2.4. The Matrix-Vector Product \(A\vect{x}\)

In this section we will introduce another interpretation/representation of a system of linear equations.
We will define the product of an \(m\times n\) matrix \(A\) with a vector \(\vect{x}\) in \(\mathbb{R}^n\). In the next chapter this will also be the stepping stone to the general matrix-matrix product.

2.4.1. Definition of the Matrix-Vector Product

Definition 2.4.1

The product \(A\mathbf{x}\) of an \(m\times n\) matrix

\[ A = [\mathbf{a}_1 \,\,\mathbf{a}_2\, \ldots\, \mathbf{a}_n] \]

with a vector

\[\begin{split} \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\in \mathbb{R}^n \end{split}\]

is defined as

\[ A\mathbf{x} = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \ldots + x_n\mathbf{a}_n. \]

So: \(A\mathbf{x}\) is the linear combination of the columns of the matrix \(A\) with the entries of the vector \(\mathbf{x}\) as coefficients.

If the size \(n\) of the vector \(\mathbf{x}\) is not equal to the number of columns of the matrix \(A\), then the product \(A\mathbf{x}\) is not defined.

Example 2.4.1

\[\begin{split} \begin{bmatrix} 2 & 3 \\ 4 & 1 \\ 3 & 5 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ -1 \end{bmatrix} = 5 \begin{bmatrix} 2 \\ 4 \\ 3 \\ 0 \end{bmatrix} + (-1) \begin{bmatrix} 3 \\ 1 \\ 5 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \\ 15 \\ 0 \end{bmatrix} + \begin{bmatrix} -3 \\ -1 \\ -5 \\ -1 \end{bmatrix} = \begin{bmatrix} 7 \\ 19 \\ 10 \\ -1 \end{bmatrix}, \end{split}\]

and

\[\begin{split} \begin{bmatrix}1 & 2 & 3 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ -2 \\ -1 \\ 3\end{bmatrix} = 1\cdot4 +2\cdot(-2)+3\cdot(-1) + 5\cdot 3 = 12. \end{split}\]

The interpretation of \(A\mathbf{x}\) as a linear combination of the columns of \(A\) is important to keep in mind. That is, to not forget it after the following slightly easier way to compute the matrix-vector product.

Proposition 2.4.1 (Row-column rule)

The product of a matrix and a vector can also be computed as follows:

\[\begin{split} \left[\begin{array}{ccccc} a_{11} & a_{12}& \ldots& \ldots& a_{1n} \\ a_{21} & a_{22}& \ldots& \ldots& a_{2n} \\ \vdots & \vdots& \ldots& \ldots& \vdots \\ a_{m1} & a_{m2}& \ldots& \ldots& a_{mn} \end{array} \right] \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} a_{11}x_1 + a_{12}x_2 + \ldots+ a_{1n}x_n \\ a_{21}x_1 + a_{22}x_2 + \ldots+ a_{2n}x_n \\ \vdots\\ a_{m1}x_1 + a_{m2}x_2 + \ldots+ a_{mn}x_n \end{bmatrix}. \end{split}\]

Proof of Proposition 2.4.1 (Row-column rule)

The vector on the left-hand side of the identity is by definition equal to the linear combination

\[\begin{split} x_1 \begin{bmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{bmatrix} + x_2 \begin{bmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{bmatrix} + \,\,\ldots\,\, + x_n \begin{bmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{bmatrix}. \end{split}\]

And this is indeed equal to the vector on the right.

Note that the entry on the \(i\)-th position of the product, which is given by

\[ a_{i1}x_1 + a_{i2}x_2 + \,\,\ldots\,\,+ a_{in}x_n,\]

is the ‘row-column product’

\[\begin{split} \begin{bmatrix} a_{i1} & a_{i2} & \ldots & \ldots & a_{in} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ \vdots \\ x_n \end{bmatrix}.\end{split}\]

Example 2.4.2

We find a matrix-vector product using the row-column rule:

\[\begin{split} \begin{bmatrix} 3 & 4 & 5 \\ 1 & 0 & -1 \\ 2 & 2 & 4 \\ 5 & -5 & 2\end{bmatrix} \begin{bmatrix} 3 \\ 1 \\ -4 \end{bmatrix} = \begin{bmatrix} 3\cdot3\!\! &+&\!\! 4\cdot1\!\! &+&\!\! 5\cdot(-4) \\ 1\cdot3\!\! &+& \!\!0\cdot1\!\! &+&\!\! (-1)\cdot(-4) \\ 2\cdot3 \!\!&+&\!\! 2\cdot1 \!\! &+&\!\! 4\cdot(-4)\\ 5\cdot3 \!\!&+& \!\!\!(-5)\cdot1 \!\!\! &+&\!\! 2\cdot(-4) \end{bmatrix} = \begin{bmatrix} -7 \\ 7 \\ -8\\ 2\end{bmatrix}.\end{split}\]

Grasple Exercise 2.4.1

To check whether \(A\vect{x}\) exists, and if so, to compute it.

