planes (not going through the apex). ::: The resulting curve is then either a _hyperbola_, a _parabola_ or an _ellipse_, with as special ellipse the _circle_. If the plane does go through the apex of the cone the conic section is called **degenerate**. ::::{exercise} :label: Exc:QuadForms:DegenerateConics Describe the (three) possible degenerate forms of conic sections. That is, what are the three different forms that result when a cone is intersected with a plane that goes through the apex? :::: ::::{admonition} Solution to {numref}`Exc:QuadForms:DegenerateConics` :class: solution, dropdown There are three essentially different situations. A plane through the apex may have the apex as unique common point with the cone. Then the conic (inter)section is just this **one point**. If (as in the central image in {numref}`Figure %s `) the plane is close to vertical, it will intersect the cone in **two lines** through the apex. The last form of a degenerate conic section is the transition between these situations, i.e. when the plane is tangent to the cone. This yields the last possible conic section, **a line**. :::{figure} Images/Fig-QuadForms-DegenerateConicSections.png :name: Fig:QuadForms:DegenerateConicSections :class: dark-light The three degenerate conic sections. ::: :::: In the plane all _non-degenerate_ conic sections may be described by a quadratic equation ::::{math} :label: Eq:ConicSec:MostGeneralFormula ax_1^2 + b x_1x_2 + cx_2^2 + dx_1 + ex_2 + f = 0, :::: where both the parameter $f$ and at least one of the parameters $a,b,c$ are not equal to zero. % When $f$ is equal to zero the conic section is sometimes called _degenerate_. ::::{prf:example} :label: Ex:ConicSec:Examples The curve given by the equation $x_1^2 + x_2^2 - 25 = 0$ is a circle with radius $5$. The equation $x_1^2 - x_2 - 2x_1 + 5 = 0$ gives a parabola with vertex ('top') at $(1, 4)$ and the line $x_1 = 1$ as axis of symmetry. :::: If the parameters $d$ and $e$ in Equation {eq}`Eq:ConicSec:MostGeneralFormula` are zero, the equation ::::{math} :label: Eq:ConicSec:CentralConic ax_1^2 + bx_1x_2 + cx_2^2 + f = 0 :::: is said to represent a **central conic**. When $b = 0$ as well, ::::{math} :label: Eq:ConicSec:StandardConic ax_1^2 + cx_2^2 + f = 0 :::: defines a central conic in **standard position**. Such a conic is symmetric with respect to both coordinate axes. If all parameters $a,c,f$ in Equation {eq}`Eq:ConicSec:StandardConic` are non-zero the equation can be rewritten in one of the two **standard forms** ::::{math} :label: Eq:QuadForms:StandardForms (I) \,\, \dfrac{x_1^2}{r_1^2} + \dfrac{x_2^2}{r_2^2} = 1, \quad\quad (II) \,\, \dfrac{x_1^2}{r_1^2} - \dfrac{x_2^2}{r_2^2} = \pm 1, :::: where we may assume that $ r_1,r_2 > 0$. In case $(I)$ the equation describes an ellipse if $r_1 \neq r_2$ and a circle if $r_1 = r_2$. In case $(II)$ the resulting curve is a hyperbola, with the lines $x_2 = \pm\dfrac{r_2}{r_1}x_1$ as asymptotes. Both curves have the coordinates axes as axes of symmetry. In this context they are also called the **principal axes**. See {numref}`Figure %s `. :::{figure} Images/Fig-QuadForms-EllipseHyperbola.svg :name: Fig:QuadForms:EllipseHyperbola :class: dark-light (Standard) Hyperbola and Ellipse. ::: ::::{prf:example} :label: Ex:QuadForms:StandardForm The equation $$ 4x_1^2 + 9x_2^2 - 25 = 0 $$ can be rewritten as $$ \dfrac{4x_1^2}{25} + \dfrac{9x_2^2}{25} = 1, $$ and a bit further to the standard form $$ \dfrac{x_1^2}{(5/2)^2} + \dfrac{x_2^2}{(5/3)^2} = 1. $$ The corresponding curve is an ellipse with the coordinate axes as principal axes and with $-5/2 \leq x_1 \leq 5/2$ and $-5/3 \leq x_2 \leq 5/3$. Likewise $$ 4x_1^2 - x_2^2 + 9 = 0 \quad \iff \quad \dfrac{x_1^2}{(3/2)^2} - \dfrac{x_2^2}{3^2} = -1. $$ When rewritten in the form $$ x_2 = \pm \sqrt{9 + 4x_1^2} = \pm2x_1\sqrt{\dfrac{9}{4x_1^2} + 1} $$ it is seen that the lines $x_2 = \pm 2x_1$ are asymptotes. Namely, if $x_1 \to \pm \infty$, then $$ \sqrt{\dfrac{9}{4x_1^2} + 1} \,\to\, \sqrt{0+1} = 1. $$ :::: If in Equation {eq}`Eq:ConicSec:CentralConic` the parameter $b$ is not equal to zero, the principal axes can be found by diagonalisation of the quadratic form $$ ax_1^2 + 2bx_1x_2 + cx_2^2 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}\begin{pmatrix} a & b \\ b & c\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. $$ The next proposition explains how.
