Let $V = P_3(\mathbb{R})$ the vector space of all polynomials in $t$ of degree at most 3.
$W = M_{2\times 2}(\mathbb{R})$ the vector space of all $2\times 2$ real matrices.
Define $T:V \rightarrow W$ by $T(a_1 + a_2t + a_3t^2 + a_4t^3) = \begin{bmatrix} \dfrac{1}{\sqrt{1}}a_1 & \dfrac{1}{\sqrt{2}}a_2 \\ \dfrac{1}{\sqrt{4}}a_4 & \dfrac{1}{\sqrt{3}}a_3 \end{bmatrix}$.
My question:
- Let $U = span\left(\left\{\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix}1 & 2\\ 0 & 3\end{bmatrix}\right\}\right)$, find $U^\bot$.
Is my attempt correct?
Using Gram-schmidt process to get a set of orthonormal basis of $M_2$ with $u_1, u_2$ as given.
Denoting $M$ as $a_{11}E_{11} + a_{12}E_{12} + a_{21}E_{21} + a_{22}E_{22}$, and denoting 2 matrices given as column vectors $(\alpha_1, \alpha_2)$, as the new matrtix $A$, solve $A'x = 0$ and orthonormalize the new basis, we get: \begin{align*} A = \begin{pmatrix} 1 & 1\\ 0 & 0\\ 0 & 2\\ 1 & 3 \end{pmatrix} \end{align*} Performing elementary row operations on $\begin{bmatrix} 1 & 0 & 0 & 1\\ 1 & 0 & 2 & 3 \end{bmatrix}$, we get the RREF: \begin{align*} \begin{bmatrix} 1 & 0 & 1 & 1\\0 & 0 & 1 & 0 \end{bmatrix} \end{align*} To find the null space, we solve the matrix equation \begin{align*} \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix} \end{align*} We take $x_2 = 2$, $x_4 = s$, then $x_1 = -s$, $x_3 = 0$. Hence, \begin{align*} \vec{x} = \begin{bmatrix} -s\\t\\0\\s \end{bmatrix} = \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix} t + \begin{bmatrix} -1\\0\\0\\1 \end{bmatrix}s \end{align*} By solving $A^Tx = 0$, the orthogonal complement $U^\bot$ is: \begin{align*} \left\{\begin{bmatrix} 0\\1\\0\\0 \end{bmatrix},\begin{bmatrix} -1\\0\\0\\1 \end{bmatrix} \right\} \end{align*}
$$u\in U\Rightarrow u=\alpha\cdot \begin{bmatrix}1&0\\0&1\end{bmatrix}+\beta\cdot\begin{bmatrix}1&2\\0&3\end{bmatrix}=\begin{bmatrix}\alpha+\beta&2\beta\\0&\alpha+3\beta\end{bmatrix}=\begin{bmatrix}a_1&\frac{a_2}{\sqrt{2}}\\\frac{a_4}{2}&\frac{a_3}{\sqrt{3}}\end{bmatrix}, \alpha, \beta \in \mathbb R$$
$$\begin{array}{rrlrlr}u\rightarrow p&=&a_1+a_2t+a_3t^2+a_4t^3\\p&=&(\alpha+\beta)+2\sqrt{2}\beta t+\sqrt{3}(\alpha+3\beta)t^2\\ q\in V, q&=&x_1+x_2t+x_3t^2+x_4t^3 \end{array}$$
$$<p,q>=0\Rightarrow (\alpha+\beta)x_1+\sqrt{2}\beta x_2+\frac{(\alpha+3\beta)x_3}{\sqrt{3}}=0$$
$$\alpha\left(x_1+\frac{x_3}{\sqrt{3}}\right)+\beta(x_1+\sqrt{2}x_2+\sqrt{3}x_3)=0$$
$$x_1=s, x_3=-s\sqrt{3}, x_2=\sqrt{2}s, x_4=r, s, r\in \mathbb R$$
$$q=s+\sqrt{2}st-s\sqrt{3}t^2+rt^3$$
$$T(q)=\begin{bmatrix}s&s\\\frac{r}{2}&-s\end{bmatrix}=s\begin{bmatrix}1&1\\0&-1\end{bmatrix}+r\begin{bmatrix}0&0\\1&0\end{bmatrix}$$
$$U^\bot=span\left(\begin{bmatrix}1&1\\0&-1\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}\right)(\therefore)$$