Question

Consider a positive definite quadratic form q on${\mathbf{R}}^{\mathbf{n}}$with symmetric matrix. We know that there exists an orthonormal eigenbasis for ${\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{n}}$for A, with associated positive eigenvalues ${\mathbf{\lambda }}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{\lambda }}_{\mathbf{n}}$. Now consider the orthogonal Eigen basis ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{n}}$, where ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{\lambda }}}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}$.

Show that $\mathbf{q}\left(c_1{\overrightarrow{w}}_1+......+c_n{\overrightarrow{w}}_n\right)\mathbf{=}{\mathbf{c}}_{\mathbf{n}}^{\mathbf{2}}$.

Short answer

The diagonalizability of quadratic form is used here to prove.

Step-by-step solution

Given information

• Let q(x) > 0 be a positive definite quadratic form which is defined by symmetric matrix$\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}>0$for all $\mathrm{x}\ne 0$.
• The orthonormal eigenbasis$\mathrm{B}=\{\overrightarrow{{\mathrm{v}}_1}\overrightarrow{{\mathrm{v}}_2}\dots ,\overrightarrow{{\mathrm{v}}_{\mathrm{n}}}\}$is and the corresponding positive eigenvalues are${\mathrm{\lambda }}_1,{\mathrm{\lambda }}_2,...,{\mathrm{\lambda }}_{\mathrm{n}}$ of A, and te quadratic form is diagonalizable as $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{\lambda }}_1{\mathrm{c}}_1^2+{\mathrm{\lambda }}_2{\mathrm{c}}_2^2+...+{\mathrm{\lambda }}_{\mathrm{n}}{\mathrm{c}}_{\mathrm{n}}^2......\left(1\right)$.
• Here, are the coordinates of with regard to the eigenbasis B.

Application

• Defining $\mathrm{C}=\left\{{\overrightarrow{\mathrm{w}}}_1,{\overrightarrow{\mathrm{w}}}_2,...,{\overrightarrow{\mathrm{w}}}_{\mathrm{n}}\right\}$, where C is the orthogonal basis of .

$$\begin{gathered} \mathrm{q}({\mathrm{c}}_1{\overrightarrow{\mathrm{w}}}_1+{\mathrm{c}}_2\overrightarrow{\mathrm{w}}+....+{\mathrm{c}}_{\mathrm{n}}{\overrightarrow{\mathrm{w}}}_{\mathrm{n}})=\mathrm{q}\left({\mathrm{c}}_1\frac{\overrightarrow{{\mathrm{V}}_1}}{\sqrt{{\mathrm{\lambda }}_1}}+{\mathrm{c}}_2\frac{\overrightarrow{{\mathrm{V}}_2}}{\sqrt{{\mathrm{\lambda }}_2}}+...{\mathrm{c}}_{\mathrm{n}}\frac{\overrightarrow{{\mathrm{V}}_{\mathrm{n}}}}{\sqrt{{\mathrm{\lambda }}_{\mathrm{n}}}}\right) \\ =\mathrm{q}\left({\mathrm{c}}_1\frac{{\mathrm{c}}_1}{\sqrt{{\mathrm{\lambda }}_1}}\overrightarrow{{\mathrm{v}}_1}+\frac{{\mathrm{c}}_2}{\sqrt{{\mathrm{\lambda }}_2}}\overrightarrow{{\mathrm{V}}_2}+...\frac{{\mathrm{c}}_{\mathrm{n}}}{\sqrt{{\mathrm{\lambda }}_{\mathrm{n}}}}\overrightarrow{{\mathrm{V}}_{\mathrm{n}}}\right) \\ ={\mathrm{\lambda }}_1{\left(\frac{{\mathrm{c}}_1}{\sqrt{{\mathrm{\lambda }}_1}}\right)}^2+{\mathrm{\lambda }}_2{\left(\frac{{\mathrm{c}}_2}{\sqrt{{\mathrm{\lambda }}_2}}\right)}^2+...+{\mathrm{\lambda }}_{\mathrm{n}}{\left(\frac{{\mathrm{c}}_{\mathrm{n}}}{\sqrt{{\mathrm{\lambda }}_{\mathrm{n}}}}\right)}^2 \\ ={\mathrm{c}}_1^2+{\mathrm{c}}_2^2...+{\mathrm{c}}_{\mathrm{n}}^2 \end{gathered}$$

Result

It is proved using the diagonalizability of quadratic form.