Question

Show that $\mathbf{Tr}\left({\text{C}}^{-1}\text{MC}\right)\mathbf{=}\mathbf{Tr}\mathbf{\text{M}}$. Hint: See $\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{13}\mathbf{)}$. Thus, show that the sum of the eigenvalues of $\mathit{M}$ is equal to $\mathbf{Tr}\mathit{M}$.

Short answer

The total of a matrix's eigen values is the matrix's trace.

Step-by-step solution

Given information from question

The matrices$M=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$and$C=\left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)$

Eigen Values

Eigenvalues are a set of scalar values that are related with a set of linear equations, and are most typically seen in matrix equations. Eigenvectors are also known as characteristic roots. Once linear transformations are applied, it's a non-zero vector that can only be changed by its scalar factor.

Calculate the trace of a matrix is equal to the sum of its eigen values

$$\begin{gathered} C^{-1}=\left(\begin{array}{cc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.& \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.\\ \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.\end{array}\right) \\ =Tr\left(\left(\begin{array}{cc}\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.& \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.\\ \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.& \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.\end{array}\right)\times \left(\begin{array}{cc}\frac{1}{\sqrt{5}}& \frac{-2}{\sqrt{5}}\\ \frac{-2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right)\times \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\right) \\ =Tr\left(\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\times \left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\right) \\ =Tr\left(\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)\right) \end{gathered}$$

$$\begin{gathered} TrD=Tr\left(C^{-1}MC\right) \\ =TrC{C}^{-1}M \\ =TrM \end{gathered}$$

But since we know D is a diagonal matrix, and its diagonal elements are the eigenvalues for the matrix $M$, therefore the trace of a matrix is equal to the sum of its eigen values.