Section 28.6 Theory
In this section.
Subsection 28.6.1 Properties of block-diagonal form
First, we'll record some of the properties of block-diagonal matrices explored in the latter part of Discovery guide 28.2. As we have sufficiently explored most of these patterns in Discovery guide 28.2, we state these properties without proof.Proposition 28.6.1.
Suppose A and B are n \times n matrices in block-diagonal form such that each has the same number of blocks and corresponding blocks are the same size. That is, assume
where A_1 and B_1 are both n_1 \times n_1\text{,} A_2 and B_2 are both n_2 \times n_2\text{,} etc..
- The product A B can be computed by taking the product of corresponding blocks, and powers (including the inverse) of A can be computed by taking powers of blocks. That is,\begin{align*} A B \amp= \begin{bmatrix} A_1 B_1 \\ \amp A_2 B_2 \\ \amp \amp \ddots \\ \amp \amp \amp A_\ell B_\ell \end{bmatrix} \text{,} \amp A^k \amp= \begin{bmatrix} A_1^k \\ \amp A_2^k \\ \amp \amp \ddots \\ \amp \amp \amp A_\ell^k \end{bmatrix} \text{,} \end{align*}where the second pattern is valid for all integer powers (assuming A is invertible in the case k \le 0).
- If \uvec{x}_1 is a vector in \R^{n_1}\text{,} \uvec{x}_2 is a vector in \R^{n_2}\text{,} etc., and \uvec{x} is the vector in \R^n formed by concatenating the vectors \uvec{x}_1,\uvec{x}_2,\dotsc,\uvec{x}_\ell together, then the product A \uvec{x} is similarly the vector in \R^n formed by concatenating the vectors A_1\uvec{x}_1,A_2\uvec{x}_2,\dotsc,A_\ell\uvec{x}_\ell together. That is,\begin{equation*} A \uvec{x} = \begin{bmatrix} A_1 \\ \amp A_2 \\ \amp \amp \ddots \\ \amp \amp \amp A_\ell \end{bmatrix} \begin{bmatrix} \uvec{x}_1 \\ \uvec{x}_2 \\ \vdots \\ \uvec{x}_\ell \end{bmatrix} = \begin{bmatrix} A_1 \uvec{x}_1 \\ A_2 \uvec{x}_2 \\ \vdots \\ A_\ell \uvec{x}_\ell \end{bmatrix}\text{.} \end{equation*}
- If \uvec{x}_j is a null space vector of block A_j\text{,} for each index j\text{,} then the vector \uvec{x} in \R^n formed by concatenating the vectors \uvec{x}_1,\uvec{x}_2,\dotsc,\uvec{x}_\ell together will be a null space vector for A\text{.}
- The rank A is the sum of the ranks of the blocks A_j\text{,} and similarly for the nullity of A\text{.}
- The determinant of A is the product of the determinants of the blocks in A\text{.} That is,\begin{equation*} \det A = (\det A_1) (\det A_2) \dotsm (\det A_\ell) \text{.} \end{equation*}
- The characteristic polynomial of A is the product of the characteristic polynomials of the blocks in A\text{.} That is,\begin{equation*} c_A(\lambda) = c_{A_1}(\lambda) c_{A_2}(\lambda) \dotsm c_{A_\ell}(\lambda) \text{.} \end{equation*}
- The eigenvalues of A are precisely the collection of eigenvalues of the blocks of A\text{.} If \uvec{x} \in \R^{n_j} is an eigenvector of block A_j corresponding to eigenvalue \lambda\text{,} then the vector\begin{equation*} \left[\begin{array}{l} \zerovec_{n_1} \\ \zerovec_{n_2} \\ \vdots \\ \uvec{x} \\ \vdots \\ \zerovec_{n_\ell} \end{array}\right] \end{equation*}is an eigenvector of A corresponding to \lambda\text{,} where each zero vector \zerovec_{n_k} has size corresponding to the size n_k of the block A_k\text{,} and vector \uvec{x} appears in the \nth[j] block.
Subsection 28.6.2 Invariant subspaces
As usual, to understand a subspace it is enough to understand a spanning set for the subspace, and this is true for invariant subspaces as well.Proposition 28.6.2.
Suppose A is an n \times n matrix, and \{ \uvec{w}_1, \uvec{w}_2, \dotsc, \uvec{w}_\ell \} is a spanning set for a subspace W of \R^n\text{.} Then W is A-invariant if and only if each of the vectors A\uvec{w}_1, A\uvec{w}_2, \dotsc, A\uvec{w}_\ell is again in W\text{.}
Proof.
Recall that with an βif and only ifβ statement, there are two things to prove.
(\(\Rightarrow\)).
Assume that \(W\) is \(A\)-invariant. Then, by definition, for every vector \(\uvec{w}\) in \(W\) the vector \(A\uvec{w}\) is again in \(W\text{.}\) Since each of the spanning vectors \(\uvec{w}_1,\uvec{w}_2,\dotsc,\uvec{w}_\ell\) lies in \(W\text{,}\) then clearly each of the vectors \(A\uvec{w}_1,A\uvec{w}_2,\dotsc,\uvec{w}_\ell\) lies again in \(W\text{.}\)
(\(\Leftarrow\)).
