# Eigenvectors for shear matrix and diagonalizing.

Here is a shear matrix $\begin{pmatrix} 1 && 0 \\ 2 && 1 \end{pmatrix}$.

The eigenvalues are 1. $\lambda^2 - 2 \lambda + 1 \to \lambda = 1$.

So now I try to find the eigenvectors.

$\begin{pmatrix} 1 -\lambda && 0 \\ 2 && 1-\lambda \end{pmatrix} \to \begin{pmatrix} 0 && 0 \\ 2 && 0 \end{pmatrix}$

$\begin{pmatrix} 0 && 0 \\ 2 && 0 \end{pmatrix} \cdot \{x_1, x_2\}$

It looks like both eigenvectors are $\{0, 0\}$.

But this is wrong! Mathematica reports the eigenvectors are $\{0, 1\}$ and $\{0, 0\}$. Where is this $\{0, 1\}$ eigenvector coming from?

The problem gives a hint that you should think about the geometric action of the shear matrix and whether this matrix is diagonalizable or not. I have no clue how that's relevant. Why is it?

Look at the second column of the given matrix. The column is $v=(0,1)^T$ and this mean that $Av=v$ so $v$ is an eigenvector associated to the eigenvalue $1$.
Since this matrix isn't diagonalizable then the eigenspace relative to the eigenvalue $1$ has the dimension $1$ and it's spanned by $v$.
• So it looks like you didn't bother with the $(A-\lambda I)v = 0$ calculation and pulled the eigenvector from the matrix directly. How did you know to do this? Is this a property of shear matrices that you happen to know or did do something else? I guess I'm puzzled that $(A-\lambda I)v = 0$ doesn't work. – Marty B. Oct 6 '14 at 16:32