# Inner Product Spaces

Definition. Let V be a vector space over F, where or . An inner product on V is a function which satisfies:

1. (Linearity) , for , .
2. (Symmetry) , for . (" " denotes the complex conjugate of x.)
3. (Positive-definiteness) If , then .

A vector space with an inner product is an inner product space. If , V is a real inner product space; if , V is a complex inner product space.

Example. Suppose V is a complex inner product space. By properties 1 and 2,

That is, the inner product is almost linear in the second variable --- constants pull out, up to complex conjugation. This is sometimes referred to as sesquilinearity.

Example. If , . In this case, the second property reads . Moreover, the result I derived in the last example becomes .

Thus, a real inner product is linear in each variable.

Why include complex conjugation in the symmetry axiom? If I had used in the complex case, then

This contradicts . That is, I can't have both pure symmetry and positive definiteness.

Example.

By symmetry, as well.

Example. If , the condition implies that .

Example. The dot product on is given by

It's easy to verify that the axioms for an inner product hold. For example,

provided that .

I can use an inner product to define lengths and angles. You can think of an inner product as an axiomatic way of introducing (metric) geometry into vector spaces.

Definition. Let V be an inner product space, and let .

1. The length of x is .
2. The distance between x and y is .
3. The angle between x and y is the smallest positive real number satisfying

Remark. The definition of the angle between x and y will make sense once I prove the Cauchy-Schwarz ineqaulity below (since I need to know that

Proposition. Let V be a real inner product space, , .

(a) . (" " denotes the absolute value of a.)

(b) if and only if .

(c) ( Cauchy- Schwarz inequality) .

(d) ( Triangle inequality) .

Proof. (a) Since ,

(b) implies , and hence . Conversely, if , then , so .

(c) First, observe that if , then

Since the last line is true for any , I may set , . Then

Replace x with ; this yields

Therefore, .

(d)

Hence, .

Example. is an inner product space using the standard dot product of vectors. The cosine of the angle between and is

Example. Let denote the real vector space of continuous functions on the interval . Define an inner product on by

Note that is integrable, since it's continuous on a closed interval.

The verification that this gives an inner product relies on standard properties of Riemann integrals. For example, if ,

Given that this is a real inner product, I may apply the preceding proposition to produce some useful results. For example, the Schwarz inequality says that

Definition. A set of vectors S in an inner product space V is orthogonal if for , .

An orthogonal set S is orthonormal if for all .

The vectors in an orthogonal set are mutually perpendicular. The vectors in an orthonormal set are mutually perpendicular unit vectors.

Notation. If I is an index set, the Kronecker delta (or ) is defined by

With this notation, a set is orthonormal if

Note that the matrix whose -th component is is the identity matrix.

Example. The standard basis for is orthonormal.

Example. Here is an orthonormal basis for :

Example. Let denote the complex-valued continuous functions on . Define an inner product by

Let . Then

It follows that the set

is orthonormal in .

Proposition. Let be an orthogonal set of vectors, for all i. Then is independent.

Proof. Suppose

Take the inner product of both sides with . Since is orthogonal, this gives . Now since , so . Similarly, for all j. Therefore, is independent.

An orthonormal set consists of vectors of length 1, so the vectors are obviously nonzero. Hence, an orthonormal set is independent, and forms a basis for the subspace it spans. A basis which is an orthonormal set is called an orthonormal basis.

It is very easy to find the components of a vector relative to an orthonormal basis.

Proposition. Let be an orthonormal basis for V, and let . Then

Note: In fact, the sum above is a finite sum --- that is, only finitely many terms are nonzero.

Proof. Since is a basis,

Let . Take the inner product of both sides with . Then

since by orthonormality if and .

Therefore,

Observe that if , then

It follows that the only nonzero terms in the sum are those for which , and

Example. Here is an orthonormal basis for :

To express in terms of this basis, take the dot product of the vector with each element of the basis:

Example. Let denote the complex inner product space of complex-valued continuous functions on , where the inner product is defined by

I showed earlier that the following set is orthonormal:

Suppose I try to compute the "components" of relative to this orthonormal set by taking inner products --- that is, using the approach of the preceding example.

For ,

Suppose . Then

There are infinitely many nonzero components! Of course, the reason this does not contradict the earlier result is that may not lie in the span of S. S is orthonormal, hence independent, but it is not a basis for .

In fact, since , a finite linear combination of elements of S must be periodic.

It is still reasonable to ask whether (or in what sense) the infinite sum

represents the function . For example, it is reasonable to ask whether the series converges uniformly to f at each point of . The answers to these kinds of questions would require an excursion into the theory of Fourier series.

Since it's so easy to find the components of a vector relative to an orthonormal basis, it's of interest to have an algorithm which converts a given basis to an orthonormal one.

The Gram-Schmidt algorithm converts a basis to an orthonormal basis by "straightening out" the vectors one by one.

