Question-92

orthogonal matrix
symmetric matrix
eigenvalue
eigenvector

Let \(\displaystyle A\) be square matrix of order \(\displaystyle 2\) that reflects vectors about a unit vector \(\displaystyle v\). \(\displaystyle v^{\perp }\) is a unit vector orthogonal to \(\displaystyle v\). Find the number of true statements from the ones given below.

\(6\)

Since \(\displaystyle A\) is a reflector about \(\displaystyle v\), it will leave \(\displaystyle v\) unchanged. That is, \(\displaystyle Av=v\), which implies that \(\displaystyle ( 1,v)\) is an eigenpair of \(\displaystyle A\). Since \(\displaystyle v^{\perp }\) is orthogonal to \(\displaystyle v\), reflecting it on \(\displaystyle v\) will result in \(\displaystyle -v^{\perp }\). That is, \(\displaystyle Av^{\perp } =-v^{\perp }\), which implies that \(\displaystyle \left( -1,v^{\perp }\right)\) is an eigenpair of \(\displaystyle A\). Using these two results and the fact that the matrix whose columns are \(\displaystyle v,v^{\perp }\) is orthogonal:

\[ \begin{equation*} \begin{aligned} A\begin{bmatrix} | & |\\ v & v^{\perp }\\ | & | \end{bmatrix} & =\begin{bmatrix} | & |\\ Av & Av^{\perp }\\ | & | \end{bmatrix}\\ & =\begin{bmatrix} | & |\\ v & -v^{\perp }\\ | & | \end{bmatrix}\\ \Longrightarrow A & =\begin{bmatrix} | & |\\ v & -v^{\perp }\\ | & | \end{bmatrix}\begin{bmatrix} - & v^{T} & -\\ - & v{^{\perp }}^{T} & - \end{bmatrix}\\ & \\ & =vv^{T} -v^{\perp } v{^{\perp }}^{T} \end{aligned} \end{equation*} \]

We see that \(\displaystyle A^{T} =A\), so \(\displaystyle A\) is symmetric. If \(\displaystyle Q_{1} =\begin{bmatrix} | & |\\ v & -v^{\perp }\\ | & | \end{bmatrix}\) and \(\displaystyle Q_{2} =\begin{bmatrix} | & |\\ v & v^{\perp }\\ | & | \end{bmatrix}\) then \(\displaystyle A=Q_{1} Q_{2}^{T}\), showing that \(\displaystyle A\) is orthogonal (product of orthogonal matrices is orthogonal). Since \(\displaystyle A\) is orthogonal, it is invertible.

All the statements given above are true.