Matrix:
\begin{bmatrix}
4& -4 &4 \\
-3& 3 &5 \\
1& -1&9
\end{bmatrix}
I want to check whether this matrix is diagonalizable and thus I want to calculate the eigenvalues. I know that this is possible with cofactor expansion and I have tried it and came to the right answer, but is there any FASTER way of getting eigenvalues of such matrix?
Best Answer
Since the first two columns are multiples of each other, so $\lambda=0$ is definitely an eigenvalue. Also $\text{trace}=16$, so sum of the other two eigenvalues is $16$.
Also observe that (by adding the second and third column) $$A\begin{bmatrix}0\\1\\1\end{bmatrix}=8\begin{bmatrix}0 \\ 1\\ 1\end{bmatrix}.$$
Thus $\lambda=8$ is a repeated eigenvalue (using the trace idea).