Question

Find the eigenvalues and eigenvectors of the real symmetric matrix

$\mathit{M}\mathbf{=}\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)$

Show that the eigenvalues are real and the eigenvectors are perpendicular.

Short answer

The eigenvalues are${\lambda }_{12}=\frac{A+B\pm \sqrt{{(A-B)}^2+4H^2}}{2}$ which are real, and the eigenvectors are$\frac{A+B\pm \sqrt{{(A-B)}^2+4H^2}}{2}x+Hy=0$ which are perpendicular.

Step-by-step solution

Given Information

The eigenvectors and eigenvalues for the real symmetric matrix

$M=\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)$

Symmetric Matrix

Eigenvalues are a unique collection of scalar values associated with a set of linear equations, most commonly found in matrix equations. Characteristic roots are another name for eigenvectors. It's a non-zero vector that can only be altered by its scalar factor once linear transformations are applied.

Eigen Values

$$\begin{gathered} Mr=\lambda r \\ foreigenvalues, \\ \left|\begin{array}{cc}A-\lambda & H\\ H& B-\lambda \end{array}\right|=0 \\ \left(A-\lambda \right)\left(B-\lambda \right)-H^2=0 \\ {\lambda }^2-(A+B)\lambda +AB-H^2=0 \end{gathered}$$

$$\begin{gathered} {\lambda }_{1,2}=\&\frac{A+B\pm \sqrt{{\left(A+B\right)}^2-4(AB-H^2)}}{2} \\ {\lambda }_{1,2}=\&\frac{A+B\pm \sqrt{{\left(A-B\right)}^2+4H^2}}{2} \end{gathered}$$

Since the matrix M is real, A , B and H are real.

The expression under the square root is positive so the eigenvalues are real.

The eigenvectors are

$\left(\begin{array}{cc}A& H\\ H& B\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)={\lambda }_{1,2}\left(\begin{array}{c}x\\ y\end{array}\right)$

$$\begin{gathered} \left(\begin{array}{cc}\frac{A-B\pm \sqrt{{\left(A-B\right)}^2+4H^2}}{2}& H\\ H& \frac{A-B\pm \sqrt{{\left(A-B\right)}^2+4H^2}}{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \\ \frac{A-B\pm \sqrt{{\left(A-B\right)}^2+4H^2}}{2}x+Hy=0 \end{gathered}$$

This is satisfied by$x=1$ and$\frac{A-B\pm \sqrt{{\left(A-B\right)}^2+4H^2}}{2}$ .

The eigenvectors are perpendicular by calculate their dot product

$$\begin{gathered} =1-\frac{A-B-\sqrt{{\left(A-B\right)}^2+4H^2}}{2H}.-\frac{A-B+\sqrt{{\left(A-B\right)}^2+4H^2}}{2H} \\ =1+\frac{{\left(A-B\right)}^2-{\left(A-B\right)}^2+4H^2}{4H^2} \\ =0 \end{gathered}$$