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Grasple Exercise 2.4.2

To check whether \(A\vect{x}\) exists, and if so, to compute it.

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Grasple Exercise 2.4.3

To check whether \(A\vect{x}\) exists, and if so, to compute it.

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Remark 2.4.1

From the above it follows that the matrix-vector equation

\[\begin{split} \left[\begin{array}{ccccc} a_{11} & a_{12}& \ldots& \ldots& a_{1n} \\ a_{21} & a_{22}& \ldots& \ldots& a_{2n} \\ \vdots & \vdots& \cdots& \cdots& \vdots \\ a_{m1} & a_{m2}& \ldots& \ldots& a_{mn} \end{array} \right] \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots\\ b_m\end{bmatrix} \end{split}\]

and the linear system

\[\begin{split} \left\{\begin{array}{ccccccccc} a_{11}x_1\! & \!+\!&\!a_{12}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{1n}x_n \! & \!=\!&\! b_1 \\ a_{21}x_1 \! & \!+\!&\!a_{22}x_2\! & \!+\!&\!\ldots\! & \!+\!&\!a_{2n}x_n \! & \!=\!&\! b_2 \\ \vdots \! & \!+\!&\! \vdots\! & \!+\!&\!\cdots\! & \!+\!&\! \vdots \! & \!=\!&\! \vdots \\ \vdots \! & \! \!&\! \vdots\! & \! \!&\!\cdots\! & \! \!&\! \vdots \! & \! \!&\! \vdots \\ a_{m1}x_1 \! & \!+\!&\!a_{m2}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{mn}x_n \! & \!=\!&\! b_m \end{array} \right. \end{split}\]

are one and the same thing!

So, we can see this linear system as

• a vector equation:

\[ x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \ldots + x_n\mathbf{a}_n = \mathbf{b} \]

or

• a matrix equation:

\[ A \mathbf{x} = \mathbf{b}. \]

As we will see later these different interpretations may lead to different insights.

Example 2.4.3

We want to write the system of equations

\[\begin{split} \left\{\begin{array}{ccccccc} 5x_1 & - & 3x_2 & -&2x_3 &=& 4 \\ 3x_1 & + & 7x_2 & -&2x_3 &=& 5 \\ 2x_1 & - &6x_2 & +&5x_3 &=& 6 \\ x_1 & & & +& x_3 &=& 8 \end{array} \right. \end{split}\]

in these two different forms.

The corresponding vector equation is

\[\begin{split} x_1 \begin{bmatrix} 5 \\ 3 \\ 2 \\ 1 \end{bmatrix} + x_2 \begin{bmatrix} -3 \\ 7 \\ -6 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -2 \\ -2 \\ 5 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \\ 6 \\ 8 \end{bmatrix}, \end{split}\]

and the corresponding matrix equation becomes

\[\begin{split} \begin{bmatrix} 5 & -3 & -2\\ 3 &7 & -2\\ 2&-6&5 \\ 1 &0&1 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \\ 6 \\ 8 \end{bmatrix}. \end{split}\]

Grasple Exercise 2.4.4

Rewriting a linear system to a matrix-vector equation

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Proposition 2.4.2

For each \(m\times n\) matrix \(A\), for all vectors \(\mathbf{x},\mathbf{y}\) in \(\mathbb{R}^n\), and for all scalars \(c\)

1. \(A\,(\mathbf{x}+\mathbf{y} ) = A\mathbf{x} + A\mathbf{y}\);

2. \(A\,(c\mathbf{x}) = c\,A\mathbf{x}\).

