7. Diagonalisation#

A very common example of a basis transformation is diagonalisation of a matrix. For this we first introduce the concept of eigenvalues and eigenvectors. These concepts are fundamentally important in quantum mechanics.

7.1. Eigenvalues and eigenvectors#

Consider an \(n\times n\)-matrix \(T\) with complex entries. We view \(T\) as a linear map on \(\CC^n\).

Definition 7.1 (Eigenvalues and eigenvectors)

An eigenvector for \(T\) is a non-zero vector \(\vv\) such that

\[ T\vv = \lambda\vv \]

for some scalar \(\lambda\). This \(\lambda\) is called the eigenvalue of \(T\) acting on \(\vv\).

Writing \(I\) for the \(n\times n\) identity matrix, we see that if \(\vv\) is an eigenvector with eigenvalue \(\lambda\), then

\[ (A-\lambda I)\vv = 0. \]

This means that \(A-\lambda I\) is not invertible; equivalently, its determinant vanishes. Thus, if \(\lambda\) is an eigenvalue of \(A\), we have

\[ \det(A-\lambda I)=0. \]

This implies that the eigenvalues of \(A\) are roots of the characteristic polynomial \(\chi_A\) (see Definition 5.6). We can factor the characteristic polynomial as

\[ \chi_A(t) = (t-\lambda_1)^{k_1}\cdots(t-\lambda_r)^{k_r} \]

where \(\lambda_1,\ldots,\lambda_r\) are the distinct roots of \(\chi_A(t)\) and each \(k_i\) is an integer called the multiplicity of \(\lambda_i\) as a root of \(\chi_A(t)\).

7.2. Diagonalising a matrix#

The idea of diagonalisation is that it is often easiest to understand a linear map \(A\) if we can find a basis of our space consisting of vectors on which \(A\) acts as multiplication by some scalar.

Definition 7.2 (Diagonalisation)

An \(n\times n\)-matrix \(A\) is diagonalisable if there exist a diagonal matrix \(D\) and an invertible \(n\times n\)-matrix \(P\) such that

\[ A = PDP^{-1}, \]

We can rewrite the above equation as \(AP=PD\). Let us write \(\vv_1,\ldots,\vv_n\) for the columns of \(P\) and \(\lambda_1,\ldots,\lambda_n\) for the diagonal entries of \(D\). Then we compute

\[ AP=\begin{pmatrix}A\vv_1& A\vv_2& \cdots & A\vv_n\end{pmatrix} \]


\[ PD=\begin{pmatrix}\lambda_1\vv_1& A\vv_2& \cdots & \lambda_n\vv_n\end{pmatrix} \]

We therefore see that the equation \(AP=PD\) comes down to the \(n\) equations \(A\vv_i=\lambda_i\vv_i\). In other words, each \(\vv_i\) is an eigenvector of \(A\) and \(\lambda_i\) is the corresponding eigenvalue.

The fact that \(P\) is invertible means that the eigenvectors \(\vv_i\) form a basis of \(\CC^n\). We obtain the following conclusion.

Theorem 7.1

An \(n\times n\)-matrix \(A\) is diagonalisable precisely when there exists a basis of \(\CC^n\) consisting of eigenvectors for \(A\).

This also gives us a way to tell whether \(A\) is diagonalisable, and if so to determine matrices \(D\) and \(P\) as above:

  1. Determine the eigenvalues \(\lambda_1,\ldots,\lambda_r\) of \(A\) by finding the roots of the characteristic polynomial \(\chi_A\).

  2. For each eigenvalue \(\lambda_i\), say with multiplicity \(k_i\), determine if there exist \(k_i\) linearly independent eigenvectors \(\vv_{\lambda_i,1},\ldots,\vv_{\lambda_i,k_i}\) with eigenvalue \(\lambda_i\).

  3. The matrix \(A\) is diagonalisable if and only if there exist \(k_i\) linearly independent eigenvectors with eigenvalue \(\lambda_i\) for all \(i\). In this case, the matrices

    \[\begin{split} D = \begin{pmatrix} \lambda_1 I_{k_1}& 0& \cdots& 0\\ 0& \lambda_2 I_{k_2}& \cdots& 0\\ \vdots& \vdots& \ddots& \vdots\\ 0& 0& \cdots& \lambda_r I_{k_r}\\ \end{pmatrix} \end{split}\]


    \[ P = \begin{pmatrix}\vv_{\lambda_1,1}& \cdots& \vv_{\lambda_1,k_1}& \vv_{\lambda_2,1}& \cdots& \vv_{\lambda_2,k_2}& \cdots& \vv_{\lambda_r,1}& \cdots& \vv_{\lambda_r, k_r}\end{pmatrix} \]

    give a diagonalisation of \(A\).

7.3. Exercises#

Exercise 7.1

Find the eigenvalues of the matrix

\[\begin{split} \begin{pmatrix}-8& 6\\ -9& 7\end{pmatrix} \end{split}\]

and for each eigenvalue find a corresponding eigenvector.

Exercise 7.2

Diagonalise the matrix \(\begin{pmatrix}a& 0\\ 0& d\end{pmatrix}\), where \(a\) and \(d\) are arbitrary scalars.

(Hint: think twice before doing unnecessary computations!)

Exercise 7.3

Consider the rotation matrix

\[\begin{split} \begin{pmatrix}\cos\phi& -\sin\phi\\ \sin\phi& \cos\phi\end{pmatrix}. \end{split}\]

(This represents a rotation around an angle \(\phi\) in the \((x,y)\)-plane.) Find the eigenvalues and eigenvectors of this matrix.

For which angles \(\phi\) are the eigenvalues real? Interpret your answer geometrically.

Exercise 7.4

  1. Assume \(A=PDP^{-1}\) with \(P\) invertible and \(D\) a scalar matrix, i.e. \(D\) is a scalar multiple of the identity matrix. Show that \(A=D\).

  2. Consider a \(2\times 2\)-matrix \(A\) with \(\tr(A)^2=4\det A\). Show that \(A\) is diagonalisable precisely when \(A\) is a scalar matrix.