For notational convenience we have denoted the coefficient of the cross term as $2b$. ::::{prf:proposition} :label: Prop:QuadForms:PrincipleAxesR2 Suppose the conic $\mathcal{C}$ is defined by the equation $$ ax_1^2 + 2bx_1x_2 + cx_2^2 = k, $$ where $a,b,c$ are not all equal to zero, and $k \neq 0$. Then the principal axes are the lines generated by the eigenvectors of the matrix $$ A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}. $$ :::: We will not give a proof of {prf:ref}`Prop:QuadForms:PrincipleAxesR2`, but instead we will give two illustrative examples. ::::{prf:example} :label: Ex:QuadForms:PrinciplesAxes1 We consider the quadratic form :::{math} :label: Eq:QuadForms:ConicExample1 x_1^2 - 4x_1x_2 + x_2^2 = 4. ::: Since $$ x_1^2 - 4x_1x_2 + x_2^2 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ -2 & 1\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, $$ {prf:ref}`Prop:QuadForms:PrincipleAxesR2` tells us we have to look for eigenvectors of the matrix $$ A = \begin{pmatrix} 1 & -2 \\ -2 & 1\end{pmatrix}. $$ The usual computations yield the following eigenvalues and eigenvectors: $$ \lambda_1 = 3,\,\vect{v}_1 = \begin{pmatrix} 1 \\ -1\end{pmatrix},\quad \lambda_2 = -1,\,\vect{v}_2 = \begin{pmatrix} 1 \\ 1\end{pmatrix}. $$ The eigenvectors are orthogonal, as they should, for a symmetric matrix. We see that $A$ can be orthogonally diagonalised as $$ A = QDQ^{-1} = QDQ^T, \quad Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}, \,\, D = \begin{pmatrix} 3 & 0 \\ 0 & -1\end{pmatrix}. $$ The substitution $\vect{x} = Q\vect{y}$ yields $$ \vect{x}^T A \vect{x} = \vect{y}^T Q^TAQ \vect{y} = \vect{y}^T D \vect{y} = 3y_1^2 - y_2^2. $$ So in the coordinates $y_1$ and $y_2$ the equation becomes $$ 3y_1^2 - y_2^2 = 4. $$ From this we can already conclude that the curve defined by Equation {eq}`Eq:QuadForms:ConicExample1` is a hyperbola. The principal axes in the $x_1$-$x_2$-plane are the lines given by $$ \mathcal{L}_1: \begin{pmatrix} x_1 \\ x_2\end{pmatrix} = c \begin{pmatrix} 1 \\ -1\end{pmatrix} \quad \text{and} \quad \mathcal{L}_2: \begin{pmatrix} x_1 \\ x_2\end{pmatrix} = c \begin{pmatrix} 1 \\ 1\end{pmatrix} $$ The asymptotes in the coordinates $y_1, y_2$ are the lines $$ y_2 = \pm\sqrt{3} y_1, \,\, \text{or} \,\,\, \begin{pmatrix} y_1 \\ y_2\end{pmatrix} = c \begin{pmatrix} 1 \\ \pm \sqrt{3}\end{pmatrix}. $$ Since $$ \begin{pmatrix} x_1 \\ x_2\end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix} y_1 \\ y_2\end{pmatrix} = \begin{pmatrix} \cos\left(-\frac14\pi\right) & -\sin\left(-\frac14\pi\right) \\ \sin\left(-\frac14\pi\right) & \cos\left(-\frac14\pi\right)\end{pmatrix}\begin{pmatrix} y_1 \\ y_2\end{pmatrix} $$ we find the asymptotes in the $x_1$-$x_2$-plane by rotating the lines $y_2 = \pm\sqrt{3}y_1$ over an angle $-\frac14\pi$. This leads to the direction vectors of the asymptotes in the $x_1$-$x_2$-plane as $$ \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix} 1 \\ 3\end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 4 \\ 2\end{pmatrix},\,\, \,\, \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix} 1 \\ -3\end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} -2 \\ -4\end{pmatrix}. $$ They can be simplified to the direction vectors $ \begin{pmatrix} 2 \\ 1\end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2\end{pmatrix}$. :::: ::::{prf:example} :label: Ex:QuadForms:PrinciplesAxes2 We consider the quadratic form :::{math} :label: Eq:QuadForms:ConicExample2 3x_1^2 + 4x_1x_2 + 6x_2^2 = 36. ::: Here we have $$ 3x_1^2 + 4x_1x_2 + 6x_2^2 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}\begin{pmatrix} 3 & 2 \\ 2 & 6\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}, $$ so now we have to look for eigenvalues and eigenvectors of the matrix $$ A = \begin{pmatrix} 3 & 2 \\ 2 & 6\end{pmatrix}. $$ They are found to be $$ \lambda_1 = 2,\,\vect{v}_1 = \begin{pmatrix} 2 \\ -1\end{pmatrix},\quad \lambda_2 = 7,\,\vect{v}_2 = \begin{pmatrix} 1 \\ 2\end{pmatrix}. $$ We orthogonally diagonalise $A$ as $$ A = QDQ^{-1} = QDQ^T, \quad Q = \frac{1}{\sqrt{5}}\begin{pmatrix} 2 & 1 \\ -1 & 2\end{pmatrix}, \,\, D = \begin{pmatrix} 2 & 0 \\ 0 & 7\end{pmatrix}. $$ The substitution $\vect{x} = Q\vect{y}$ yields the quadratic form $$ 2y_1^2 + 7y_2^2 = 36, $$ or $$ \dfrac{y_1^2}{(3\sqrt{2})^2} + \dfrac{y_2^2}{(6/\sqrt{7})^2} = 1. $$ This is an ellipse in the $y_1$-$y_2$-plane with long axis $6\sqrt{2}$, the length of the line segment from $(-3\sqrt{2},0)$ to $(3\sqrt{2},0)$, and short axis $\dfrac{12}{\sqrt{7}}$. For the ellipse in the $x_1$-$x_2$-plane we find the principal axes $$ \begin{pmatrix} x_1 \\ x_2\end{pmatrix} = c\vect{v}_1 = c\begin{pmatrix} 2 \\ -1\end{pmatrix} \quad \text{and}\quad \begin{pmatrix} x_1 \\ x_2\end{pmatrix} = c\vect{v}_2 = c\begin{pmatrix} 1 \\ 2\end{pmatrix}. $$ See Figure {numref}`Fig:QuadForms:Ellipses`. ```{applet} :url: symmetric_matrices/quadratic_forms :fig: Images/Fig-QuadForms-Ellipses(2).svg :name: Fig:QuadForms:Ellipses :class: dark-light The ellipse with its principal axes. ``` :::: ::::{exercise} :label: Exc:QuadForms:DegenerateStandardForm What happens if in Equation {eq}`Eq:ConicSec:CentralConic` the coefficient $f$ is equal to zero? (There are actually three cases to consider!) :::: ::::{admonition} Solution to {numref}`Exc:QuadForms:DegenerateStandardForm` :class: solution, dropdown We have to look at the solutions $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$ of the equation $ax_1^2 + bx_1x_2 + cx_2^2 = 0$, where not all three coefficients $a,b,c$ are zero. We can rewrite the equation as $$ \vect{x} ^TA\vect{x} = \vect{0},\text{ with }\vect{x}=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\text{ and }A=\begin{pmatrix} a & b \\ b & c \end{pmatrix}. $$ As $A$ is symmetric, it can be orthogonally diagonalised as $A=QDQ^T$, with $D$ a diagonal matrix with the eigenvalues $\lambda_1, \lambda_2$ of $A$ on the diagonal and $Q$ an orthogonal matrix with the corresponding eigenvectors as columns. The substitution $\vect{x} = Q\vect{y}$ transforms the equation to $$ \vect{y}^TD\vect{y} = 0, $$ or equivalently :::{math} :label: Eq:QuadForms:DegenerateStandardForm \lambda_1y_1^2 + \lambda_2y_2^2 = 0. ::: We now consider three cases: $\lambda_1\lambda_2 > 0$, $\lambda_1\lambda_2 < 0$ and $\lambda_1\lambda_2 = 0$. For $\lambda_1\lambda_2 > 0$, the only solution to {eq}`Eq:QuadForms:DegenerateStandardForm` is $\vect{y} = \vect{0}$, which means that the only solution to the original equation is $\vect{x} = \vect{0}$, i.e. the conic section is the point $(x_1,x_2)=(0,0)$. If $\lambda_1\lambda_2 < 0$, the two solutions to {eq}`Eq:QuadForms:DegenerateStandardForm` are $$ y_2 = \pm \sqrt{-\frac{\lambda_1}{\lambda_2}}y_1, $$ which are are with respect to the $(y_1,y_2)$-coordinates two intersecting lines through the origin, with slopes $\pm \sqrt{-\frac{\lambda_1}{\lambda_2}}$. As the transformation $\vect{x} = Q\vect{y}$ is invertible, the two intersecting lines in the $(y_1,y_2)$-plane are transformed to two intersecting lines through the origin in the $(x_1,x_2)$-plane. Finally, consider the case $\lambda_1\lambda_2 = 0$. Then one of the eigenvalues is zero, say $\lambda_1 = 0$, and the other is non-zero, say $\lambda_2 \neq 0$. The equation {eq}`Eq:QuadForms:DegenerateStandardForm` then becomes $$ \lambda_2y_2^2 = 0, $$ which has the