# The Unapologetic Mathematician

## Real Inner Products

Now that we’ve got bilinear forms, let’s focus in on when the base field is $\mathbb{R}$. We’ll also add the requirement that our bilinear forms be symmetric. As we saw, a bilinear form $B:V\otimes V\rightarrow\mathbb{R}$ corresponds to a linear transformation $B_1:V\rightarrow V^*$. Since $B$ is symmetric, the matrix of $B_1$ must itself be symmetric with respect to any basis. So let’s try to put it into a canonical form!

We know that we can put $B$ into the almost upper-triangular form

$\displaystyle\begin{pmatrix}A_1&&*\\&\ddots&\\{0}&&A_m\end{pmatrix}$

but now all the blocks above the diagonal must be zero, since they have to equal the blocks below the diagonal. On the diagonal, the $1\times1$ blocks are fine, but the $2\times2$ blocks must themselves be symmetric. That is, they must look like

$\displaystyle\begin{pmatrix}a&b\\b&d\end{pmatrix}$

which gives a characteristic polynomial of $X^2-(a+d)X+(ad-b^2)$ for the block. But recall that we could only use this block if there were no eigenvalues. And, indeed, we can check

\displaystyle\begin{aligned}\tau^2-4\delta&=(a+d)^2-4(ad-b^2)\\&=a^2+2ad+d^2-4ad+4b^2\\&=a^2-2ad+d^2+b^2\\&=(a-d)^2+b^2\geq0\end{aligned}

The discriminant is positive, and so this $2\times2$ block will break down into two $1\times1$ blocks. Thus any symmetric real matrix can be diagonalized, which means that any symmetric real bilinear form has a basis with respect to which its matrix is diagonal.

Let $\left\{e_i\right\}$ be such a basis. To be explicit, this means that $\langle e_i,e_j\rangle=b_i\delta_{ij}$, where the $b_i$ are real numbers and $\delta_{ij}$ is the Kronecker delta${1}$ if its indices match, and ${0}$ if they don’t. But we still have some freedom. If I multiply $e_i$ by a scalar $c$, we find $\langle ce_i,ce_i\rangle=c^2b_i$. We can always find some $c$ so that $c^2=\frac{1}{|b_i|}$, and so we can always pick our basis so that $b_i$ is ${1}$, $-1$, or ${0}$. We’ll call such a basis “orthonormal”.

The number of diagonal entries $b_i$ with each of these three values won’t depend on the orthonormal basis we choose. The form is nondegenerate if and only if there are no ${0}$ entries on the diagonal. If not, we can decompose $V$ as the direct sum of the subspace $\bar{V}$ on which the form is nondegenerate, and the remainder $W$ on which the form is completely degenerate. That is, $\langle w_1,w_2\rangle=0$ for all $w_1,w_2\in W$. We’ll only consider nondegenerate bilinear forms from here on out.

We write $p$ for the number of diagonal entries equal to ${1}$, and $q$ for the number equal to $-1$. Then the pair $(p,q)$ is called the signature of the form. Clearly for nondegenerate forms, $p+q=d$, the dimension of $V$. We’ll have reason to consider some different signatures in the future, but for now we’ll be mostly concerned with the signature $(d,0)$. In this case we call the form positive definite, since we can calculate

$\displaystyle\langle v,v\rangle=v^iv^j\langle e_i,e_j\rangle=v^iv^j\delta_{ij}=\sum\limits_{i=1}^d\left(v^i\right)^2$

The form is called “positive”, since this result is always nonnegative, and “definite”, since this result can only be zero if $v$ is the zero vector.

This is what we’ll call an inner product on a real vector space $V$ — a nondegenerate, positive definite, symmetric bilinear form $\langle\underbar{\hphantom{X}},\underbar{\hphantom{X}}\rangle:V\otimes V\rightarrow\mathbb{R}$. Notice that choosing such a form picks out a certain class of bases as orthonormal. Conversely, if we choose any basis $\left\{e_i\right\}$ at all we can create a form by insisting that this basis be orthonormal. Just define $\langle e_i,e_j\rangle=\delta_{ij}$ and extend by bilinearity.

April 15, 2009 Posted by | Algebra, Linear Algebra | 14 Comments