# Find an inner product such that A and B are orthogonal

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.

• If memory serves, $\left<A, B\right> = \text{tr}(A^T JB)$ defines an inner-product for any positive-definite matrix $J$. So a potential solution would be to find a $J$ such that this inner product vanishes for your given matrices. May 13 at 21:19
• @infinitylord thank you, I will try with your inner-product. May 13 at 21:27

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