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Q35E

Expert-verifiedFound in: Page 324

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

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

We have proved that similar matrices have the same eigenvalues.

**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**.

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:

${\mathrm{S}}^{-1}\mathrm{AS}\hat{\mathrm{v}}={\mathrm{\lambda v}}_{\mathrm{r}}^{\mathrm{r}}{\mathrm{SS}}^{-1}\phantom{\rule{0ex}{0ex}}\mathrm{ASv}={\mathrm{S\lambda}}_{\mathrm{r}}^{\mathrm{r}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}(\mathrm{Multiplyby}\mathrm{S}\mathrm{fromLeft})\phantom{\rule{0ex}{0ex}}\mathrm{ASv}=\mathrm{\lambda S}\stackrel{\mathrm{r}}{\mathrm{r}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\left(\mathrm{M}\right)$

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.

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