Proof of Proposition 2.4.2

We will prove the first of the two statements; the other statement goes in a similar fashion. There are several ways to derive the formula. Via the linear combination idea it may be the easiest. So assume

\[\begin{split} A = [\,\mathbf{a}_1\,\,\,\mathbf{a}_2\,\,\,\ldots\,\,\,\mathbf{a}_n\,], \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ \vdots \\ x_n \end{bmatrix}, \quad \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ \vdots \\ y_n \end{bmatrix}. \end{split}\]

Then

\[\begin{split} A\,(\mathbf{x}+\mathbf{y} ) = A\, \begin{bmatrix} x_1+y_1 \\ x_2+y_2 \\ \vdots \\ \vdots \\ x_n+y_n \end{bmatrix} = (x_1+y_1 )\mathbf{a}_1 + (x_2+y_2 )\mathbf{a}_2 + \ldots + (x_n+y_n )\mathbf{a}_n. \end{split}\]

Changing the order of the terms, putting the terms involving \(x_i\) to the front, shows that the last expression is equal to

\[ \big(x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \ldots + x_n\mathbf{a}_n\big)+ \big(y_1\mathbf{a}_1 + y_2\mathbf{a}_2 + \ldots + y_n\mathbf{a}_n\big). \]

The last sum of two vectors can be identified as being

\[ A\mathbf{x} + A\mathbf{y}. \]

Exercise 2.4.1

Prove statement (ii) of the previous proposition.

Solution to Exercise 2.4.1

Assume

\[\begin{split} A = [\,\mathbf{a}_1\,\,\,\mathbf{a}_2\,\,\,\ldots\,\,\,\mathbf{a}_n\,], \quad \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ \vdots \\ x_n \end{bmatrix}, \end{split}\]

and let \(c\) be any real number.

Then

\[\begin{split} A\,(c\mathbf{x}) = A\, \begin{bmatrix} cx_1 \\ cx_2 \\ \vdots \\ \vdots \\ cx_n \end{bmatrix} = cx_1\mathbf{a}_1 + cx_2\mathbf{a}_2 + \ldots + cx_n\mathbf{a}_n. \end{split}\]

In the last expression we can take the common factor \(c\) out to the front, and we see that it becomes equal to

\[ c(x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \ldots + x_n\mathbf{a}_n) = cA\mathbf{x}. \]

Using the above rules we can give shorter proofs of statements concerning linear systems. We illustrate this by having a second look at Proposition 2.3.1:

Example 2.4.4

The contents of that proposition: suppose \((c_{1},...,c_{n})\) is a solution of a linear system. Then \((c_{1}',...,c_{n}')\) is also a solution of the linear system if and only if there exists a solution \((d_{1},...,d_{n})\) of the associated homogeneous system such that \(c'_{i}=c_{i}+d_{i}\) for all \(i\).

Using the matrix-vector product we can derive this property as follows:  we can consider the solutions in vector form,

\[\begin{split} \mathbf{c} = \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ \vdots \\ c_n \end{bmatrix}, \quad \mathbf{c'} = \begin{bmatrix} c'_1 \\ c'_2 \\ \vdots \\ \vdots \\ c'_n \end{bmatrix}, \end{split}\]

and let \(A\) and \(\mathbf{b}\) have the obvious meanings.

It is then given that both

\[ A\mathbf{c} = \mathbf{b} \quad \text{and} \quad A\mathbf{c'} = \mathbf{b}. \]

From the rules just found it follows that

\[ A(\mathbf{c} -\mathbf{c'}) = A\mathbf{c} -A\mathbf{c'} = \mathbf{b} - \mathbf{b} = \mathbf{0},\]

which show that the vector

\[ (\mathbf{c} -\mathbf{c'}) = \mathbf{d} \]

is a solution of the homogeneous system. Note that

\[ (\mathbf{c} -\mathbf{c'}) = \mathbf{d} \iff \mathbf{c} = \mathbf{c'}+\mathbf{d} \,\,\iff \,\,c_i = c'_i + d_i,\,i=1,\ldots, n.\]

On the other hand, if \(\mathbf{c'}\) is a solution of the linear system

\[ A\mathbf{x} = \mathbf{b} \]

and \(\mathbf{d}\) is a solution of the homogeneous system

\[ A\mathbf{x} = \mathbf{0}, \]

then

\[ A(\mathbf{c'}+\mathbf{d}) = A\mathbf{c'}+A\mathbf{d} = \mathbf{b} + \mathbf{0} = \mathbf{b}, \]

so

\[ \mathbf{c} = \mathbf{c'} + \mathbf{d} \]

is a solution of the system

\[ A\mathbf{x} = \mathbf{b}. \]

The proof is basically the same as before, but using the matrix-vector product it can be written more concisely.

