Bases

A basis of a vector space is one of the most useful concepts in linear algebra. It often provides a concise, finite description of an infinite vector space.

1. A= \begin{bmatrix} -3 & 0 & 1\\ 1 & 2 & 1\\ 5 & 1 & 6 \end{bmatrix}
2. Write the vector $\vect{w}=\colvector{6\\6\\15}$ as a linear combination of the columns of the matrix $A$ above. How many ways are there to answer this question?
3. Why is an orthonormal basis desirable?

We now have all the tools in place to define a basis of a vector space.

So, a basis is a linearly independent spanning set for a vector space. The requirement that the set spans $V$ insures that $S$ has enough raw material to build $V$, while the linear independence requirement insures that we do not have any more raw material than we need. As we shall see soon in , a basis is a minimal spanning set.

You may have noticed that we used the term basis for some of the titles of previous theorems ( , , ) and if you review each of these theorems you will see that their conclusions provide linearly independent spanning sets for sets that we now recognize as subspaces of $\complex{m}$. Examples associated with these theorems include , and . As we will see, these three theorems will continue to be powerful tools, even in the setting of more general vector spaces.

Furthermore, the archetypes contain an abundance of bases. For each coefficient matrix of a system of equations, and for each archetype defined simply as a matrix, there is a basis for the null space, three bases for the column space, and a basis for the row space. For this reason, our subsequent examples will concentrate on bases for vector spaces other than $\complex{m}$.

Notice that does not preclude a vector space from having many bases, and this is the case, as hinted above by the statement that the archetypes contain three bases for the column space of a matrix. More generally, we can grab any basis for a vector space, multiply any one basis vector by a nonzero scalar and create a slightly different set that is still a basis. For important vector spaces, it will be convenient to have a collection of nice bases. When a vector space has a single particularly nice basis, it is sometimes called the standard basis though there is nothing precise enough about this term to allow us to define it formally it is a question of style. Here are some nice bases for important vector spaces.

The bases described above will often be convenient ones to work with. However a basis does not have to obviously look like a basis.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

We have seen several examples of bases in different vector spaces. In this subsection, and the next (), we will consider building bases for $\complex{m}$ and its subspaces.

Suppose we have a subspace of $\complex{m}$ that is expressed as the span of a set of vectors, $S$, and $S$ is not necessarily linearly independent, or perhaps not very attractive. says that row-equivalent matrices have identical row spaces, while says the nonzero rows of a matrix in reduced row-echelon form are a basis for the row space. These theorems together give us a great computational tool for quickly finding a basis for a subspace that is expressed originally as a span.

provides another example of this flavor, though now we can notice that $X$ is a subspace, and that the resulting set of three vectors is a basis. This is such a powerful technique that we should do one more example.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

A quick source of diverse bases for $\complex{m}$ is the set of columns of a nonsingular matrix.

Perhaps we should view the fact that the standard unit vectors are a basis () as just a simple corollary of ? (See .)

With a new equivalence for a nonsingular matrix, we can update our list of equivalences.

We can easily illustrate our latest equivalence for nonsingular matrices.

We learned about orthogonal sets of vectors in $\complex{m}$ back in , and we also learned that orthogonal sets are automatically linearly independent (). When an orthogonal set also spans a subspace of $\complex{m}$, then the set is a basis. And when the set is orthonormal, then the set is an incredibly nice basis. We will back up this claim with a theorem, but first consider how you might manufacture such a set.

Suppose that $W$ is a subspace of $\complex{m}$ with basis $B$. Then $B$ spans $W$ and is a linearly independent set of nonzero vectors. We can apply the Gram-Schmidt Procedure () and obtain a linearly independent set $T$ such that $\spn{T}=\spn{B}=W$ and $T$ is orthogonal. In other words, $T$ is a basis for $W$, and is an orthogonal set. By scaling each vector of $T$ to norm 1, we can convert $T$ into an orthonormal set, without destroying the properties that make it a basis of $W$. In short, we can convert any basis into an orthonormal basis. , followed by , illustrates this process.

Unitary matrices () are another good source of orthonormal bases (and vice versa). Suppose that $Q$ is a unitary matrix of size $n$. Then the $n$ columns of $Q$ form an orthonormal set () that is therefore linearly independent (). Since $Q$ is invertible (), we know $Q$ is nonsingular (), and then the columns of $Q$ span $\complex{n}$ (). So the columns of a unitary matrix of size $n$ are an orthonormal basis for $\complex{n}$.

Why all the fuss about orthonormal bases? told us that any vector in a vector space could be written, uniquely, as a linear combination of basis vectors. For an orthonormal basis, finding the scalars for this linear combination is extremely easy, and this is the content of the next theorem. Furthermore, with vectors written this way (as linear combinations of the elements of an orthonormal set) certain computations and analysis become much easier. Here is the promised theorem.

A slightly less intimidating example follows, in three dimensions and with just real numbers.

Not only do the columns of a unitary matrix form an orthonormal basis, but there is a deeper connection between orthonormal bases and unitary matrices. Informally, the next theorem says that if we transform each vector of an orthonormal basis by multiplying it by a unitary matrix, then the resulting set will be another orthonormal basis. And more remarkably, any matrix with this property must be unitary! As an equivalence () we could take this as our defining property of a unitary matrix, though it might not have the same utility as .

Condensed — the full section is in Beezer, A First Course in Linear Algebra.