Question-77 – GATE-DA

A square matrix \(\displaystyle A\) of order \(\displaystyle n\) is diagonalizable if \(\displaystyle \mathbb{R}^{n}\) has a basis made up of the eigenvectors of \(\displaystyle A\). Which of the following are true?

Answer

The characteristic polynomial of a diagonalizable matrix has \(\displaystyle n\) distinct roots.
If a matrix has \(\displaystyle n\) distinct eigenvalues, it is diagonalizable.
If \(\displaystyle A\) is diagonalizable, \(\displaystyle A+cI\) is diagonalizable for any \(\displaystyle c\in \mathbb{R}\).
A diagonalizable matrix is symmetric.
If a matrix is diagonalizable, it has to be a diagonal matrix.

Solution

Option-1

This is not necessary. Consider the identity matrix, which is diagonalizable. It has only one eigenvalue, namely \(\displaystyle 1\), repeated \(\displaystyle n\) times.

Option-2

Eigenvectors corresponding to distinct eigenvalues are linearly independent. We can use this fact to show that we can find a basis of eigenvectors for \(\displaystyle \mathbb{R}^{n}\).

Option-3

If \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle A\), \(\displaystyle ( \lambda +c,v)\) is an eigenpair of \(\displaystyle A+cI\):

\[ \begin{equation*} \begin{aligned} ( A+cI) v & =Av+cv\\ & =\lambda v+cv\\ & =( \lambda +c) v \end{aligned} \end{equation*} \]

It follows that \(\displaystyle A\) and \(\displaystyle A+cI\) have the same set of eigenvectors. It follows that \(\displaystyle A+cI\) is also diagonalizable.

Option-4

Symmetric matrices are diagonalizable. But all diagonalizable matrices need not be symmetric. For instance, consider:

\[ \begin{equation*} A=\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} \end{equation*} \]

\(\displaystyle A\) has two distinct eigenvalues, hence we can choose two linearly independent eigenvectors, which form a basis for \(\displaystyle \mathbb{R}^{2}\).

Option-5

The previous example can serve as a counter-example here as well.