Find an inner product such that A and B are orthogonal

Question

I don't how to find an inner product such that $A$ and $B$ are orthogonal. I was thinking about working with canonical space and transporting it by changing the base, but I don't know if that's right. I need some help please.

The problem:

Let M$_{2\times2}(\mathbb{R})$ the $\mathbb{R}$ vector space of $2\times2$ matrices with real entries. Find an inner product defined in $M_{2\times2}(\mathbb{R})$ such that:

\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}

and

\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}

are orthogonal.

Answer 1

Define an ordered basis $\mathcal{B} = \left\lbrace\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right\rbrace$. For any $A \in M_{2\times 2}(\mathbb{R})$, let $[A]$ denote the coordinates of $A$ with respect to the basis $\mathcal{B}$.

Now, the standard inner product (also called the Frobenius inner product) is given by $\left<A, B\right> = \text{tr}(A^T B)$ for all $A, B \in M_{2\times 2}(\mathbb{R})$. Consider instead the inner product defined by $\left<A, B\right> =\text{tr}([A]^T [B])$. It trivially follows that $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$ are orthogonal.