Quadratic form and inner product

Question

In the context of inner products and norms I need help with the following exercise :

First, some notation. In the vector space $\mathbb R^2$, we write $\mathbf x = (x_1, x_2), \mathbf y = (y_1, y_2)$ ect. for points $\mathbf x, \mathbf y$ in terms of a given base. For simplicity we assume that we are working over the field $\mathbb R$. Here's the exercise :

$(1)$ To which inner product corresponds the quadratic form

$$q(x_1, x_2) = 4x_1^2 - 6x_1x_2 + 5x_2^2$$

Find a vector perpendicular to $(2, 3)$ with respect to this inner product. Also find the angle between $(-2, 3)$ and $(1, 2)$.

Here's a simpler exercise which I was able to solve :

$(2)$ To which inner product corresponds the bilinear, symmetric function

$$f(\mathbf x, \mathbf y)= x_1y_1-x_2y_1-x_1y_2+4x_2y_2.$$

I used the following proposition : $\langle x, y \rangle$ is an inner product on $\mathbb R^n$ if and only if $\langle x, y \rangle = x^TAy$, where $A$ is a symmetric matrix whose eigenvalues are strictly positive.

So in this case the matrix is $A=\begin{pmatrix}1&-1\\-1&4\end{pmatrix}$ and the inner product is given by $\langle x,y\rangle_f=x^TAy$.

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I wanted to use a similar argument for $(1)$ but I was not able to. One of the problems that I'm having is that in $(2)$ we are given a bilinear, symmetric function $f(\mathbf x, \mathbf y)$ which depends on $\mathbf x$ and $\mathbf y$ while in $(1)$ the quadratic form $q$ only depends on $\mathbf x$. Also, any help on the sub-questions would be appreciated.

Answer 1

Let $$f(\mathbf x) = \sum_{j=i}^n\sum_{i=1}^n a_{ij}x_ix_j = x^TAx$$ be a quadratic form. Then for $1\leq i \leq n$, $A_{ii}$ is given by the coefficient of $x_i^2$. For $i\neq j$, $A_{ij}=A_{ji}$ is given by one half of the coefficient of $x_ix_j$. For example, in your exercise, if $f(\mathbf x ) = 4x_1^2-6x_1x_2+5x_2^2$, then we have $$A = \begin{bmatrix} 4\,\,\,\,\,\,&-3 \\-3\,\,\,\,\,&5\end{bmatrix}$$ Then $\langle \mathbf x, \mathbf y\rangle = \mathbf x^T A\mathbf y$ is the inner product you're looking for.

To find a vector perpendicular to $(2,3)$, since the system of equations given by $(2,3)\cdot A\cdot\mathbf x =0$. Finally, the angle between two vectors $\mathbf u$ and $\mathbf v$ is given by $$\theta = \cos^{-1}\left(\frac{\langle \mathbf u, \mathbf v\rangle}{\sqrt{\langle \mathbf u, \mathbf u\rangle\langle \mathbf v, \mathbf v\rangle}} \right)$$