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Q67E

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Found in: Page 403

### Linear Algebra With Applications

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

# Consider a quadratic form q on with symmetric matrix A, with rank A = r. Suppose that A has p positive eigenvalues, if eigenvalues are counted with their multiplicities. Show that there exists an orthogonal basis ${\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{n}}}{\mathbf{}}{\mathbf{of}}{\mathbf{}}{{\mathbf{R}}}^{{\mathbf{n}}}$ such that ${\mathbf{q}}{\mathbf{\left(}}{{\mathbf{c}}}_{{\mathbf{1}}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{1}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{n}}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{n}}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbf{c}}}_{{\mathbf{p}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{p}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{p}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{c}}}_{\mathbf{p}\mathbf{+}\mathbf{1}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{-}}{{\mathbf{c}}}_{{\mathbf{r}}}^{{\mathbf{2}}}{\mathbf{.}}$. Hint: Modify the approach outlined in and 65.

The diagonalizability of a quadratic form and definiteness property are used here to prove this.

See the step by step solution

## Step 1 0f 2: Given information

• q(x) is the quadratic form, defined by symmetric matrix $\mathrm{A}\in {\mathrm{R}}^{\mathrm{n}×\mathrm{n}}$ , that is $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}$ with rank (A) = r and it has p positive eigen values.
• As rank (A) = r , and (n - r ) eigenvalues are zero and since p of the r nonzero eigen values are positive and (r - p) eigenvalue are negative.
• Let ${\lambda }_{1},{\lambda }_{2},.....,{\lambda }_{p}$ be the positive eigenvalues and the negative eigenvalues are role="math" localid="1659700981027" ${\lambda }_{p+1},{\lambda }_{p+2},.....,{\lambda }_{r}$ .
• ${\lambda }_{r+1},{\lambda }_{p+2},.....,{\lambda }_{r}$ are all zero eigenvalues.
• As per theorem 8.2.2, we know that for orthonormal eigenbasis $B=\left\{{\stackrel{\to }{v}}_{1},{\stackrel{\to }{v}}_{2},...{\stackrel{\to }{v}}_{n}\right\}$ and corresponding eigenvalue ${\mathrm{\lambda }}_{1},{\mathrm{\lambda }}_{2},...,{\mathrm{\lambda }}_{\mathrm{p}},{\mathrm{\lambda }}_{\mathrm{p}+1},....,\mathrm{\lambda r},{\mathrm{\lambda }}_{\mathrm{r}+1},....,{\mathrm{\lambda }}_{\mathrm{n}}$ of A, the above quadratic form is diagonalizable as $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{\lambda }}_{1}{\mathrm{c}}_{1}^{2}+{\mathrm{\lambda }}_{2}{\mathrm{c}}_{2}^{2}+...+{\mathrm{\lambda }}_{\mathrm{p}}{\mathrm{c}}_{\mathrm{p}}^{2}+{\mathrm{\lambda }}_{\mathrm{p}+1}{\mathrm{c}}_{\mathrm{p}+1}^{2}+...+{\mathrm{\lambda }}_{\mathrm{r}}{\mathrm{c}}_{\mathrm{r}}^{2}+{\mathrm{\lambda }}_{\mathrm{r}+1}{\mathrm{c}}_{\mathrm{r}+1}^{2}+...+{\mathrm{\lambda }}_{\mathrm{n}}{\mathrm{c}}_{\mathrm{n}}^{2}$, $\therefore \mathrm{q}\left(\mathrm{x}\right)={\mathrm{\lambda }}_{1}{\mathrm{c}}_{1}^{2}+{\mathrm{\lambda }}_{2}{\mathrm{c}}_{2}^{2}+...+{\mathrm{\lambda }}_{\mathrm{p}}{\mathrm{c}}_{\mathrm{p}}^{2}+{\mathrm{\lambda }}_{\mathrm{p}+1}{\mathrm{c}}_{\mathrm{p}+1}^{2}+...+{\mathrm{\lambda }}_{\mathrm{r}}{\mathrm{c}}_{\mathrm{r}}^{2}$ where ${\mathrm{\lambda }}_{\mathrm{i}}=0\mathrm{for}\mathrm{i}=\mathrm{r}+1,...\mathrm{n}$
• Here ${\mathrm{c}}_{\mathrm{i}}\mathrm{s}$ are the coordinates of x with respect to the eigenbasis B
• Let us define $C=\left\{{\stackrel{\to }{w}}_{1},{\stackrel{\to }{w}}_{2},....,{\stackrel{\to }{w}}_{n}\right\}$ where ${\stackrel{\to }{\mathrm{w}}}_{\mathrm{i}}=\frac{{\stackrel{\to }{\mathrm{v}}}_{\mathrm{i}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{i}}\right|}}\mathrm{for}\mathrm{i}=1,2...\mathrm{r}\mathrm{and}{\stackrel{\to }{\mathrm{w}}}_{\mathrm{j}}={\stackrel{\to }{\mathrm{v}}}_{\mathrm{j}}$ for $\mathrm{j}=\mathrm{r}+1,...,\mathrm{n}$ then, is orthogonal basis of as defined.

