Gram-Schmidt and Inner Product Spaces

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.

• Just perform Gram Schmidt using the inner product given above (also use it to compute the relevant norm $\|x\|_A = x^T Ax$). Mar 19, 2015 at 5:02

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

• If $\vec{u}=\frac{\vec{a}}{2}$, isn't $||\vec{u}||^2=\bigg(\sqrt{(\frac{1}{2})^2+(\frac{0}{2})^2}\bigg)^2=\frac{1}{4}?$ Mar 20, 2015 at 1:52
• check your calculus because $||u||^2=||\dfrac{a}{||a||}||^2=\dfrac{1}{||a||}^2 ||a||^2=1$
– L F
Mar 20, 2015 at 2:41
• Similarly, isn't $\sqrt{\begin{pmatrix}2&1\end{pmatrix}\cdot\begin{pmatrix}1\\0\end{pmatrix}}= \sqrt{2}$? Mar 20, 2015 at 3:25