# How do you get the orthogonal complement given the span?

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*}

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. $$\eta_1 = \left( {\begin{array}{*{20}{c}} -1\\ 0 \\ 0 \\ 1 \end{array}} \right)$$ $$\eta_2 = \left( {\begin{array}{*{20}{c}} 0\\ 1 \\ 0 \\ 0 \end{array}} \right)$$

• As far as I know, with Gram-Schmidt process, I solve these with a given equation, then I work my way by setting them to $x_1$ to get $c_1, c_2, ... c_n$ but there is no equation given, so how do I start?
– muw
Jun 3, 2021 at 11:29
• Take elementary matrices as basis and convert the problem. Jun 3, 2021 at 11:31
• Please check my attempt if it's correct.
– muw
Jun 3, 2021 at 11:51
• No,you should get the solution of $A'x=0$ to get $U^{\perp}$ and use Gram-Schmidt process. Jun 3, 2021 at 12:01
• your attempt is to orthonormalize the basis of $U$, and it's not enough. Jun 3, 2021 at 12:02