Find orthogonal complement and its basis

Question

Let $U=\mbox{span}\left\{ \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\right\}$. What is a basis of orthogonal complement $U^{\perp}$?

So far I have deduced from the theorem $$\mbox{dim}(U)+\mbox{dim}(U^{\perp})=\mbox{dim}(M_{2\times2})$$ that $\mbox{dim}(U^{\perp})=1$. Morover, if I pick a matrix $A\in M_{2\times 2}$ and conider $$\langle A, M\rangle=0\qquad \mbox{for all}\, M\in U$$ I may assume that $M$ is linear combination of elements from $U$ (span) for some constants. I cannot however deduce the general form of the orthogonal complement from the latter.

Answer 1

$X \in U^{\perp} \iff \langle X, Y \rangle=0 \> \> \forall \> Y \in U $

$\langle X, Y \rangle=\mbox{tr}(X^T Y)$

Let's take: $X=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $

\begin{equation}\text{tr} \Biggl( \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}\Biggl) = a + b = 0 \end{equation} \begin{equation}\text{tr} \Biggl( \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} 1 & 0\\ 1 & 0 \end{pmatrix}\Biggl) = a + c = 0 \end{equation} \begin{equation}\text{tr} \Biggl( \begin{pmatrix} a & c \\ b & d \end{pmatrix} \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}\Biggl) = a + c + d = 0 \end{equation} If you solve you'll get $d=0 \land b=-a \land c=b$.

So $X=\begin{pmatrix} a & -a \\ -a & 0 \end{pmatrix}, \> a \neq 0$.