To determine the properties of the matrix \( A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \), we will check if it is unitary, orthogonal, nilpotent, or involutory.
### Step 1: Check if \( A \) is Unitary
A matrix \( A \) is unitary if \( A^* A = I \), where \( A^* \) is the conjugate transpose of \( A \).
1. Find the transpose of \( A \):
\[
A^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}
\]
2. Since \( A \) has real entries, \( A^* = A^T \).
3. Calculate \( A^T A \):
\[
A^T A = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}
\]
Performing the multiplication:
- First row, first column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} + \frac{1}{2} = 1 \)
- First row, second column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} + \frac{1}{2} = 1 \)
- Second row, first column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} + \frac{1}{2} = 1 \)
- Second row, second column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} + \frac{1}{2} = 1 \)
Thus,
\[
A^T A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \neq I
\]
### Step 2: Check if \( A \) is Orthogonal
A matrix \( A \) is orthogonal if \( A^T A = I \). From the previous step, we found that \( A^T A \neq I \).
### Step 3: Check if \( A \) is Nilpotent
A matrix \( A \) is nilpotent if \( A^n = 0 \) for some positive integer \( n \).
1. Calculate \( A^2 \):
\[
A^2 = A \cdot A = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}
\]
Performing the multiplication:
- First row, first column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} - \frac{1}{2} = 0 \)
- First row, second column: \( \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{2} - \frac{1}{2} = 0 \)
- Second row, first column: \( -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
- Second row, second column: \( -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + (-\frac{1}{\sqrt{2}}) \cdot (-\frac{1}{\sqrt{2}}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
Thus,
\[
A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\]
Since \( A^2 = 0 \), \( A \) is nilpotent.
### Step 4: Check if \( A \) is Involutory
A matrix \( A \) is involutory if \( A^2 = I \). Since we found \( A^2 = 0 \), it is not involutory.
### Conclusion
The matrix \( A \) is nilpotent.
