Is $\langle u,v\rangle $ $=$ $4x_1y_1 - 6x_1y_2 - 6x_2y_1 + 11x_2y_2$ an inner product in $\mathbb R ^2$

Question

I already know that $\langle u,v\rangle$ is already in the form of $x^T$$Ay$, but I'm not sure if this is correct or not. Should the value of $A$ suppose to be the same?

Answer 1

As you already note that the expression can be written as $$\Big \langle \dbinom{x_1}{x_2},\dbinom{y_1}{y_2} \Big \rangle = \big( x_1 \ \ x_2) \begin{pmatrix} 4 & -6 \\ -6 & 11 \end{pmatrix} \dbinom{y_1}{y_2} \tag{1}$$ or more compact, $\langle \mathbf{x},\mathbf{y} \rangle = \mathbf{x}^tA\mathbf{y}$, where $A$ is the above matrix, the linearity in the first entry is easy to prove.

Also, since $A$ is symmetric, and $\mathbf{x}^tA\mathbf{y}$ is equal to its own transpose (since it is a scalar) it follows that $$\langle \mathbf{x},\mathbf{y} \rangle = \mathbf{x}^tA\mathbf{y} = (\mathbf{x}^tA\mathbf{y})^t = \mathbf{y}^tA^t\mathbf{x} = \langle \mathbf{y},\mathbf{x} \rangle.$$

Finally, we need to prove that $\langle X,X \rangle > 0$ for $X \in \mathbb{R}^2 \setminus \{\mathbf0\}$, or equivalently, we need to verify that $\langle X,X \rangle \geq 0$ for all $X \in \mathbb{R}^2$ and the equality only when $X = \mathbf0$. I think is more easy to check the second. So, let $$X = \dbinom{x}{y} \mathbb{R}^2$$ and then

\begin{align} \langle X,X \rangle &= 4x^2 - 12xy + 11y^2 \\ &= (2x-3y)^2 + 2y^2. \end{align} Now, it is easy to see that $\langle X,X \rangle \geq 0$ and $\langle X,X \rangle = 0$ only when $2x-3y=0$ and $y=0$, that is, only when $X$ is the zero vector.

Answer 2

Hint: $\langle x,y \rangle = x^TAy$ is an inner product iff $A$ is symmetric and positive definite iff $A$ is symmetric and has positive eigenvalues.