# 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

## 3 Answers

"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 $P$.

• if a matrix have distict eigen values , it is diagonalizable? this is what you said right? @user165670 – srishti Jul 24 '14 at 6:15
• @srishti Yes, an $(n \times n)$-matrix with $n$ distinct eigenvalues is always diagonalizable. – user165670 Jul 24 '14 at 6:17
• what if it has 1,2,2 as eigen values? it may or may not diagonalize.am i right? – srishti Jul 24 '14 at 6:31
• @srishti That's right – user165670 Jul 24 '14 at 6:34

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.

• It should be: "each of its rows and columns is a linear combinations of the others" – enzotib Jul 24 '14 at 8:12
• @enzotib Thank you, I updated the first sentence to be clearer about that. – user164367 Jul 24 '14 at 8:15

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.

• can you suggest me a book? i couldn't find these topics anywhere – srishti Jul 24 '14 at 8:59
• I studied that condition in this book: Linear Algebra by Jin Ho Kwak and Sungpyo Hong – Saravanan Jul 24 '14 at 9:19