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Found in: Page 324

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Show that similar matrices have the same eigenvalues. Hint:$\stackrel{\mathbf{\to }}{\mathbf{v}}$ If is an eigenvector of ${{\mathbf{S}}}^{\mathbf{-}\mathbf{1}}{\mathbf{AS}}$, then role="math" localid="1659529994406" ${\mathbf{S}}\stackrel{\mathbf{\to }}{\mathbf{v}}{\mathbf{}}$ is an eigenvector of A.

We have proved that similar matrices have the same eigenvalues.

See the step by step solution

## Step 1: Definition of the Eigenvectors

Eigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.

## Step 2: Find eigenvalue

Assume, that is an Eigen vector for${\mathrm{S}}^{1}\mathrm{AS}$ .

Therefore, by definition:

${S}^{-1}A{S}^{r}v=\lambda {v}^{r}$

Now manipulate the above equation as shown below:

${\mathrm{S}}^{-1}\mathrm{AS}\stackrel{^}{\mathrm{v}}={\mathrm{\lambda v}}_{\mathrm{r}}^{\mathrm{r}}{\mathrm{SS}}^{-1}\phantom{\rule{0ex}{0ex}}\mathrm{ASv}={\mathrm{S\lambda }}_{\mathrm{r}}^{\mathrm{r}} \left(\mathrm{Multiplyby}\mathrm{S}\mathrm{fromLeft}\right)\phantom{\rule{0ex}{0ex}}\mathrm{ASv}=\mathrm{\lambda S}\stackrel{\mathrm{r}}{\mathrm{r}} \left(\mathrm{M}\right)$

From the above, the Eigen value of $A$ is $\lambda$and ${S}_{v}^{r}$ is the Eigenvector.

Similarly, Eigen vector of A is ${S}_{w}^{u}$ then ${{S}^{-1}}_{w}^{u}$ is an Eigen vector for ${S}^{-1}AS.$

Hence, similar matrices have same eigen values.