The matrix A = begin{bmatrix} frac{1}{sqrt{2} | vedclass

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(C) To determine the type of matrix $A$,we calculate $A^2$: $A^2 = A 	imes A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\

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Nilpotent

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(C) To determine the type of matrix $A$,we calculate $A^2$: $A^2 = A 	imes A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ $A^2 = \begin{bmatrix} (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) & (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) \ (-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) & (-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) \end{bmatrix}$ $A^2 = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} & \frac{1}{2} - \frac{1}{2} \ -\frac{1}{2} + \frac{1}{2} & -\frac{1}{2} + \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$ Since $A^2 = O$ (the null matrix),the matrix $A$ is a nilpotent matrix.

The matrix $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ is

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