# 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 \(\CC^n\).

(Eigenvalues and eigenvectors)

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

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

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

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

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.

(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

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

and

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.

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:

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

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 \(\lambda_i\).

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

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

\[ P = \begin{pmatrix}\vv_{\lambda_1,1}& \cdots& \vv_{\lambda_1,k_1}& \vv_{\lambda_2,1}& \cdots& \vv_{\lambda_2,k_2}& \cdots& \vv_{\lambda_r,1}& \cdots& \vv_{\lambda_r, k_r}\end{pmatrix} \]give a diagonalisation of \(A\).

## 7.3. Exercises#

Find the eigenvalues of the matrix

and for each eigenvalue find a corresponding eigenvector.

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!)

Consider the rotation matrix

(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.

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 \(A=D\).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.