The picture shows the first step in the straightening process. Given vectors and , I want to replace with a vector perpendicular to . I can do this by taking the component of perpendicular to , which is

Lemma. ( Gram-Schmidt algorithm) Let is a set of nonzero vectors in an inner product space V. Suppose , ..., are pairwise orthogonal. Let

Then is orthogonal to , ..., .

Proof. Let . Then

Now for , so the right side collapses to

Suppose that I start with an independent set . Apply the Gram-Schmidt procedure to the set, beginning with . This produces an orthogonal set . In fact, is a nonzero orthogonal set, so it is independent as well.

To see that each is nonzero, suppose

Then

which contradicts the independence of .

In general, if the algorithm is applied iteratively to a set of vectors, the span is preserved at each state. That is,

This is trivially true at the start, since . Assume inductively that

The equation

shows that .

Conversely,

It follows that , so , as claimed.

To summarize: If you apply Gram-Schmidt to a set of vectors, the algorithm produces a new set of vectors with the same span as the old set. If the original set was independent, the new set is independent (and orthogonal) as well.

So, for example, if Gram-Schmidt is applied to a basis for an inner product space, it will produce an orthogonal basis for the space.

Finally, you can always produce orthonormal set from a orthogonal set (of nonzero vectors) --- merely divide each vector in the orthogonal set by its length.

Example. ( Gram-Schmidt) Apply Gram-Schmidt to the set

(A common mistake here is to project onto , , ... . I need to project onto the vectors that have already been orthogonalized. That is why I projected onto and rather than and .)

The set

is orthogonal.

The correponding orthonormal set is

Example. ( Gram-Schmidt) Find an orthonormal basis for the subspace spanned by the vectors

I'll use , , to denote the orthonormal basis.

To simplify the computations, you should fix the vectors so they're mutually perpendicular first. Then you can divide each by its length to get vectors of length 1.

First,

Next,

You can check that , so the first two are perpendicular.

Finally,

Thus, the orthogonal set is

To get an orthonormal basis, divide each of these vectors by its length:

For the next result, I'll use the following convention. Vectors will be understood to be column vectors; that is, v will refer to an n-dimensional column vector

If I need an n-dimensional row vector, I'll take the transpose. Thus,

Lemma. Let A be an invertible matrix with entries in . Then

defines an inner product on .

Proof. I have to check linearity, symmetry, and positive definiteness.

First, if , then

This proves that the function is linear in the first slot.

Next,

The second equality comes from the fact that for matrices. The third inequality comes from the fact that is a matrix, so it equals its transpose.

This proves that the function is symmetric.

Finally,

Now is an vector --- I'll label its components this way:

Then

That is, the inner product of a vector with itself is a nonnegative number. All that remains is to show that if the inner product of a vector with itself is 0, them the vector is .

But with the notation above,

implies , i.e.

Finally, I'll use the fact that A is invertible:

This proves that the function is positive definite, so it's an inner product.

Example. The previous lemma provides lots of examples of inner products on besides the usual dot product. All I have to do is take an invertible matrix A and form , defining the inner product as above.

For example,

Now

(Notice that will always be symmetric.) The inner product defined by this matrix is

For example, under this inner product,

Definition. A matrix A in is orthogonal if .

Proposition. Let A be an orthogonal matrix.

(a) .

(b) --- in other words, .

(c) The rows of A form an orthonormal set. The columns of A form an orthonormal set.

(d) A preserves dot products --- and hence, lengths and angles --- in the sense that

Proof. (a) If A is orthogonal,

Therefore, .

(b) Since , the determinant is certainly nonzero, so A is invertible. Hence,

But , so as well.

(c) The equation implies that the rows of A form an orthonormal set of vector. Likewise, shows that the same is true for the columns of A.

(d) The ordinary dot product of vectors and can be written as a matrix multiplication:

(Remember the convention that vectors are column vectors.)

Suppose A is orthogonal. Then

In other words, orthogonal matrices preserve dot products. It follows that orthogonal matrices will also preserve lengths of vectors and angles between vectors, because these are defined in terms of dot products.

Example. Find real numbers a and b such that the following matrix is orthogonal:

Since the columns of A must form an orthonormal set, I must have

(Note that already.) The first equation gives

The easy way to get a solution is to swap 0.6 and 0.8 and negate one of them; thus, and .

Since , I'm done. (If the a and b I chose had made , then I'd simply divide by its length.)

Example. Orthogonal matrices represent rotations of the plane about the origin or reflections across a line through the origin.

Rotations are represented by matrices

You can check that this works by considering the effect of multiplying the standard basis vectors and by this matrix.

Multiplying a vector by the following matrix product reflects the vector across the line L that makes an angle with the x-axis:

Reading from right to left, the first matrix rotates everything by radians, so L coincides with the x-axis. The second matrix reflects everything across the x-axis. The third matrix rotates everything by radians. Hence, a given vector is rotated by and reflected across the x-axis, after which the reflected vector is rotated by . The net effect is to reflect across L.

Many transformation problems can be easily accomplished by doing transformations to reduce a general problem to a special case.