Why are the following statements true?

Question

Suppose we have a matrix $A \in \mathbb{R}^{n \times n}$

Consider a vector $ v \in R^n$. Why it is the case that

$\langle Av, v \rangle = \langle v, A^Tv\rangle $

for the inner product?

Where $A^T$ is the transpose of A

Answer 1

Notice that for all $(x,y)\in\mathbb{R}^n\times\mathbb{R}^n$, one has the following equality: $$\langle x,y\rangle ={}^\intercal xy.$$ Indeed, if $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$ are column vectors, one has: $${}^\intercal xy=\sum_{i=1}^nx_iy_i.$$ Whence the result, since: $$\langle Av,v\rangle={}^\intercal(Av)v={}^\intercal v{}^\intercal Av=\langle v,{}^\intercal Av\rangle.$$

Answer 2

In general for vector spaces over $\mathbb{C}$ we have that $\langle Av,w \rangle = \langle v, A^*w \rangle$ and $A^*$ is the adjoint of $A$. In fact you get $A^*$ by taking the transpose of the matrix and then the conjugate of each term. But as all entries in the matrix $A$ are real in your case, the conjugation doesn't change anything. Hence we have $A^* = A^T$.

To prove the fact in the first line you can find some orthonormal basis of the vector space and then express $v$ and $w$ using it, expand and you would be able to find the entries in the matrix $B$ such that $\langle Av,w \rangle = \langle v, Bw \rangle$. This will lead you to conclude that $B = A^*$