# Generalized Eigenvector: Determining the eigenspace

Let $T(z_1, z_2, z_3) = (4z_2 , 0 , 5z_3)$ and let $T$ be a linear map in $\mathbb{C}^3$. Find the eigenvectors of $T$ and the eigenspace of $T$ associated with each eigenvalue.

## Below, I show what I have done. According to the video, the eigenspace calculated when $\lambda = 0$ is incorrect. Work below.

$T(z_1, z_2, z_3) = (4z_2 , 0 , 5z_3)$. Let $A$ be the matrix of $T$.

$A= \begin{bmatrix} 0 & 0 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 5 \end{bmatrix}$

$A- \lambda I$, where $I$ is the identity matrix in $\mathbb{C}^3$, is defined below:

$A-\lambda I = \begin{bmatrix} 0 - \lambda & 0 & 0 \\ 4 & 0 - \lambda & 0 \\ 0 & 0 & 5 - \lambda \end{bmatrix}$

$\det (A- \lambda I) = (- \lambda)(- \lambda * (5 - \lambda)) - 0 - 0 = (- \lambda)^2 (5- \lambda)$, so $0$ and $5$ are eigenvalues of A.

To find the eigenspace, $E( \lambda , A)$, we find the vectors in the $null (A - \lambda I)$. So,

$0=(A-\lambda I)(v) = \begin{bmatrix} 0 - \lambda & 0 & 0 \\ 4 & 0 - \lambda & 0 \\ 0 & 0 & 5 - \lambda \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$

So, when $\lambda = 0$, $-\lambda v_1 = 0$ $\Rightarrow$ $v_1 = t$, $t \in \mathbb{C}$; $4v_1 + (-\lambda v_2) = 0 \Rightarrow v_1 = 0$, $v_2 = s, s\in \mathbb{C}$; and $(5- \lambda)v_3 = 0 \Rightarrow v_3 = 0$.

So, $E ( 0 , T ) = \{(0 , v_2 , 0): v_2 \in \mathbb{C}\}$

Similarly, when $\lambda = 5$, $E(5,T)=\{(0 , 0 , v_3): v_3 \in \mathbb{C}$.

The eigenspace when $\lambda = 5$ is right, but the eigenspace when $\lambda = 0$ is wrong. (in the video, Sheldon Axler states that the eigenspace associated with $\lambda = 0$ is $E ( 0 , T ) = \{(v_1 , 0, 0): v_1 \in \mathbb{C}\}$)

Reference: Linear Algebra Done Right, Sheldon Axler: Generalized Eigenvectors. Video: https://www.youtube.com/watch?v=xyhaYHGZN-w, at 5:57.

• $T(1,0,0)=(0,0,0)$ so $(1,0,0)$ is an eigenvector corresponding to the eigenvalue $0$ – Lozenges Jun 5 '17 at 14:53
• Your matrix is the transpose of the one you should be using. – amd Jun 5 '17 at 19:41

You wrote the matrix incorrectly. This would map the column vector $(z_1, z_2, z_3)^T$ onto $$\begin{pmatrix} 0&0&0 \\ 4&0&0 \\ 0&0&5 \end{pmatrix} \begin{pmatrix} z_1\\z_2\\z_3 \end{pmatrix} = \begin{pmatrix} 0\\4z_1\\5z_3 \end{pmatrix}$$ instead of $(4z_2,0,5z_3)^T$ as asked. Try again with $$A = \begin{pmatrix} 0&4&0 \\ 0&0&0 \\ 0&0&5 \end{pmatrix}$$ and you'll see that both Axler and your work are correct.