If the matrix, $$A = \begin{pmatrix}-3&0&0 \\\ -4&-7&8\\-4&-4&5 \end{pmatrix}$$
Find the characteristic polynomial of A and, hence, find all the eigenvalues of A.
Parametrizing (as a subset of $\mathbb{R}^3)$ the eigenspace of A that corresponds to the eigenvalue $\lambda = -3$. Is it possible to diagonalize A?
So far I have taken the determinant of $(A-\lambda I)$ and got the characteristic polynomial $-\lambda^{2} -5\lambda^{2} -3\lambda +9$. Then calculated the eigenvalues to be $\lambda = 1, -3$. After this, I am unsure how to proceed.
First, you should not expand the determinant of $A-\lambda I$, but try to factor it: \begin{align}\begin{vmatrix} -3-\lambda &&0 \\ -4&-7-\lambda &8 \\ -4&-4&5-\lambda\end{vmatrix}&=\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&3+\lambda&-3-\lambda\end{vmatrix}=(3+\lambda)\begin{vmatrix} -3-\lambda &&0\\-4&-7-\lambda &8 \\ 0&1&-1\end{vmatrix}\\[1ex] &=(3+\lambda)\Bigl[(-3-\lambda)(\lambda+7-8)\Bigr]=-(\lambda+3)^2(\lambda-1) \end{align}
Second, to determine if the matrix is diagonalisable, you have to check whether the geometric multiplicity is equal to the algebraic multiplicity of each of the eigenvalues.
For $1$, which is a simple eigenvalue, there is no problem. For the double eigenvalue $-3$, this means you have to check whether the corresponding eigenspace $\ker(A+3I)$ has dimension $2$.