# Finding an orthonormal basis of V

1. Apr 10, 2013

### Unredeemed

I've done most of this question apart from the very last bit. I have an answer to the very last bit, but it doesn't use any of my previously proved statements and I think they probably mean me to deduce from what I already have.

1. The problem statement, all variables and given/known data
Let V be the finite-dimensional vector space of 2 x 2 matrices with real entries.

State, without proof, the dimension of V as a real vector space. (done)

A real-valued function V x V is defined by:

<A|B> = tr((A^t)B) where A^t is the transpose of the matrix A.

Show that <A|B> defines an inner product on V. (done)

Show that, with respect to this inner product, any symmetric matrix is orthogonal to any antisymmetric matrix, and that the identity matrix is orthogonal to any matrix whose trace is zero. (done)

Find an orthonormal basis of V. (bit I can't do)

2. Relevant equations

3. The attempt at a solution

I thought I could just write down:

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

\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}

\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}

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

And then show that it's an orthonormal basis. But, this obviously doesn't use what I've already proved. How can I deduce it from what I already have?

Cheers.

2. Apr 10, 2013

### LCKurtz

Why do you have to deduce it from the previous steps? And note that, while your examples are orthogonal, they aren't orthonormal, although that's easy to fix. What about the four matrices like this one$$\begin{pmatrix} 1 & 0\\0 & 0\end{pmatrix}$$which might be slightly easier to show are a basis.

3. Apr 10, 2013

### vela

Staff Emeritus
But you will need to use the function you showed was an inner product to show your choice of matrices form an orthonormal basis. Plus it looks like you already did use some of what you proved in choosing those matrices.

4. Apr 11, 2013

### Unredeemed

Oh yeah, good point, they are only orthogonal.

So, as LCKurtz suggested, I should just use the standard basis for V and show that <A|B>=0 for A≠B and 1 for A=B?

5. Apr 11, 2013

### LCKurtz

Yes. Although your original matrices can be normalized to work too.