Exercise 2.4.2

(To practice with a proof as in the previous proposition.)

Suppose the linear system

\[\begin{split} \left\{\begin{array}{ccccccccc} a_{11}x_1\! & \!+\!&\!a_{12}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{1n}x_n \! & \!=\!&\! p_1 \\ a_{21}x_1 \! & \!+\!&\!a_{22}x_2\! & \!+\!&\!\ldots\! & \!+\!&\!a_{2n}x_n \! & \!=\!&\! p_2 \\ \vdots \! & \!+\!&\! \vdots\! & \!+\!&\!\cdots\! & \!+\!&\! \vdots \! & \!=\!&\! \vdots \\ \vdots \! & \! \!&\! \vdots\! & \! \!&\!\cdots\! & \! \!&\! \vdots \! & \! \!&\! \vdots \\ a_{m1}x_1 \! & \!+\!&\!a_{m2}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{mn}x_n \! & \!=\!&\! p_m \\ \end{array} \right. \end{split}\]

is consistent and the linear system

\[\begin{split} \left\{\begin{array}{ccccccccc} a_{11}x_1\! & \!+\!&\!a_{12}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{1n}x_n \! & \!=\!&\! q_1 \\ a_{21}x_1 \! & \!+\!&\!a_{22}x_2\! & \!+\!&\!\ldots\! & \!+\!&\!a_{2n}x_n \! & \!=\!&\! q_2 \\ \vdots \! & \!+\!&\! \vdots\! & \!+\!&\!\cdots\! & \!+\!&\! \vdots \! & \!=\!&\! \vdots \\ \vdots \! & \! \!&\! \vdots\! & \! \!&\!\cdots\! & \! \!&\! \vdots \! & \! \!&\! \vdots \\ a_{m1}x_1 \! & \!+\!&\!a_{m2}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{mn}x_n \! & \!=\!&\! q_m \\ \end{array} \right. \end{split}\]

is inconsistent. Show that the system

\[\begin{split} \left\{\begin{array}{ccccccccc} a_{11}x_1\! & \!+\!&\!a_{12}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{1n}x_n \! & \!=\!&\! r_1 \\ a_{21}x_1 \! & \!+\!&\!a_{22}x_2\! & \!+\!&\!\ldots\! & \!+\!&\!a_{2n}x_n \! & \!=\!&\! r_2 \\ \vdots \! & \!+\!&\! \vdots\! & \!+\!&\!\cdots\! & \!+\!&\! \vdots \! & \!=\!&\! \vdots \\ \vdots \! & \! \!&\! \vdots\! & \! \!&\!\cdots\! & \! \!&\! \vdots \! & \! \!&\! \vdots \\ a_{m1}x_1 \! & \!+\!&\!a_{m2}x_2\! & \!+\!&\! \ldots\! & \!+\!&\!a_{mn}x_n \! & \!=\!&\! r_m \\ \end{array} \right. \end{split}\]

where \(r_i = p_i - q_i,\,i=1,\ldots,\,m\) , is inconsistent.

Solution to Exercise 2.4.2

We start with some notations.

\[\begin{split} A = \left[\begin{array}{cccc} a_{11} & a_{12}& \ldots& a_{1n} \\ a_{21} & a_{22}& \ldots& a_{2n} \\ \vdots & \vdots& \cdots& \vdots \\ a_{m1} & a_{m2}& \ldots& a_{mn} \end{array} \right], \quad \mathbf{p} = \left[\begin{array}{c} p_1 \\ p_2 \\ \vdots \\ p_n \end{array}\right], \quad \mathbf{q} = \left[\begin{array}{c} q_1 \\ q_2 \\ \vdots \\ q_n \end{array}\right] \quad \mathbf{r} = \left[\begin{array}{c} r_1 \\ r_2 \\ \vdots \\ r_n \end{array}\right]. \end{split}\]

In matrix-vector form the assumptions are that the system \(A\vect{x}=\vect{p}\) is consistent and that the system \(A\vect{x}=\vect{q}\) is inconsistent. Morever, \(\vect{r} = \vect{p} - \vect{q}\).

We have to show that the system \(A\vect{x}=\vect{r}\) cannot be consistent.

