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.

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April 15, 2009 - Posted by | Algebra, Linear Algebra

14 Comments »

  1. [...] Today I want to present a deceptively simple fact about spaces equipped with inner products. The Cauchy-Schwarz inequality states [...]

    Pingback by The Cauchy-Schwarz Inequality « The Unapologetic Mathematician | April 16, 2009 | Reply

  2. [...] Products and Angles We again consider a real vector space with an inner product. We’re going to use the Cauchy-Schwarz inequality to give geometric meaning to this [...]

    Pingback by Inner Products and Angles « The Unapologetic Mathematician | April 17, 2009 | Reply

  3. [...] Products and Lengths We’re still looking at a real vector space with an inner product. We used the Cauchy-Schwarz inequality to define a notion of angle between two [...]

    Pingback by Inner Products and Lengths « The Unapologetic Mathematician | April 21, 2009 | Reply

  4. [...] Polarization Identities If we have an inner product on a real or complex vector space, we get a notion of length called a “norm”. It turns out that [...]

    Pingback by The Polarization Identities « The Unapologetic Mathematician | April 23, 2009 | Reply

  5. [...] Gram-Schmidt Process Now that we have a real or complex inner product, we have notions of length and angle. This lets us define what it means [...]

    Pingback by The Gram-Schmidt Process « The Unapologetic Mathematician | April 28, 2009 | Reply

  6. [...] as a vector space over the reals: . If we ask that this natural basis be orthonormal, we get a real inner product on complex numbers, which in turn gives us lengths and angles. In fact, this notion of length is [...]

    Pingback by Complex Numbers and the Unit Circle « The Unapologetic Mathematician | May 26, 2009 | Reply

  7. [...] and inner product spaces. I want to look at the matrix of a linear map between finite-dimensional inner product [...]

    Pingback by Matrix Elements « The Unapologetic Mathematician | May 29, 2009 | Reply

  8. [...] for our new bilinear form to be an inner product it must be symmetric (or conjugate-symmetric). This is satisfied by picking our transformation to be symmetric (or [...]

    Pingback by Positive-Definite Transformations « The Unapologetic Mathematician | July 13, 2009 | Reply

  9. [...] of all, we’re going to assume our space comes with a positive-definite inner product, but it doesn’t really matter which one. We’re choosing a positive-definite form with [...]

    Pingback by Hard Choices « The Unapologetic Mathematician | September 22, 2009 | Reply

  10. [...] Let’s focus back in on a real, finite-dimensional vector space and give it an inner product. As a symmetric bilinear form, the inner product provides us with an isomorphism . Now we can use [...]

    Pingback by Tensor Algebras and Inner Products I « The Unapologetic Mathematician | October 29, 2009 | Reply

  11. [...] Let’s focus back in on a real, finite-dimensional vector space and give it an inner product. As a symmetric bilinear form, the inner product provides us with an isomorphism . Now we can use [...]

    Pingback by Tensor Algebras and Inner Products « The Unapologetic Mathematician | October 30, 2009 | Reply

  12. [...] of posts, I’d like to talk a bit about reflections in a real vector space equipped with an inner product . If you want a specific example you can think of the space consisting of -tuples of real numbers [...]

    Pingback by Reflections « The Unapologetic Mathematician | January 18, 2010 | Reply

  13. [...] fixed point at will be important. In particular, we will see that this norm is associated with an inner product on , and that Hölder’s inequality actually implies the Cauchy-Schwarz [...]

    Pingback by Hölder’s Inequality « The Unapologetic Mathematician | August 26, 2010 | Reply

  14. [...] metric? We now add the assumption that is not just a bilinear form, but that it’s an inner product. That is, is symmetric, nondegenerate, and positive-definite. We can let the last condition slip a [...]

    Pingback by (Pseudo-)Riemannian Metrics « The Unapologetic Mathematician | September 20, 2011 | Reply


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