Inner Product Space of Symmetric Matrices

Question

How do we show that when A is symmetric i.e the transpose of A is equal to A itself where A is a 2x2 matrix with λ2>λ1>0 THEN =a.(A(b)) is an inner product ?

I think I need to work in the basis of eigenvectors in order to prove this.I just don't know how to process any further. All the best

Answer 1

If we want to define our inner product.

$\langle \mathbf x,\mathbf y\rangle = \mathbf x^TA\mathbf y$

We need to show that by this definition our inner product has:

Symmetry: (if our vectors are real, conjugate symmetry if they are complex)

$\langle \mathbf x,\mathbf y\rangle = \langle \mathbf y,\mathbf x\rangle$

Since $\langle \mathbf x,\mathbf y\rangle = \mathbf x^TA\mathbf y$ is a real $1\times 1$ matrix

$\mathbf x^TA\mathbf y = (\mathbf x^TA\mathbf y)^T \\ \langle \mathbf x,\mathbf y\rangle = \mathbf x^TA\mathbf y = (\mathbf x^TA\mathbf y)^T = \mathbf y^TA^T\mathbf x = \mathbf y^TA\mathbf x = \langle \mathbf y,\mathbf x\rangle$

Linearity:

$\langle \mathbf x+\mathbf z,\mathbf y\rangle = \langle \mathbf x,\mathbf y\rangle + \langle \mathbf z,\mathbf y\rangle$

$\mathbf (\mathbf x+ \mathbf z)^TA\mathbf y =\mathbf (\mathbf x^T+ \mathbf z^T)A\mathbf y = \mathbf x^TA\mathbf y + \mathbf z^TA\mathbf y$

$\langle a\mathbf x,\mathbf y\rangle = a\langle \mathbf x,\mathbf y\rangle$

and postive definite:

$\|\mathbf x\|^2 = \langle \mathbf x,\mathbf x\rangle > 0$ for all $x \ne \mathbf 0$

The last one is the one that requires the eigenvectors to be greater than $0$

let $\mathbf x = a\mathbf v_1 + b\mathbf v_2$ where $\mathbf v_1,\mathbf v_2$ are normalized eigenvectors

$\langle \mathbf x,\mathbf x\rangle = a^2\lambda_1 + b^2\lambda_2$