Diagonalisability

Eigenvalues and eigenvectors: Eigenvalues and eigenvectors

Diagonalisability

The eigenvalues of a diagonal matrix are, almost by definition, the eigenvalues of such a matrix. Conversely, you might wonder whether the eigenvalues of a matrix also say something about the diagonalisability of the matrix, i.e., whether there is a similar diagonal matrix with the eigenvalues as non-zero elements. This is based on the following theorem.

Similar square matrices have the same eigenvalues.

If \(\lambda\) is an eigenvalue of the square matrix \(A\) and \(T\) an invertible matrix having the same size as that of \(A\), then \(\lambda\) is also an eigenvalue of \(T^{-1}\!AT\). Indeed, for an eigenvector \(\vec{v}\) corresponding to eigenvalue \(\lambda\) we have: \[\begin{aligned}(T^{-1}\!A\,T)(T^{-1}\vec{v})&= (T^{-1}\!A\,T\,T^{-1})\vec{v}\\ &= (T^{-1}\!A)\vec{v}\\ &=T^{-1}(A\vec{v}) \\ &=T^{-1}(\lambda\vec{v})=\lambda (T^{-1}\vec{v})\end{aligned}\] In other words, \(T^{-1}\vec{v}\) is an eigenvector of \(T^{-1}\!A\,T\) corresponding to the eigenvalue \(\lambda\).

The following theorem about diagonalisability follows from the above proposition.

Diagonalisability Let \(A\) be an \(n\times n\) matrix. Then \(A\) is diagonalisable if \(A\) has \(n\) distinct eigenvalues.

Furthermore, if \(T\) is the transformation matrix which has the eigenvectors corresponding to the distinct eigenvalues as columns, then \(T^{-1}A\,T\) is the diagonal matrix with the eigenvalues on the main diagonal.

We can rephrase the proposition above so that eigenvalues may be equal, but count more than once in the so-called eigendecomposition of a matrix.

Eigendecomposition of a matrix Let \(A\) be an \(n\times n\) matrix that has \(k\) distinct eigenvalues \(\lambda_1, \ldots, \lambda_k\) with the dimensions of the corresponding eigenspace equal to \(d_1,\ldots, d_k\). Then \(A\) is diagonalisable if and only if \(d_1+\cdots + d_k=n\).

In other words, an \(n\times n\) matrix is diagonalisable if and only is \(A\) has \(n\) linearly independent eigenvectors.

We already know that not every square matrix can be diagonalized; the matrix \(\matrix{1& 1\\ 0 & 1}\) is an example of a non-diagonalisable matrix. All the more remarkable is the following property of symmetric matrices.

Spectral theorem for symmetric matrices Any symmetric matrix is diagonalisable.

The spectral theorem is in fact even stronger by the assertion that the transformation matrix \(T\) which has eigenvectors as columns can be chosen such that all eigenvectors are of length 1 and are orthogonal to each other with respect to the standard dotproduct. In other words, the transformation matrix \(T\) can be chosen such that \(T\, T^{\top}=I\), that is, \(T\) is an orthogonal matrix. Furthermore, all eigenvalues are real for a real symmetric matrix.

In fact, the spectral theorem can be rephrased as follows:

Let \(A\) be an \(n\times n\) real matrix. The \(A\) is symmetric if and only if there exists an orthogonal matrix \(T\) (that is, with the property \(T\, T^{\top}=I\)) such that \(T^{\top}\!A\,T\) is a diagonal matrix.