Determine if the function $B:\mathbb{R}^3\times\mathbb{R}^3 \to \mathbb{R}$ defines an inner product.

Question

I have to determine that this function:

$$ \text{B(x,y)} = \begin{bmatrix}x_1&x_2&x_3\end{bmatrix} \begin{bmatrix}1&1&1\\1&3&1\\1&1&5\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix} $$ defines an inner product.

So I'm pretty sure i have to show that this function satisfies bilinearity, symmetry and positive definite. I'm watching a solution to this on online lectures, but I could not follow the lecturer on the bilinearity part. He used the fact that the function $B=x^TBy$ because of the distributivity law of matrix multiplication.

Can anyone expand on showing that this function is bilinear?

Answer 1

The bilinearity part:

1. $B(x_1+x_2,y)=(x_1+x_2)^TBy=(x_1^T+x_2^T)By=x_1^tBy+x_2^TBy)=B(x_1,y)+B(x_2,y)$

and for $t \in \mathbb R$ we have

$B(tx,y)=(tx)^TBy=tx^TBy=tB(x,y)$.

2. It is now your turn to show that $B$ is linear in the second argument $y$.

Answer 2

$$B(x, y+y') = x^TB(y+y') = x^T(B(y+y')) = x^T(By+By') = x^T(By) + x^TBy'$$

if any one of the steps above is unclear, please ask.

You can prove the other bilinearity properties in a similar way. Just remember that $(x^T B)y=x^T(By)$.