# Let A and B be Hermitian matrices. Answer true or false for each of the statements that follow. In each case, explain or prove your answer. The eigenvalues of AB are all real.

Let A and B be Hermitian matrices. Answer true or false for each of the statements that follow. In each case, explain or prove your answer. The eigenvalues of AB are all real.
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Step 1
The given statement is False
Step 2
Counter Example:
Let $A=\left[\begin{array}{cc}2& -2i\\ 2i& 0\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 2i\\ -2i& 0\end{array}\right]$
A and B are Hermitian matrices
Now consider the product AB
$AB=\left[\begin{array}{cc}2& -2i\\ 2i& 0\end{array}\right]\left[\begin{array}{cc}2& 2i\\ -2i& 0\end{array}\right]=\left[\begin{array}{cc}4+4{i}^{2}& 4i-0\\ 4i-0& 4{i}^{2}\end{array}\right]$
$⇒AB=\left[\begin{array}{cc}4-4& 4i\\ 4i& -4\end{array}\right]=\left[\begin{array}{cc}0& 4i\\ 4i& -4\end{array}\right]$
Consider characteristic equation for the matrix AB to get eigenvalue
$det\left(A-\lambda I\right)=|\begin{array}{cc}0-\lambda & 4i\\ 4i& -4-\lambda \end{array}|=|\begin{array}{cc}-\lambda & 4i\\ 4i& -4-\lambda \end{array}|=0$
$⇒-\lambda \left(-4-\lambda \right)-\left(16{i}^{2}\right)=0$
$⇒4\lambda +-{\lambda }^{2}+16=0$
$⇒{\lambda }^{2}+4\lambda +16=0$
$⇒\lambda =\frac{-4±\sqrt{{4}^{2}-4\left(16\right)}}{2}$
$⇒\lambda =\frac{-4±\sqrt{16-64}}{2}$
$⇒\lambda =\frac{-4±\sqrt{48i}}{2}=-2±2\sqrt{3}i$
Clearly the eigenvalues are not real numbers.
Step 3
The given statement is False.
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Jeffrey Jordon

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