3
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ May 13 at 21:19
  • $\begingroup$ @infinitylord thank you, I will try with your inner-product. $\endgroup$
    – Ángel
    May 13 at 21:27

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.