# show that $\langle u, v \rangle := u\cdot Av$ is an inner product

In $$\Bbb R^n,$$ if vectors are written as column-vectors (i.e., $$n \times 1$$ column-matrices) like $$u=\begin{bmatrix}{c}u_{1}\\\vdots\\ u_{n}\end{bmatrix}$$ then the standard dot-product in $$\mathbb{R}^{n}$$ can be written as $$u\cdot v=v^Tu$$ by using the meaning of "transpose of a matrix' and multiplication of matrices.

(a) Let $$A \in \mathcal{M}_{n \times n}$$ be an $$n \times n$$ square-matrix. Using the properties of 'transpose of a matrix' and the definition given above, show that $$Au\cdot v=u\cdot A^T v\forall u,v \in \Bbb R^n$$

(b) Now, let $$A$$ be the $$2 \times 2$$ symmetric-matrix given by $$A=\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}.$$ Show that the real-valued function $$\langle\cdot \cdot \cdot\rangle: \Bbb R^2\times \Bbb R^2\to\Bbb R$$ defined by $$\langle u, v\rangle:=u\cdot A v$$ is an inner-product on $$\Bbb R^2$$

How to prove (b) is an inner product? I proved $$\langle u,v\rangle=\langle v,u \rangle$$. How to prove $$\langle u+v,w\rangle =\langle u,w\rangle +\langle v+w\rangle$$, $$\langle ku,v\rangle =k\langle u,v\rangle$$ and $$\langle v,v\rangle \geq 0$$ ??

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– amd
Commented Dec 1, 2016 at 19:37

The properties $\langle u+v, w \rangle = \langle u, w \rangle + \langle v,w \rangle$ and $\langle ku,v \rangle = k \langle u,v \rangle$ follow trivially from linearity of the dot product. For example, $$\langle u+v, w \rangle = (u+v)\cdot Aw = u\cdot Aw + v\cdot Aw = \langle u, w \rangle + \langle v,w \rangle.$$ For positive definiteness, you need to use the specific matrix given. If $v = \binom{v_1}{v_2}$, then $$\langle v,v\rangle = v \cdot \left( \begin{matrix} 1 & 1 \\ 1 & 2\end{matrix} \right)v = \binom{v_1} {v_2} \cdot \binom{v_1 + v_2}{v_1 + 2v_2} = v_1^2 + 2v_2 + 2v_2^2 = (v_1 + v_2)^2 + v_2^2.$$

Inner product satisfies that following

1. Symmetry: $\newcommand{\inp}[1]{\left\langle #1 \right\rangle}$ $\inp{u, v} = \inp{v, u}$

2. Linearity: $\inp{k u, v} = k \inp{u, v}$.

3. Positive-definite: $\inp{u, u} \ge 0$ with equality if and only if $u = 0$.

So the question is prove $\inp{u, v} = v^TA^Tu$ is an inner product for the given $A$.

1. Symmetry: Trivial (you proved it as well).
2. Linearity: $\inp{k u, v} = v^TA^T k u = k.v^TA^Tu = k.\inp{u, v}$
3. Positive-definite: $\inp{u, u} = u^TA^T u > 0$ since the determinanats of the leading principal minors of $A$ are positive definite ($1 > 0$ and $det(A) > 0$)