A Second Course in Linear Algebra

Consider the linear transformation \(\ltdefn{T}{\complex{4}}{\complex{4}}\) defined by \(\lteval{T}{\vect{x}}=A\vect{x}\) where \(A\) is given by

Define (with zero motivation),

and set \(W=\spn{\set{\vect{w}_1,\,\vect{w}_2}}\text{.}\) We verify that \(W\) is an invariant subspace of \(\complex{4}\) with respect to \(T\text{.}\) By the definition of \(W\text{,}\) any vector chosen from \(W\) can be written as a linear combination of \(\vect{w}_1\) and \(\vect{w}_2\text{.}\) Suppose that \(\vect{w}\in W\text{,}\) and then check the details of the following verification,

So, by Definition IS, \(W\) is an invariant subspace of \(\complex{4}\) relative to \(T\text{.}\)

In an entirely similar manner we construct another invariant subspace of \(T\text{.}\)With zero motivation, define

and set \(X=\spn{\set{\vect{x}_1,\,\vect{x}_2}}\text{.}\) We verify that \(X\) is an invariant subspace of \(\complex{4}\) with respect to \(T\text{.}\) By the definition of \(X\text{,}\) any vector chosen from \(X\) can be written as a linear combination of \(\vect{x}_1\) and \(\vect{x}_2\text{.}\) Suppose that \(\vect{x}\in X\text{,}\) and then check the details of the following verification,

So, by Definition IS, \(X\) is an invariant subspace of \(\complex{4}\) relative to \(T\text{.}\)

There is a bit of magic in each of these verifications where the two outputs of \(T\) happen to equal linear combinations of the two inputs. But this is the essential nature of an invariant subspace. We'll have a peek under the hood later in Example Example 3.1.8, and it will not look so magical after all.

Verify that \(B=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{x}_1,\,\vect{x}_2}\) is linearly independent, and hence a basis of \(\complex{4}\text{.}\) Splitting this basis in half, Theorem Theorem 1.2.3 tells us that \(\complex{4}=W\ds X\text{.}\) To see exactly why a decomposition of a vector space into a direct sum of invariant subspaces might be interesting work Exercise Checkpoint 1.4.3 now.

Define the linear transformation \(\ltdefn{S}{M_{22}}{M_{22}}\) by

Build a matrix representation of \(S\) relative to the standard basis (Definition MR) and compute eigenvalues and eigenspaces of \(S\) with the computational techniques of Chapter E in concert with Theorem EER. Then

So by Theorem Theorem 1.4.5, both \(\eigenspace{S}{1}\) and \(\eigenspace{S}{2}\) are invariant subspaces of \(M_{22}\) relative to \(S\text{.}\)

However, Theorem Theorem 1.4.5 provides even more invariant subspaces. Since \(\eigenspace{S}{1}\) has dimension 1, it has no interesting subspaces, however \(\eigenspace{S}{2}\) has dimension 2 and has a plethora of subspaces. For example, set

and define \(U=\spn{\set{\vect{u}}}\text{.}\) Then since \(U\) is a subspace of \(\eigenspace{S}{2}\text{,}\) Theorem Theorem 1.4.5 says that \(U\) is an invariant subspace of \(M_{22}\) (or we could check this claim directly based simply on the fact that \(\vect{u}\) is an eigenvector of \(S\)).

Consider the \(10\times 10\) matrix \(A\) below as defining a linear transformation \(\ltdefn{T}{\complex{10}}{\complex{10}}\text{.}\) We also provide a Sage version of the matrix for use online.

The matrix \(A\) has rank \(9\) and so \(T\) has a nontrivial kernel. But it gets better. \(T\) has been constructed specially so that the nullity of \(T^k\) is exactly \(k\text{,}\) for \(0\leq k\leq 10\text{.}\) This is an extremely unusual situation, but is a good illustration of a very general theorem about kernels of null spaces, coming next. We compute the invariant subspace \(\krn{T^5}\text{,}\) you can practice with others.

We work with the matrix, recalling that null spaces and column spaces of matrices correspond to kernels and ranges of linear transformations once we understand matrix representations of linear transformations (Section MR).

As an illustration of \(\krn{T^5}\) as a subspace invariant under \(T\text{,}\) we form a linear combination of the five basis vectors (named \(\vect{z}_i\text{,}\) in order), which will be then be an element of the invariant subspace. We apply \(T\text{,}\) so the output should again be in the subspace, which we verify by giving an expression for the output as a linear combination of the basis vectors for the subspace.

In Exercise Checkpoint 1.4.4 you verified that the subspaces \(U, V\subseteq\complex{5}\) are invariant with respect to the linear transformation \(\ltdefn{R}{\complex{5}}{\complex{5}}\) defined by \(\lteval{R}{\vect{x}}=B\vect{x}\text{.}\)

It is a simple matter to define two new linear transformations, \(\restrict{R}{U}, \restrict{R}{V}\text{,}\)

It should not look like we have accomplished much. Worse, it might appear that \(R=\restrict{R}{U}=\restrict{R}{V}\) since each is described by the same rule for computing the image of an input. The difference is that the domains are all different: \(\complex{5},U,V\text{.}\) Since \(U\) and \(V\) are invariant subspaces, we can then use these subspaces for the codomains of the restrictions.

We will frequently need the matrix representations of linear transformation restrictions, so let's compute those now for this example. Let

be the bases for \(U\) and \(V\text{,}\) respectively.

For \(U\)

Applying Definition MR, we have

For \(V\)

Applying Definition MR, we have

It is no accident that these two square matrices are the diagonal blocks of the matrix representation you built for \(R\) relative to the basis \(C\cup D\) in Exercise Checkpoint 1.4.4.