Step 1

The given statement is False

Step 2

Counter Example:

Let \(A=\begin{bmatrix}2 & -2i \\2i & 0 \end{bmatrix}\) and \(B=\begin{bmatrix}2 & 2i \\-2i & 0 \end{bmatrix}\)

A and B are Hermitian matrices

Now consider the product AB

\(AB=\begin{bmatrix}2 & -2i \\2i & 0 \end{bmatrix} \begin{bmatrix}2 & 2i \\-2i & 0 \end{bmatrix}=\begin{bmatrix}4+4i^2 & 4i-0 \\4i-0 & 4i^2 \end{bmatrix}\)

\(\Rightarrow AB=\begin{bmatrix}4-4 & 4i \\4i & -4 \end{bmatrix}=\begin{bmatrix}0 & 4i \\4i & -4 \end{bmatrix}\)

Consider characteristic equation for the matrix AB to get eigenvalue

\(det(A-\lambda I)=\begin{vmatrix}0-\lambda & 4i \\4i & -4-\lambda \end{vmatrix}=\begin{vmatrix}-\lambda & 4i \\4i & -4-\lambda \end{vmatrix}=0\)

\(\Rightarrow -\lambda(-4-\lambda)-(16i^2)=0\)

\(\Rightarrow 4\lambda+-\lambda^2+16=0\)

\(\Rightarrow \lambda^2+4\lambda+16=0\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{4^2-4(16)}}{2}\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{16-64}}{2}\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{48i}}{2}=-2\pm2 \sqrt{3}i\)

Clearly the eigenvalues are not real numbers.

Step 3

Answer:

The given statement is False.

The given statement is False

Step 2

Counter Example:

Let \(A=\begin{bmatrix}2 & -2i \\2i & 0 \end{bmatrix}\) and \(B=\begin{bmatrix}2 & 2i \\-2i & 0 \end{bmatrix}\)

A and B are Hermitian matrices

Now consider the product AB

\(AB=\begin{bmatrix}2 & -2i \\2i & 0 \end{bmatrix} \begin{bmatrix}2 & 2i \\-2i & 0 \end{bmatrix}=\begin{bmatrix}4+4i^2 & 4i-0 \\4i-0 & 4i^2 \end{bmatrix}\)

\(\Rightarrow AB=\begin{bmatrix}4-4 & 4i \\4i & -4 \end{bmatrix}=\begin{bmatrix}0 & 4i \\4i & -4 \end{bmatrix}\)

Consider characteristic equation for the matrix AB to get eigenvalue

\(det(A-\lambda I)=\begin{vmatrix}0-\lambda & 4i \\4i & -4-\lambda \end{vmatrix}=\begin{vmatrix}-\lambda & 4i \\4i & -4-\lambda \end{vmatrix}=0\)

\(\Rightarrow -\lambda(-4-\lambda)-(16i^2)=0\)

\(\Rightarrow 4\lambda+-\lambda^2+16=0\)

\(\Rightarrow \lambda^2+4\lambda+16=0\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{4^2-4(16)}}{2}\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{16-64}}{2}\)

\(\Rightarrow \lambda=\frac{-4\pm \sqrt{48i}}{2}=-2\pm2 \sqrt{3}i\)

Clearly the eigenvalues are not real numbers.

Step 3

Answer:

The given statement is False.