How do you get the orthogonal complement given the span?

Question

I know how it is done if the vectors are given but how is it done for a span?

The problem:

Let $H = span\left\{\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\right\}$. Find a basis for $H^{\bot}$.

My attempt:

\begin{align*} c_1 \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} + c_2 \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} &= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}\\\\ \begin{bmatrix} c_1 + 0c_2 & c_1 + c_2\\ 0c_1 + 0c_2 & c_1 + 0c_2 \end{bmatrix} &= \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}\\\\ \left\{\begin{bmatrix} c_1 + 0c_2 &= 0\\ 0c_1 + 0c_2 &= 0\\ c_1 + c_2 &= 0\\ c_1 + 0c_2 &= 0 \end{bmatrix}\right\} \Rightarrow \begin{bmatrix} 1 & 0\\ 0 & 0\\ 1 & 1\\ 1 & 0 \end{bmatrix} &= \begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix} \end{align*} By performing Gram-schmidt process on to the two vectors, we get the set of the orthonormal vectors: \begin{align*} \left\{\begin{bmatrix} \dfrac{\sqrt{3}}{3}\\ 0\\ \dfrac{\sqrt{3}}{3}\\ \dfrac{\sqrt{3}}{3} \end{bmatrix}\right\}, \left\{\begin{bmatrix} -\dfrac{\sqrt{6}}{6}\\ 0\\ \dfrac{\sqrt{6}}{3}\\ -\dfrac{\sqrt{6}}{6} \end{bmatrix}\right\} \end{align*}

Answer 1

Using Gram-schmidt process to get a set of orthonormal basis of $\mathcal{M}_2$ with $u_1,u_2 $ as given.

Also you can denote $M$ as $a_{11}E_{11}+a_{12}E_{12}+a_{21}E_{21}+a_{22}E_{22}$, and denote 2 matrices given as column vectors $(\alpha_1,\alpha_2)$, as the new matrix $A$,solve $A'x=0$,and orthonormalize the new basis you get.

EDIT:$A$= \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 1 & 1 \\ 1 & 0 \end{pmatrix} By solving $A^{T}x=0$，we get the basis of the solution space. $\eta_1 = \left( {\begin{array}{*{20}{c}} -1\\ 0 \\ 0 \\ 1 \end{array}} \right)$ $\eta_2 = \left( {\begin{array}{*{20}{c}} 0\\ 1 \\ 0 \\ 0 \end{array}} \right)$