1. ## Eigenvectors/values

Let A = $\displaystyle \begin{bmatrix}1&-1&0\\-1&1&0\\0&0&1\end{bmatrix}$ and $\displaystyle T: R^3 \rightarrow R^3$ by T(X) = AX.
Find eigenpairs for A, find a basis for Ker(T) using the eigenvectors previously found, find a basis for Im(T) using eigenvectors previously found.

The eigenpairs I got were...
(0, $\displaystyle \begin{bmatrix}1\\1\\0\end{bmatrix}$), (1, $\displaystyle \begin{bmatrix}0\\0\\1\end{bmatrix}$), and (2, $\displaystyle \begin{bmatrix}-1\\1\\0\end{bmatrix}$

I wasn't sure about the Kernel or Image parts but for the kernel I got
t$\displaystyle \begin{bmatrix}1\\1\\0\end{bmatrix}$
And for the image I got
$\displaystyle [r\begin{bmatrix}0\\0\\1\end{bmatrix}, s\begin{bmatrix}-1\\1\\0\end{bmatrix}]$

Could someone please verify if this is correct or not? I assume the kernel and image are not so could someone also please help me with that?

Thanks

2. Yes, that is correct. Your matrix is 3 by 3 so the base space is 3 dimensional and you have three independent eigenvectors. Since there are 3 and they are independent (eigevectors corresponding to distinct eigenvalues are always independent), they form a basis for the space. That means that any vector v, can be written as $\displaystyle a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$.

Applying A to that, $\displaystyle Av= A(a\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix})$$\displaystyle = aA\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ bA\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ cA\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$$\displaystyle = a(0)\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}+ b(1)\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ c(2)\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$.

That is, Av= 0 if and only if $\displaystyle v= a\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ so $\displaystyle \{\begin{bmatrix}1 \\ 1\\ 0\end{bmatrix}\}$ is a basis for the kernel of A and $\displaystyle Av= b\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}+ 2c\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}$ so $\displaystyle \{\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, \begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}\}$ is a basis for the image.