How to show that $\langle A,B \rangle = a_{11}b_{11}+a_{12}b_{12}+a_{21}b_{21}+a_{22}b_{22}$ is an inner product on $M_{2x2}$?

Question

Let $$\langle A,B \rangle = a_{11}b_{11}+a_{12}b_{12}+a_{21}b_{21}+a_{22}b_{22}.$$ Show that this in an inner product on the vector space $M_{2x2}$?

I just do not get how to prove this with matrices.

Answer 1

Here is one way:

Define the invertible linear map $\phi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^4$ as $\phi(A) = ([A]_{11}, [A]_{21}, [A]_{12}, [A]_{22})^T$, and note that $\langle A, B \rangle_* = \langle \phi(A), \phi(B) \rangle_{\mathbb{R}^4}$, where $\langle \cdot, \cdot \rangle_*$ is the function in the question.

Since $\langle \cdot, \cdot \rangle_{\mathbb{R}^4}$ is an inner product, it follows that $\langle \cdot, \cdot \rangle_*$ is an inner product.

Answer 2

We will use the properties of inner product.

1. Antisymmetry Consider $<B,A>=b_{11}a_{11}+b_{12}a_{12}+b_{21}a_{21}+b_{22}a_{22}=<A,B>$ so in particular we have symmetry.
2. Positive-definiteness Let us consider $<A,A>=a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2\geq0$ and equality holds iff $a_{11}=a_{12}=a_{21}=a_{22}=0$ i.e. $A=0$
3. Summability in first argument Can you show that $<\alpha A+\beta B,C>=\alpha<A,C>+\beta<B,C>$ for scalars $\alpha,\beta$? This is a simple exercise.

Therefore this is an inner product!