The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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