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*}