Prove that a formula defines a dot product

Question

Let V be vector space on the field ℝ of all two-dimensional column vectors. Choose whether or not a transformation $$\left\langle { - | - } \right\rangle :V \times V \to ℝ,$$ defined by a formula $$\left\langle {u\left| v \right.} \right\rangle = {u^T} \cdot A \cdot v$$ where A is equal to\begin{bmatrix}2&-1\\-1&2\end{bmatrix} defines a dot product.

I know that these conditions must be true:$$\left\langle {u|v} \right\rangle = \left\langle {v|u} \right\rangle $$ $$\left\langle {u + v|w} \right\rangle = \left\langle {u|w} \right\rangle + \left\langle {v|w} \right\rangle $$ $$\left\langle {u|u} \right\rangle \ge 0 \wedge \left\langle {u|u} \right\rangle = 0 \Leftrightarrow u = {0_V}.$$

I got stuck on the first one: $$\left\langle {u\left| v \right.} \right\rangle = {u^T} \cdot A \cdot v$$ $$\left\langle {v\left| u \right.} \right\rangle = {v^T} \cdot A \cdot u$$ but $${u^T} \cdot A \cdot v \ne {v^T} \cdot A \cdot u$$ And I was also wondering if A has to be positive definite. Also, how do I prove the second part of the third condition?

Edit: There is another condition I forgot (but trivial to prove): $$\left\langle {\alpha \cdot u\left| v \right.} \right\rangle = \alpha \cdot \left\langle {u\left| v \right.} \right\rangle $$

Answer 1

For the first condition, you can use that $A$ is a symmetric matrix.

$$\langle u|v\rangle = u^TA v = (u^TAv)^T = v^TA^T(u^T)^T = v^TAu = \langle v|u\rangle$$

Where we have used that for any scalar $r$ that $r = r^T$, $A = A^T$, and the rule for the transpose of a product.

For the second one we have

$$\langle u + v|w\rangle = (u + v)Aw^T = (u + v)(Aw^T) = uAw^T + vAw^T = \langle u|w\rangle + \langle v|w\rangle$$

This follows from the distributive property of matrix multiplication.

For the third one, we need to be a bit more careful.

Let $u = [a,b]^T$. Then we get

$$\langle u,u\rangle = \begin{bmatrix}a & b \end{bmatrix}\begin{bmatrix}2 &-1\\ -1 &2\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix} = \begin{bmatrix}2a-b & 2b-a\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix} \\ = (2a^2 - ab) + (2b^2 - ab) = a^2 + b^2 + (a-b)^2 \geq 0$$

From the expanded expression it is clear that only when $a=0, b=0$ will $<u|u>= 0.$

Since we have satisfied all three conditions the transformation does define the dot product.

To answer your question about positive definite, yes for this to work for a matrix it must be positive definite since otherwise, some vectors could have negative dot products with themselves. Positive semidefinite also doesn't work since there may be non-zero vectors which have zero dot product.

Edit (tomashauser):

The last condition: $$\left\langle {\alpha u\left| v \right.} \right\rangle = {\left( {\alpha u} \right)^T}Av = {\alpha ^T}{u^T}Av = \alpha {u^T}Av = \alpha \left\langle {u\left| v \right.} \right\rangle $$

Answer 2

In general $u^TAv \neq v^TAv$ but this will be true if the row space and the column space are the same. That's always the case for symmetric matrices real valued matrices and $A$ is such a matrix. To see this we just note that $u^TAv=(u^TAv)^T = v^TA^Tu=v^TAu$ with the first equality being true because the transpose of a scalar is a scalar, the second equality by the usual properties of the transpose and the final equality because $A$ is symmetric.

To prove the case $\left\langle {u|u} \right\rangle = 0 $ we note that $A$ is invertible, so it represents and injective (bijective) map from $\mathbb{R}^2$ to itself and so $Au = 0$ if and only if $u=0$. The same argument holds for $v^TA$ with a row vector instead of a column vector and that show $\left\langle {u|u} \right\rangle = 0$ only when $u=0$.