# Prove $\left\langle v | u \right\rangle = v\cdot (A^tAu)$ is an inner product

Let $v,u\in\mathbb{R}^n$ and $A\in M_n\left(\mathbb{R}\right)$ an invertible matrix.

Prove $\left\langle v | u \right\rangle = v\cdot (A^tAu)$ is an inner product over the reals.

I was able to (allegedly) prove both symmetry, linearity and positive-definiteness without using the invertibility of $A$ at all... Linearity is straightforward; symmetry I showed as follows: $$\left\langle v | u \right\rangle = v\cdot (A^tAu) = (A^tAu)\cdot v = (A^tAu)^t v = (u^tA^t(A^t)^t)v$$ $$= u^t(A^tAv) = u\cdot(A^tAv) = \left\langle u | v \right\rangle$$ And finally for positive-definiteness I identified $A^tA$ as a symmetric matrix $M$ with non-negative entries in the main diagonal: $$A^tA \equiv M \begin{bmatrix}a^2_{11} & a_{12} & \dots & a_{1n}\\a_{12} & a^2_{22} & \dots & a_{1n}\\ \vdots & & \ddots & \vdots\\a_{1n}& \dots & & a^2_{nn}\end{bmatrix}$$ yielding in short $$\left\langle v | v \right\rangle = \sum_{i=1}^na_{ii}^2v_i^2 + 2\sum_{i=1}^n\sum_{j\neq i}a_{ij}v_iv_j = \left(\sum_{i,j=1}^na_{ij}v_iv_j\right)^2\geq0$$ Is there anything I'm missing in my proof?

Edit: I think I got it - A needs to be invertible for $\left\langle v | u \right\rangle = 0$ to hold iff $v=0$, otherwise it can be the zero matrix. Is there another requirement for the invertibility?

• The non-negative entries in the main diagonal does not ensure positive-definiteness. For example $\begin{bmatrix} 1&-2\\ -2&1\end{bmatrix}$ is not positive-definite (though it's probably not a $A^\top A$ with $A$ invertible and real since it's det is negative). – anderstood Jun 14 '16 at 22:22
• And I don't think $\langle v|v\rangle = (\sum a_{ij}v_iv_j)^2$. For example, where's the $a_{12}v_1^2$ term in your sum? – anderstood Jun 14 '16 at 22:32
• I understand your first comment, you are correct. Concerning your second- $a_{12}v_1^2$ shouldn't be in my sum, $a_{12}$ should only multiply $v_1v_2$, and specifically twice. But nevertheless the calculation is indeed false, as I now see... mehh... – Yoni Jun 14 '16 at 22:38
• I assumed $a_{12}$ was the element (1,2) of $A$. If so, it should be in the sum defined by $v^\top A^\top A v$. You can easily check this with $A\in\mathbb R^{2\times 2}$, for example. – anderstood Jun 14 '16 at 22:58
• I probably should've wrote the entries as $m_{ij}$, I marked the entries on the main diagonal a squared to show explicitly they're positive. But I see your point by the example... Thanks – Yoni Jun 15 '16 at 7:37

You need to show that if $v$ is not zero then $\langle v, v\rangle > 0$.
This requires the assumption that the null space of $A$ is trivial.
(Note that $\langle v,v \rangle = v^T A^T Av = (Av)^T Av = \| Av \|_2^2$.)