Diagonalization with the given eigenvalue and its vector Let $-3$ be an eigenvalue of a $3\times3$ singular matrix $P$ and 
$$P\begin{bmatrix}
5\\ 
3\\ 
-2
\end{bmatrix}=\begin{bmatrix}
-20\\ 
-12\\ 
8
\end{bmatrix}.$$
Then find whether $P$ is diagonalizable or not. Also find whether $P^2+3P$ is diagonalizable.
My attempt
Given $-3$ is one eigenvalue
 and from 
$$P\begin{bmatrix}
5\\ 
3\\ 
-2
\end{bmatrix}=\begin{bmatrix}
-20\\ 
-12\\ 
8
\end{bmatrix},$$ 
we know $Ax=\lambda x$. So we get $-4$ is another eigenvalue .
Also 
$$\begin{bmatrix}
5\\ 
3\\ 
-2
\end{bmatrix}$$ 
is an eigenvector corresponding to the eigen value $-4$.
I am struck now don't know how to find another eigenvalue and how to proceed further.
This was the question asked for NET exam for PhD entrance in India
 A: "Singular" means that $0$ is a third eigenvalue. With three distinct eigenvalues $0,-3,-4$, $P$ must be diagonalizable.
$P^2 + 3P$ is also diagonalizable with respect to the same basis as $P$.
A: A singular matrix is one which is not full rank, meaning that there is at least one row (or column) that is a linear combination of the others, since a matrix represents a linear basis-space.
If one of the eigenvalues has been given, then as shown, you can find a second eigenvalue of 4. Since the first two "discovered" eigenvalues are non-zero and the matrix is singular, thus not full-rank, then there must be a zero-valued eigenvalue.
You can also understand that a singular matrix has a determinant of 0. Since the determinant of a square matrix is equal to the product of its eigenvalues, the only way to have a zero-valued determinant is for at least one of those eigenvalues to be 0.
Once you've diagonalized a matrix, then you can now use the features of matrix calculus by applying functions of a matrix instead to its eigenmatrix.
$$f\left( \underline{\underline{A}} \right) = \underline{\underline{Q}} f\left( \underline{\underline{\Lambda}} \right) \underline{\underline{Q}}^{-1}
$$
Since the function is being applied to the diagonal matrix, it only affects the diagonal values, thus the result is still a diagonalizable matrix. This is a more practical view of understanding that the operations being applied to a diagonalizable matrix are equivalent to simply operating upon the underlying space's bases.
A: To diagonalize a matrix, only "distinct eigen values" is not enough...
We must consider the field over which, we are going to diagonalize.
You can easily get an example of a matrix, which has distinct eigen values but "not diagonalizable over Real numbers, But diagonalizable over complex numbers". 
Also there is a necessary and sufficient condition for diagonalizable, which is based on eigen vectors, not on eigen values... You can get that from any good book on linear algebra.
