1. ## eigenvectors

Find a basis of the linear space V of all 2x2 matrices A for which
[1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

Attempt at a solution:
So A[1; -3] = (lambda)[1; -3].
[a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?

2. Originally Posted by noles2188
Find a basis of the linear space V of all 2x2 matrices A for which
[1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

Attempt at a solution:
So A[1; -3] = (lambda)[1; -3].
[a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?
Yes, that's a good direction. Now do the multiplication on the left to get two equations is a, b, c, d, and $\lambda$.

You should get $a- 3b= \lambda$ and $c- 3d= -3\lambda$. We need to get rid of $\lambda$ to have a condition on a, b, c, and d only so I would try dividing one equation by the other: $\frac{c-3d}{a-3b}= -3$ so c-3d= -3a+ 9b. You can choose a value of any 3 of those and solve for the remaining one. You can find a basis by taking each of the 3 values equal to 1 in turn and the other two 0.

3. Originally Posted by noles2188
Find a basis of the linear space V of all 2x2 matrices A for which
[1; -3] is an eigenvector, and thus determine the dimension of V (; denotes a new row).

Attempt at a solution:
So A[1; -3] = (lambda)[1; -3].
[a b; c d][1; -3] = (lambda)[1; -3]. Here I'm stuck--is this even the right direction to go in?

Yes, the direction looks right....let us continue a little more:

$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\ left(\begin{array}{c}1\\\!\!\!\!-3\end{array}\right)=\lambda\left(\begin{array}{c}1 \\\!\!\!\!-3\end{array}\right)\Longrightarrow \left(\begin{array}{c}a-3b\\c-3d\end{array}\right)=\left(\begin{array}{c}\lambda \\\!\!\!-3\lambda\end{array}\right)$ $\Longrightarrow\,3\lambda=3a-9b=3d-c=3\lambda\Longrightarrow\,3a-9b-3d+c=0$

So any matrix $\left(\begin{array}{cc}x&y\\w&z\end{array}\right)$ whose (ordered!) entries belong to the hyperplane $\left\{\left(\begin{array}{c}x\\y\\z\\w\end{array} \right)\in \mathbb{R}^4\,\,\slash \,3x-9y-3z+w=0\right\}$ will have the given vector $\left(\begin{array}{c}1\\\!\!\!\!-3\end{array}\right)$ as eigenvector
wrt the eigenvalue $\lambda=a-3b$

Tonio