Let \(\vect{x}_1\) be a solution for the (consistent) system \(A\vect{x}=\vect{p}\). If \(\vect{x}_2\) would be a solution for the system \(A\vect{x}=\vect{r}\), i.e., if \(A\vect{x}_2=\vect{r}\), then \(A(\vect{x}_1 -\vect{x}_2) = A\vect{x}_1 - A\vect{x}_2 = \vect{p} - \vect{r} = \vect{q}\),
so then the system \(A\vect{x}=\vect{q}\) would be consistent. This is in clear contradiction we the assumption that the last system is inconsistent.
So a solution for the system \(A\vect{x}=\vect{r}\) cannot exist.

We now revisit the question we left open at the end of Section 2.2: when is the span of a set of vectors in \(\R^{n}\) all of \(\R^{n}\)?

Proposition 2.4.3

Let \(\mathbf{v}_1, \ldots, \mathbf{v}_k\) be vectors in \(\mathbb{R}^n\). Define the matrix \(A\) such that

\[ A= \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \ldots & \mathbf{v}_k \end{bmatrix}. \]

The collection \(\Span{\mathbf{v}_1, \ldots, \mathbf{v}_k}\) is equal to \(\mathbb{R}^n\) if and only if the equation \(A \mathbf{x}=\mathbf{b}\) has a solution for each \(\mathbf{b}\) in \(\mathbb{R}^n\).

Proof of Proposition 2.4.3

If \(\Span{\mathbf{v}_1, \ldots, \mathbf{v}_k}\) is equal to \(\mathbb{R}^n\), then each vector \(\mathbf{b}\) is a vector in the span of the vectors \(\mathbf{v}_1, \ldots, \mathbf{v}_k\). This means that we can write \(\mathbf{b}\) as a linear combination

\[ \mathbf{b}=x_1\mathbf{v}_1+ \ldots + x_k\mathbf{v}_k. \]

Define the vector \(\mathbf{x}\) such that

\[\begin{split} \mathbf{x}= \begin{bmatrix} x_1 \\ \vdots \\ x_k \end{bmatrix}. \end{split}\]

By definition of the matrix-vector product we now have

\[ A\mathbf{x} = x_1\mathbf{v}_1+ \ldots + x_k\mathbf{v}_k = \mathbf{b}. \]

The proof of the other implication is similar.

Proposition 2.4.4

The equation \(A \mathbf{x}=\mathbf{b}\) has a solution for each \(\mathbf{b}\) in \(\mathbb{R}^n\) if and only if \(A\) has a pivot position in each row.

Proof of Proposition 2.4.4

Suppose that \(A\) does not contain a pivot position in each row. By definition of the reduced echelon form we know that the last row of \(A\) does not have a pivot position. If \(E\) is the reduced echelon form of \(A\), then this means that the bottom row of \(E\) contains only zeros. Let \(\mathbf{e}_n\) be again the vector of which the last entry is equal to 1 and all other entries are equal to zero.

Since \(E\) is the reduced form of \(A\) we can find a sequence of elementary row operations \(R_1, \ldots , R_m\) that transform in \(A\) into \(E\). Now take the augmented matrix \([E \, | \, \mathbf{e}_n]\) and perform the row operations \(R_m^{-1}, \ldots , R_1^{-1}\), where \(R_i^{-1}\) is the inverse row operation of \(R_i\). We obtain a matrix \([A \, | \, \mathbf{b}]\), where \(\mathbf{b}\) is a vector in \(R^n\). Because \([E \, | \, \mathbf{e}_n]\) is the reduced echelon form of the augmented matrix \([A \, | \, \mathbf{b}]\) and \([E \, | \, \mathbf{e}_n]\) has a pivot in the last column, we know that \([A \, | \, \mathbf{b}]\) is inconsistent. This means that \(A\mathbf{x}=\mathbf{b}\) does not have a solution.

On the other hand, if we assume that \(A\mathbf{x}=\mathbf{b}\) does not have a solution for some \(\mathbf{b}\) in \(\mathbb{R}^n\), then the reduced echelon form \([E \, | \, \mathbf{c}]\) of the augmented matrix \([A \, | \, \mathbf{b}]\) has a pivot in the last column. Let us assume that this pivot is located in row \(m\). The matrix \(E\) cannot have a pivot in row \(m\), but \(E\) is also the reduced echelon form of \(A\). This means that \(A\) has no pivot position in row \(m\).

