Question

Show that $\mathit{d}\mathit{e}\mathit{t}\left({\text{C}}^{-1}\text{MC}\right)\mathbf{=}\mathit{d}\mathit{e}\mathit{t}$ M. Hints: See (6.6). What is the product of $\mathit{d}\mathit{e}\mathit{t}\left({\text{C}}^{-1}\right)$ and det $\mathit{C}$ ? Thus, show that the product of the eigenvalues of $\mathit{M}$ is equal to $\mathit{d}\mathit{e}\mathit{t}\mathbf{\text{M}}$.

Short answer

The determinants of a matrix is equal to the product of its eigen values

Step-by-step solution

Given information from question

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

Eigen Values

Eigenvalues are a set of scalar values associated to a set of linear equations, and they are 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.

Calculate the determinants of a matrix is equal to the product of its eigen values

$$\begin{gathered} detD=det\left(C^{-1}MC\right) \\ =det{C}^{-1}\times detM\times detC \\ =\frac{1}{detC}\times detM\times detC \\ =detM \end{gathered}$$

But since, we know D is a diagonal matrix, then its determinant equals the product of its diagonal elements, and also its diagonal elements are the eigenvalues for the matrix $M$, therefore the determinants of a matrix is equal to the product of its eigen values.