Find the values for which the matrix is diagonalizable

Question

I have this matrix A= $$\begin{bmatrix} 2&0&3\\0&L&0\\1&0&4\end{bmatrix}$$ and for find characteristic matrix I do $\lambda I - A$ so I've got $$\begin{bmatrix} \lambda - 2&0&-3\\0&\lambda-L&0\\-1&0&\lambda-4\end {bmatrix}$$ and here I'm in stuck, I think it's something simple but I can't to calcolate this: $(\lambda-2)(\lambda-L)(\lambda-4)-(3\lambda-L)=\lambda^3-6\lambda^2-L\lambda^2+6\lambda+6L\lambda-8L$ and it's not correct. Is here someone who can explain to me whot I'm doing wrong or whot rule or theorem I must to study. Here I used Sarrus for find determinant of the matrix.

Answer 1

Just use the fact that the middle column has two zeros:$$\begin{vmatrix}\lambda-2&0&-3\\0&\lambda-L&0\\-1&0&\lambda-4\end{vmatrix}=(\lambda-L)\begin{vmatrix}\lambda-2&-3\\-1&\lambda-4\end{vmatrix}.$$So, the eigenvalues are $L$ and the roots of the quadratic equation$$\begin{vmatrix}\lambda-2&-3\\-1&\lambda-4\end{vmatrix}=0.$$

Answer 2

You can find the eigenvalues of this matrix by inspection by looking for simple eigenvectors. The second column only has a non-zero entry in its second element, so $(0,1,0)^T$ is an eigenvector with eigenvalue $L$. Adding up the first and third columns gives you $(5,0,5)^T = 5(1,0,1)^T$, but this column addition is equivalent to multiplying by $(1,0,1)^T$, so that’s an eigenvector with eigenvalue $5$. You get the last eigenvalue “for free” by using the trace: $(L+6)-L-5=1$. The characteristic polynomial is therefore $(\lambda-L)(\lambda-5)(\lambda-1)$, so you’ll need to investigate what happens when $L=5$ or $L=1$.