Question-79

Let \(\displaystyle A\) be a \(\displaystyle 2\times 2\) matrix of real numbers. Which of the following are true?

• \(\displaystyle \text{det}\left( A^{2}\right)\) is non-negative
• \(\displaystyle \text{trace}\left( A^{2}\right)\) is non-negative
• All the elements of \(\displaystyle A^{2}\) are non-negative
• All the eigenvalues of \(\displaystyle A^{2}\) are non-negative
• If \(\displaystyle A\) has two distinct real eigenvalues, then so does \(\displaystyle A^{2}\)

Answer

• \(\displaystyle \text{det}\left( A^{2}\right)\) is non-negative
• \(\displaystyle \text{trace}\left( A^{2}\right)\) is non-negative
• All the elements of \(\displaystyle A^{2}\) are non-negative
• All the eigenvalues of \(\displaystyle A^{2}\) are non-negative
• If \(\displaystyle A\) has two distinct real eigenvalues, then so does \(\displaystyle A^{2}\)

Solution

Option-1

\[ \begin{equation*} \begin{aligned} \text{det}\left( A^{2}\right) & =\left[\text{det}( A)\right]^{2}\\ & \geqslant 0 \end{aligned} \end{equation*} \]

Option-2, 3, 4

These three options are false. Interestingly, all three can be addressed with the help of a single counter-example. The key is to look for a matrix whose eigenvalues are complex. Here is one such example:

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

We see that the elements of \(\displaystyle A^{2}\) could also be negative. The eigenvalues of \(\displaystyle A^{2}\) are both negative and its trace is also negative.

Option-5

This is true. Let the real eigenvalues of \(\displaystyle A\) be \(\displaystyle \lambda _{1} ,\lambda _{2}\):

\[ \begin{equation*} \begin{aligned} Av_{1} & =\lambda _{1} v_{1}\\ \Longrightarrow A^{2} v_{1} & =\lambda _{1}^{2} v_{1} \end{aligned} \end{equation*} \]

We see that \(\displaystyle \lambda _{1}^{2} ,\lambda _{2}^{2}\) are distinct eigenvalues of \(\displaystyle A^{2}\).