Examples from Diagonalisation

A real matrix with complex eigenvalues

For

the characteristic polynomial is

Its eigenvalues are $\lambda=2+i$ and $\lambda=2-i$. For $\lambda=2+i$,

and solving $(A-(2+i)I)v=0$ gives

The conjugate eigenvalue has eigenvector $\begin{pmatrix}-i\\1\end{pmatrix}$. These two eigenvectors are independent over $\mathbb C$, so the matrix is diagonalisable over $\mathbb C$ although it has no real eigenvalues. Discarding the complex roots of a real matrix would miss this diagonalisation.

A repeated eigenvalue without enough modes

Let

Then

The only eigenvalue is $2$. Solving

gives

The algebraic multiplicity is $2$, but the geometric multiplicity is $\dim(E_2)=1$. There is only one independent eigenvector, so no eigenbasis of $\mathbb R^2$ exists and $A$ is not diagonalisable over either $\mathbb R$ or $\mathbb C$.

A triangular matrix with an eigenbasis

For

the eigenvalues are the diagonal entries $1$, $4$ and $6$. They are distinct, so the corresponding eigenvectors are independent. Solving $(A-\lambda I)v=0$ gives $v_1=(1,0,0)^{\mathsf T}$ for $\lambda=1$. For $\lambda=4$, the last two equations give $z=0$ and the first gives $3x=2y$, so choose $v_2=(2,3,0)^{\mathsf T}$. For $\lambda=6$, the second equation gives $2y=5z$; taking $z=10$ gives $y=25$ and the first equation then gives $x=16$. Thus one choice is

With

the column checks $Av_j=\lambda_jv_j$ give $AP=PD$ and hence

Orthonormal eigenvectors

For

the eigenvectors

are perpendicular and have length $\sqrt2$. Normalising them gives

Thus $Q=(\widehat u_1\ \widehat u_2)$ is orthogonal and