Find a basis for the eigenspace corresponding to the eigenvalue

Question

The given matrix and eigenvalue are $$A = \begin{bmatrix}10 & -9 \\ 4 & -2 \end{bmatrix} \space \lambda=4$$ Applying the eigenvalue to the matrix $$\begin{bmatrix}10-4 & -9 \\ 4 & -2-4 \end{bmatrix}=\begin{bmatrix}6 & -9 \\ 4 & -6 \end{bmatrix}$$ Then taking the determinant $$\begin{vmatrix}6 & -9 \\ 4 & -6 \end{vmatrix}=-36-(-36)=0$$ This is where I'm lost, does this mean the eigenspace is the zero vector or that the eigenspace doesn't exist?

Answer 1

If $\lambda$ is an eigenvalue of $A$, then $\det(A-\lambda I)$ will always be zero. That's almost the definition of eigenvalue.

To find the eigenspace, you have to find all vectors $v$ for which $(A-\lambda I)v=0$. In your case, that is all $(x,y)$ for which $$\begin{bmatrix}6 & -9 \\ 4 & -6 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix}=0$$ Can you take it from there?