0
$\begingroup$

Consider the matrix $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$.

We define a new inner product over $\mathbb{R}^2$ given by $\langle\vec{v},\vec{w}\rangle=\vec{v}^T\cdot A\cdot \vec{w}$.

Find an orthonormal basis of the inner product space $(\mathbb{R}^2,\langle\cdot,\cdot\rangle)$ by applying Gram-Schmidt process to the set of vectors

$\left\{\vec{a}=\begin{pmatrix}1\\0\end{pmatrix},\vec{b}=\begin{pmatrix}0\\1\end{pmatrix}\right\}$


I haven't used Gram-Schmidt in conjunction with inner spaces yet, but I'd like to get ahead a bit. Any suggestions?


This is my best attempt so far:

$\vec{u_1} = \frac{\vec{a}}{\|\vec{a}\|}$, where $\|\vec{a}\|$ = $\sqrt{\begin{pmatrix}1&0\end{pmatrix}\cdot\begin{pmatrix}2&1\\1&2\end{pmatrix}\cdot\begin{pmatrix}1\\0\end{pmatrix}}=\sqrt{2}$;

$\therefore\,\vec{u_1}= \begin{pmatrix}\frac{1}{\sqrt{2}}\\0\end{pmatrix}$.

$\vec{b}^\perp = \vec{b}-(\vec{u_1}\cdot \vec{b})\,\vec{u_1}=\begin{pmatrix}0\\1\end{pmatrix}-\bigg(\vec{u_1}^T\cdot A\cdot \vec{b}\bigg)\vec{u_1}=\begin{pmatrix}0\\1\end{pmatrix}-\bigg[\begin{pmatrix}\frac{1}{\sqrt{2}}&0\\\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\bigg]\vec{u_1}\\=\begin{pmatrix}0\\1\end{pmatrix}-\bigg[\begin{pmatrix}\frac{2}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\bigg]\vec{u_1}=\begin{pmatrix}0\\1\end{pmatrix}-\frac{\vec{u_1}}{2}=\begin{pmatrix}-\frac{1}{2}\\1\end{pmatrix}$.

$\therefore \|\vec{b}^{\perp}\|=\vec{b}^{\perp T}\cdot A\cdot \vec{b}^\perp\\=\begin{pmatrix}-\frac{1}{2}&1\end{pmatrix}\cdot \begin{pmatrix}2&1\\1&2\end{pmatrix}\cdot\begin{pmatrix}-\frac{1}{2}\\1\end{pmatrix}=\frac{3}{2}$

So, an orthonormal basis of the product space in question is given by the $R$-matrix $R=\begin{pmatrix}\|\vec{a}\|&\vec{b}\cdot\vec{u_1}\\0&\|\vec{b}^\perp\|\end{pmatrix}=\begin{pmatrix}\sqrt{2}&\frac{1}{\sqrt{2}}\\0&\frac{3}{2}\end{pmatrix}$.


I'm not convinced with my own calculations here. Am I right to apply $\langle\vec{v},\vec{w}\rangle=\vec{v}^T\cdot A\cdot \vec{w}\,$ every time I use a scalar product? I'm especially concerned with the formula for $\vec{b}^\perp$, which involves (at least one?) scalar product. Input would be greatly appreciated.

$\endgroup$
  • 1
    $\begingroup$ Just perform Gram Schmidt using the inner product given above (also use it to compute the relevant norm $\|x\|_A = x^T Ax$). $\endgroup$ – copper.hat Mar 19 '15 at 5:02
1
$\begingroup$

OK, lets see this: let be $a=\begin{pmatrix}1\\0\end{pmatrix}$ and $b=\begin{pmatrix}0\\1\end{pmatrix}.$ Then we need $u,v$ such $\operatorname{span}\{u,v\}=\mathbb{R}^2$ and $\|u\|=\|v\|=1$, so: $$u=\dfrac{a}{\|a\|}$$ $$v'=b-\dfrac{\langle b,u\rangle}{\|u\|^2}u$$

If we make counts, $\|a\|=\sqrt{ \begin{pmatrix}1\\0\end{pmatrix}^T \begin{pmatrix}2&1\\1&2\end{pmatrix} \begin{pmatrix}1\\0\end{pmatrix}}=\sqrt{\begin{pmatrix}2&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}}=2$ So $u=a/2.$ Then $$v'=\begin{pmatrix}0\\1\end{pmatrix}-0.5\dfrac{\left\langle \begin{pmatrix}0\\1\end{pmatrix},u\right\rangle}{\|u\|^2}u$$

$$\Rightarrow\|u\|^2=1$$ $$\Rightarrow \langle b,u\rangle=0.5$$ So $v'=b-\dfrac{1}{4}u=\begin{pmatrix}-\dfrac{1}{8}\\1\end{pmatrix}.$ Remember that $\|\cdot\|=1$ is taken by the new norm induced by the inner product. Every inner product can be expresed by a matrix like in this problem.

Then $v=\dfrac{v'}{\|v'\|}$

$\endgroup$
  • $\begingroup$ If $\vec{u}=\frac{\vec{a}}{2}$, isn't $||\vec{u}||^2=\bigg(\sqrt{(\frac{1}{2})^2+(\frac{0}{2})^2}\bigg)^2=\frac{1}{4}?$ $\endgroup$ – Benjamin Loya Mar 20 '15 at 1:52
  • $\begingroup$ check your calculus because $||u||^2=||\dfrac{a}{||a||}||^2=\dfrac{1}{||a||}^2 ||a||^2=1$ $\endgroup$ – Luis Felipe Mar 20 '15 at 2:41
  • $\begingroup$ Similarly, isn't $\sqrt{\begin{pmatrix}2&1\end{pmatrix}\cdot\begin{pmatrix}1\\0\end{pmatrix}}= \sqrt{2}$? $\endgroup$ – Benjamin Loya Mar 20 '15 at 3:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.