Question

Given matrix $A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right]$ is type of

• A. Unitary
• B. Orthogonal
• C. Nilpotent
• D. Involuntary

Answer 1

The correct option is C

Nilpotent

The explanation for the correct option

For option C

The given matrix $A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right]$

$\Rightarrow A=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$.

Thus, $A^2=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]\times \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$

$$\begin{gathered} \Rightarrow A^2=\frac{1}{2}\left[\begin{array}{cc}\left(1\right)+\left(-1\right)& \left(1\right)+\left(-1\right)\\ \left(-1\right)+\left(1\right)& \left(-1\right)+\left(1\right)\end{array}\right] \\ \Rightarrow A^2=\frac{1}{2}\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

As per the definition, the given matrix $A$ is a nilpotent matrix.

The explanation for the incorrect options

For option A

It is known that, $A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\ne \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow A^2\ne I_2$, where $I_2$ is a $2\times 2$ identity matrix.

As per the definition, the given matrix $A$ is not a unitary matrix.

For option B

As $A=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]$, therefore the transpose of the matrix can be given by, $A^T=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]$.

Thus, $A\times A^T=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& -1\end{array}\right]\times \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]$

$$\begin{gathered} \Rightarrow A\times A^T=\frac{1}{2}\left[\begin{array}{cc}1+1& -1-1\\ -1-1& 1+1\end{array}\right] \\ \Rightarrow A\times A^T=\frac{1}{2}\left[\begin{array}{cc}1+1& -1-1\\ -1-1& 1+1\end{array}\right] \\ \Rightarrow A\times A^T=\frac{1}{2}\left[\begin{array}{cc}2& -2\\ -2& 2\end{array}\right] \\ \Rightarrow A\times A^T=\frac{1}{2}\times 2\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right] \\ \Rightarrow A\times A^T=\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right] \\ \Rightarrow A\times A^T\ne I_2 \end{gathered}$$

Thus, $A$ is not an orthogonal matrix.

For option D

It is known that, $A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\ne \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow A^2\ne I_2$, where $I_2$ is a $2\times 2$ identity matrix.

Thus, as per the definition, $A$ is not an involuntary matrix.

Hence, (C) is the correct option.