## Step 2 of 2: Application

$\mathrm{q}\left({\mathrm{c}}_{1}{\stackrel{\to }{\mathrm{w}}}_{1}+{\mathrm{c}}_{2}{\stackrel{\to }{\mathrm{w}}}_{2}+...+{\mathrm{c}}_{\mathrm{n}}{\stackrel{\to }{\mathrm{w}}}_{\mathrm{n}}\right)$

• role="math" localid="1659702408287" $={\mathrm{c}}_{1}\left({\mathrm{c}}_{1}\frac{{\stackrel{\to }{\mathrm{v}}}_{1}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}+{\mathrm{c}}_{2}\frac{{\stackrel{\to }{\mathrm{v}}}_{2}}{\sqrt{\left|{\mathrm{\lambda }}_{2}\right|}}+...+{\mathrm{c}}_{\mathrm{r}}\frac{{\stackrel{\to }{\mathrm{v}}}_{\mathrm{r}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{r}}\right|}}+{\mathrm{c}}_{\mathrm{r}+1}{\stackrel{\to }{\mathrm{v}}}_{\mathrm{r}+1}+...+{\mathrm{c}}_{\mathrm{n}}{\stackrel{\to }{\mathrm{v}}}_{\mathrm{n}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{{\mathrm{c}}_{1}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}{\stackrel{\to }{\mathrm{v}}}_{1}+...+\frac{{\mathrm{c}}_{\mathrm{p}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{p}}\right|}}{\stackrel{\to }{\mathrm{v}}}_{\mathrm{p}}+\frac{{\mathrm{c}}_{\mathrm{p}+1}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{p}+1}\right|}}{\mathrm{v}}_{\mathrm{p}+1}+\frac{{\mathrm{c}}_{\mathrm{r}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{r}}\right|}}{\stackrel{\to }{\mathrm{v}}}_{\mathrm{r}}+{\mathrm{c}}_{\mathrm{r}+1}{\mathrm{v}}_{\mathrm{r}+1}+{\mathrm{c}}_{\mathrm{n}}{\stackrel{\to }{\mathrm{v}}}_{\mathrm{n}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{\lambda }}_{1}{\left(\frac{{\mathrm{c}}_{1}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}\right)}^{2}+...+{\mathrm{\lambda }}_{\mathrm{p}}{\left(\frac{{\mathrm{c}}_{\mathrm{p}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{p}}\right|}}\right)}^{2}+{\mathrm{\lambda }}_{\mathrm{p}+1}{\left(\frac{{\mathrm{c}}_{\mathrm{p}+1}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{p}+1}\right|}}\right)}^{2}+...+{\mathrm{\lambda }}_{\mathrm{r}}{\left(\frac{{\mathrm{c}}_{\mathrm{r}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{r}}\right|}}\right)}^{2}+0.{\mathrm{c}}_{\mathrm{r}+1}^{2}+...+0.{\mathrm{c}}_{\mathrm{n}}^{2}$

, where ${\mathrm{\lambda }}_{1}=0\mathrm{for}\mathrm{i}=\mathrm{r}+1,...,\mathrm{n}$

• $={\mathrm{c}}_{1}^{2}+{\mathrm{c}}_{2}^{2}+...+{\mathrm{c}}_{\mathrm{p}}^{2}-{\mathrm{c}}_{\mathrm{p}+1}^{2}-{\mathrm{c}}_{\mathrm{p}+2}^{2}-...-{\mathrm{c}}_{\mathrm{r}}^{2},\mathrm{where}{\mathrm{\lambda }}_{\mathrm{i}}>0\mathrm{for}\mathrm{i}=1,....\mathrm{p}\mathrm{and}{\mathrm{\lambda }}_{\mathrm{j}}<0\phantom{\rule{0ex}{0ex}}\mathrm{j}=\mathrm{p}+1,...\mathrm{r}$

Result:

It is proved using diagonalizability of a quadratic form and definiteness property.