Proposition 2.4.5

Let \(\mathbf{v}_1, \ldots, \mathbf{v}_k\) be vectors in \(\mathbb{R}^n\). Define the matrix \(A\) by

\[ A= \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \ldots & \mathbf{v}_k \end{bmatrix}. \]

The following statements are equivalent:

1. The set \(\Span{\vect{v}_1, \ldots, \vect{v}_k}\) is equal to \(\R^n\).

2. The equation \(A \vect{x}=\vect{b}\) has a solution for each \(\vect{b}\) in \(\R^n\).

3. The matrix \(A\) has a pivot position in each row.

Proof of Proposition 2.4.5

This follows from Proposition 2.4.3 and Proposition 2.4.4.

Example 2.4.5

Is the span of the following three vectors equal to \(\mathbb{R}^3\)?

\[\begin{split}\mathbf{v}_1= \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \quad \mathbf{v}_2= \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \quad \mathbf{v}_3= \begin{bmatrix} 3 \\5 \\-1 \end{bmatrix}\end{split}\]

We can use Proposition 2.4.5 to solve this problem. We will first use these vectors as the columns of a matrix \(A\).

\[\begin{split}A= \begin{bmatrix} 1 & 0 & 3 \\ 1 & 1 & 5 \\ -1 & 1 & -1 \end{bmatrix}\end{split}\]

The three given vectors span the entire space \(\mathbb{R}^3\) if and only if the matrix \(A\) has three pivot positions. Using elementary row operations we find that A has the following reduced echelon form:

\[\begin{split}A= \begin{bmatrix} 1 & 0 & 3 \\ 1 & 1 & 5 \\ -1 & 1 & -1 \end{bmatrix}\sim \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}. \end{split}\]

Since there are only two pivots in the reduced echelon matrix, we know that \(\mathbf{v}_1\), \(\mathbf{v}_2\) and \(\mathbf{v}_3\) do not span the space \(\mathbb{R}^3\).

Proposition 2.4.6

If \(\mathbf{v}_1, \dots ,\mathbf{v}_k\) are vectors in \(\mathbb{R}^n\) and \(k<n\), then the span of \(\mathbf{v}_1, \dots ,\mathbf{v}_k\) is not equal to \(\mathbb{R}^n\).

Proof of Proposition 2.4.6

Use the vectors \(\mathbf{v}_1, \dots ,\mathbf{v}_k\) as the columns for a matrix \(A\). By definition, the matrix \(A\) is an \(n\times k\) matrix. Let \(E\) be the reduced echelon form of \(A\). Since \(E\) has \(k\) columns we know that \(E\) can have at most \(k\) pivots. Because \(k<n\) this means that the number of pivots is less than \(n\). Therefore, we find that the number of pivots is less than the number of rows in \(E\). This implies that it is impossible for \(E\) to have a pivot in each row. Proposition 2.4.5 now tells us that the span of the vectors \(\mathbf{v}_1, \dots ,\mathbf{v}_k\) cannot be equal to \(\mathbb{R}^n\).

2.4.2. Grasple Exercises

Grasple Exercise 2.4.5

For a given matrix \(A\), does the equation \(A\vect{x}=\vect{b}\) have a solution for every \(\vect{b}\)?

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Grasple Exercise 2.4.6

For a given matrix \(A\), does the equation \(A\vect{x}=\vect{b}\) have a solution for every \(\vect{b}\)?

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Grasple Exercise 2.4.7

For a given matrix \(A\), does the equation \(A\vect{x}=\vect{b}\) have a solution for every \(\vect{b}\)?

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Grasple Exercise 2.4.8

Using a vector equation to find solution of a linear system.

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Grasple Exercise 2.4.9

A statement concerning two systems \(A\vect{x}=\vect{p}\), \(A\vect{x}=\vect{q}\).

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Grasple Exercise 2.4.10

If \(A\vect{x}=\vect{b}\) has a unique solution, what about \(A\vect{x}=\vect{0}\)?

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Grasple Exercise 2.4.11

What if the zero vector is a solution to \(A\vect{x}=\vect{b}\)?

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Grasple Exercise 2.4.12

About the ‘sum’ of two systems \(A_1\vect{x}=\vect{b}_1\), \(A_2\vect{x}=\vect{b}_2\).

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Grasple Exercise 2.4.13

About the ‘stack’ of two systems \(A_1\vect{x}=\vect{b}_1\), \(A_2\vect{x}=\vect{b}_2\).

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Grasple Exercise 2.4.14

What if \(A\vect{v}= A\vect{w} = \vect{b}\)?

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Grasple Exercise 2.4.15

Given \( A\vect{v} = A\vect{w} = \vect{b}\), how to create more solutions.

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