solution $y_2 = 0$. The other variable $y_1$ can take any value, so the solution set is the line $y_2 = 0$ in the $(y_1,y_2)$-plane. The transformation $\vect{x} = Q\vect{y}$ is still invertible, so the solution set in the $(x_1,x_2)$-plane is also a line, namely the line through the origin with direction vector $\vect{v}_1$, the eigenvector corresponding to the zero eigenvalue. :::: ## Grasple exercises ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/b0668dff-a174-447b-8008-09b242a804fb?id=87448 :label: grasple_exercise_8_2_1 :dropdown: :description: To write down the matrix of a quadratic form in three variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/eaec14bd-fe7a-4c7a-a269-68e1d369bc2b?id=90207 :label: grasple_exercise_8_2_2 :dropdown: :description: To write down the matrix of a quadratic form in three variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/7044809f-28ca-4caf-b3fc-139010112ca1?id=90052 :label: grasple_exercise_8_2_3 :dropdown: :description: To perform a change of variables for a quadratic form $\vect{x}^TA\vect{x}$ in two variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/b71d8b9f-a3e8-48f6-b236-58f85a4818a6?id=90997 :label: grasple_exercise_8_2_4 :dropdown: :description: To classify a $3\times3$-matrix of which the characteristic polynomial is given. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/f0f9e677-eb21-4f84-9850-039b24ee0999?id=93112 :label: grasple_exercise_8_2_5 :dropdown: :description: To classify a quadratic form in two variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/f4657e73-219d-4ac9-bf17-81b240ddac96?id=93113 :label: grasple_exercise_8_2_6 :dropdown: :description: To classify a quadratic form in two variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/21ad829b-77f5-4e14-9808-4fbbb901c9b4?id=93119 :label: grasple_exercise_8_2_7 :dropdown: :description: To classify two quadratic forms in two variables. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/03704333-e9db-46f0-b292-eb235aee6b22?id=91025 :label: grasple_exercise_8_2_8 :dropdown: :description: For which value of a parameter $\beta$ is a quadratic form in two variables indefinite? :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/8b851997-932f-4ee6-8dfd-785bd7908e1c?id=91091 :label: grasple_exercise_8_2_9 :dropdown: :description: To describe three central conic sections geometrically. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/7c47dc31-b7d8-409e-9334-8a5de188c928?id=91912 :label: grasple_exercise_8_2_10 :dropdown: :description: Natural sequel to previous exercise. :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/ee9c377f-5150-4264-8d3e-1150c482fd7f?id=93116 :label: grasple_exercise_8_2_11 :dropdown: :description: For which parameter $a$ is a conic section $\vect{x}^TA\vect{x} =1$ an ellipse/hyperbola/something else? :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/51f56e96-3761-44c5-8d20-4cf0047a1ea4?id=93115 :label: grasple_exercise_8_2_12 :dropdown: :description: Maximising $\vect{x}^TA\vect{x}$ under the restriction $\norm{\vect{x}}=1$, for a $2\times2$-matrix $A$. :::: The following exercises are a more theoretical. ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/01a3d009-b2e0-4f2b-9d49-fcca955d6c5d?id=91048 :label: grasple_exercise_8_2_13 :dropdown: :description: True or false? If $A$ is a positive definite matrix, then the diagonal of $A$ is positive (v.v.). :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/7fea2ed7-c54f-4665-8ac8-611a8b0f6c5e?id=93114 :label: grasple_exercise_8_2_14 :dropdown: :description: If $A$ and $B$ are symmetric matrices with positive eigenvalues, what about $A+B$? :::: ::::{grasple} :iframeclass: dark-light :url: https://embed.grasple.com/exercises/78b55ab4-4f27-4d32-90ec-9bb28bb8f7b8?id=91021 :label: grasple_exercise_8_2_15 :dropdown: :description: Two True/False questions about vectors $\vect{x}$ for which $\vect{x}^TA\vect{x} = 0$. ::::