Inner Product Space of Symmetric Matrices 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
 A: 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
$\langle \mathbf x,\mathbf x\rangle = a^2\lambda_1 + b^2\lambda_2$
