Diagonalization with the given eigenvalue and its vector

Question

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

Answer 1

"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 $P$.

Answer 2

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.

Answer 3

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.