<s>[INST] Find bases for eigenspaces of A $$A = \begin{pmatrix} 6 & 4 \\ -3 & -1\end{pmatrix}$$
Find the bases for eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$ of $A$.
I don't really know where to start on this problem.
 [/INST] First step: find the eigenvalues, via the characteristic polynomial $$\det(A - \lambda I) = \begin{vmatrix} 6 - \lambda & 4  \\ -3 & -1-\lambda \end{vmatrix} = 0 \implies \lambda^2 - 5\lambda + 6 = 0.$$
One of the eigenvalues is $\lambda_1 = 2$. You find the other one. 
Second step: to find a basis for $E_{\lambda_1}$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda_1 I)\mathbf{v} = \mathbf{0}$, in this case, we go for:
$$(A-2I)\mathbf{v} = 
\begin{pmatrix}
4 & 4 \\ -3 & -3
\end{pmatrix}
\begin{pmatrix} v_1 \\ v_2
\end{pmatrix} =
\begin{pmatrix}
0 \\ 0
\end{pmatrix} \implies
\begin{cases}
4v_1 + 4v_2 = 0 \\ -3v_1 -3v_2 = 0
\end{cases}
\implies v_1 = -v_2.$$
So, $\mathbf{v} = (v_1,v_2) = (v_1,-v_1) = v_1(1,-1)$, so $(1,-1)$ is a basis for that eigenspace with eigenvalue $\lambda_1$. Try to find a basis for the other one.
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