Question-89
Let \(\displaystyle A\) be an orthogonal matrix. Find the number of true statements.
- \(\displaystyle A\) cannot have a zero eigenvalue.
- If \(\displaystyle \lambda\) is an eigenvalue of \(\displaystyle A\), \(\displaystyle \frac{1}{\lambda}\) is an eigenvalue of \(\displaystyle A^{T}\).
- If \(\displaystyle \lambda\) is a real eigenvalue of \(\displaystyle A\), then \(\displaystyle |\lambda |=1\).
\(3\)
For an orthogonal matrix, we have:
\[ \begin{equation*} A^{T} A=AA^{T} =I \end{equation*} \]
Statement-1
We see that \(\displaystyle A\) is invertible with its inverse being \(\displaystyle A^{T}\). So it doesn’t have a zero eigenvalue.
Statement-2
Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of \(\displaystyle A\). Then:
\[ \begin{equation*} \begin{aligned} Av & =\lambda v\\ A^{T} Av & =\lambda A^{T} v\\ \lambda A^{T} v & =v \end{aligned} \end{equation*} \]
Now, we can divide by \(\displaystyle \lambda\) only if \(\displaystyle \lambda \neq 0\), which is the case as shown above. Therefore, we divide both sides by \(\displaystyle \lambda\) to get:
\[ \begin{equation*} \begin{aligned} A^{T} v & =\frac{1}{\lambda } v \end{aligned} \end{equation*} \]
This shows that \(\displaystyle \frac{1}{\lambda }\) is an eigenvalue of \(\displaystyle A^{T}\).
Statement-3
If \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle A\), consider:
\[ \begin{equation*} \begin{aligned} ||Av||^{2} & =( Av)^{T}( Av)\\ & =v^{T} A^{T} Av\\ & =v^{T} v\\ & =||v||^{2} \end{aligned} \end{equation*} \]
Since \(\displaystyle Av=\lambda v\), \(\displaystyle ||Av||^{2} =\lambda ^{2} ||v||^{2}\). It follows that \(\displaystyle |\lambda |=1\).