I'm learning linear algebra, specifically inner products on $\mathbb{R}^n$, and need help with the following exercise :
We give the quadratic form in $\mathbb{R}^3$ (where $\mathbf{x} = (x_1, x_2, x_3)$) :
$$q(\mathbf{x}) = 2x_1^2 – 2x_1x_2 + ax_2^2 + 2x_1x_3 – 4x_2x_3 + 3x_3^2.$$
$(1)$ Write $q$ in matrix form and find the values of $a$ for which $q$ is positive definite. $(2)$ Define an inner product on $\mathbb{R}^3$ such that the norm defined by the inner product is given by $\|x\|_q = \sqrt{q(\mathbf{x})}$.
There are many things which I don't understand in this exercise. From the theory, I know that $\langle \mathbf x, \mathbf y \rangle$ is an inner product on $\mathbb R^n$ if and only if $\langle \mathbf x, \mathbf y \rangle = \mathbf x^T A \mathbf y$, where $A$ is a symmetric matrix whose eigenvalues are strictly positive, i.e. $A$ is positive definite. Also, an inner product on $\mathbb{R}^n$ is a function, say $f$, such that $f : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$. But in the present exercise, $q$ is defined as $q : \mathbb{R}^3 \to \mathbb{R}$, so how can it define an inner product?
Even though it's wrong, I'm sharing the matrix form of $q$ which I found :
$$q = \begin{pmatrix}x_1 \\x_2\\x_3 \end{pmatrix}^T \begin{pmatrix} 2 & -2 & 2\\0 & a & -4\\0 & 0 & 3\end{pmatrix} \begin{pmatrix}x_1 \\x_2\\x_3 \end{pmatrix}.$$
If I develop I get back $q(\mathbf{x})$ but the matrix I found is clearly not symmetric.
Possibly someone can help me understand this exercise and my errors. For $(2)$ I have no idea.
Best Answer
The matrix is $$A=\begin{pmatrix} 2 & -1 & 1\\ -1 & a & -2\\ 1 & -2 & 3 \end{pmatrix}$$ which is symmetric! Reason: Each $x_ix_j$-term for $i\neq j$ in $q$ will be obtained in $x^tAx$ by the $a_{ij}$ and $a_{ji}$ entry of $A$. So if we have a $\beta x_ix_j$-term in $q$, you have $a_{ij}=a_{ji}=\beta/2$.
Now determine $a$ such that $A$ is positive definite. For this $a$, the quadratic form $q$ is positive definite. Define $f\colon\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R},(x,y)\mapsto x^t Ay$, then you have $f(x,x)=q(x)$ and since $q$ is positve definite, $f$ is an inner product.