Assume that each of the vectors \(A\uvec{w}_1,A\uvec{w}_2,\dotsc,\uvec{w}_\ell\) lies in \(W\text{.}\) We must prove that for every vector \(\uvec{w}\) in \(W\text{,}\) the vector \(A\uvec{w}\) will again lie in \(W\text{.}\) Since \(\{\uvec{w}_1,\dotsc,\uvec{w}_\ell\}\) is a spanning set for \(W\text{,}\) a vector \(\uvec{w}\) in \(W\) can be expressed as a linear combination of these spanning vectors, say
Then,
Above we have expressed \(A \uvec{w}\) as a linear combination of the vectors \(A \uvec{w}_1, A \uvec{w}_2, \dotsc, A \uvec{w}_\ell\text{,}\) each of which, by assumption, lies in \(W\text{.}\) Therefore, since \(W\) is a subspace, it is closed under vector operations, and so we also have \(A\uvec{w}\) in \(W\text{.}\)
Proposition 28.6.3.
Suppose \lambda is an eigenvalue of the n \times n matrix A\text{.} Then the eigenspace E_\lambda (A) is an A-invariant subspace of \R^n\text{.}
Proof idea.
The point that must be verified is that if \(\uvec{w}\) is a vector in the eigenspace \(E_\lambda (A)\text{,}\) then the transformed vector \(\uvec{u} = A \uvec{w}\) is also in that eigenspace. Use the definition of eigenvector (i.e. \(A \uvec{x} = \lambda \uvec{x}\)), both applied to \(\uvec{x} = \uvec{w}\text{,}\) and as the condition to be verified in the case \(\uvec{x} = \uvec{u}\text{.}\)
Subsection 28.6.3 Independent subspaces
Here is a special case of testing independence of subspaces, in the case of two subspaces.Theorem 28.6.4.
Subspaces W_1,W_2 of a finite dimensional vector space are independent if and only if W_1 \cap W_2 = \{\zerovec\}\text{.}
Proof.
Again, with an βif and only ifβ statement, there are two things to prove. In both cases, begin with bases
for \(W_1\) and \(W_2\text{,}\) respectively, where \(d_1 = \dim W_1\) and \(d_2 = \dim W_2\text{.}\)
(β)
Assume that \(W_1\) and \(W_2\) are independent. By definition, this means that the combined set
is linearly independent.
We must prove that \(W_1 \cap W_2 = \{\zerovec\}\text{.}\) So suppose \(\uvec{v}\) lies in the intersection \(W_1 \cap W_2\text{.}\) If we are to prove that \(W_1\cap W_2\) consists of only the zero vector, then we must prove that \(\uvec{v} = \zerovec\text{.}\)
Since \(\uvec{v}\) is in \(W_1\text{,}\) there exist scalars \(a_1, a_2, \dotsc,a_{d_1}\) such that
But \(\uvec{v}\) also lies in \(W_2\text{,}\) and so there exist scalars \(b_1, b_2, \dotsc, b_{d_2}\) such that
Now, from the vector space identity \(\uvec{v} - \uvec{v} = \zerovec\text{,}\) we can subtract these two linear combination expressions for \(\uvec{v}\) to get
Linear independence of the combined set of basis vectors says that all of the scalars above are zero, so that
as desired.
(β)
Assume that \(W_1 \cap W_2 = \{\zerovec\}\text{.}\) We must prove that \(W_1,W_2\) are independent. That is, we need to verify that the combined collection \(\{ \uvec{u}_1, \dotsc, \uvec{u}_{d_1}, \uvec{w}_1, \dotsc, \uvec{w}_{d_2} \}\) remains linearly independent.
Setting up the test for linear independence, we assume that
Set
Then \(\uvec{v}\) is in \(W_1\text{.}\) But by moving all the \(\uvec{w}_j\) terms to the other side of (\(\star\)), we obtain a second expression for \(\uvec{v}\text{,}\)
so that \(\uvec{v}\) is also in \(W_2\text{.}\) Therefore, \(\uvec{v} \in W_1 \cap W_2\text{.}\) By assumption, this intersection contains only the zero vector, so we must have \(\uvec{v} = \zerovec\text{.}\) Our two different expressions for \(\uvec{v}\) as linear combinations lead to
These expressions are essentially the test for linear independence, one for basis \(\basisfont{B}_1\) and one for basis \(\basisfont{B}_2\text{.}\) As each of these bases must be an independent set, these two tests for independence lead to the conclusions
So, in fact, all of the coefficients in (\(\star\)) must be zero, and the test for independence has been confirmed.
Warning 28.6.5.
The condition W_1 \cap W_2 = \{\zerovec\} is not saying that the intersection of W_1 and W_2 is empty! In fact, it is impossible for two subspaces of any vector space to be disjoint (i.e. have empty intersection), since every subspace of a vector space contains the zero vector. The condition W_1 \cap W_2 = \{\zerovec\} is saying that the intersection of W_1 and W_2 contains only the zero vector.
Proposition 28.6.6. Complement spaces.
For every subspace W of a vector space V\text{,} there exists a complement subspace W' of V so that W,W' form a complete set of independent subsets.
Proof idea.
Choose a basis for \(W\text{,}\) and use Proposition 20.5.4 to enlarge it to a basis for all of \(V\text{.}\) Take \(W'\) to be the span of the new vectors that were used to enlarge the initial basis for \(W\text{,}\) and verify that the pair \(W,W'\) satisfies the definition of complete set of independent subspaces. It might be convenient to use Theorem 28.6.4 to check independence of the pair.
Corollary 28.6.7.
If W_1,W_2,\dotsc,W_m form a collection of independent subspaces of a vector space V\text{,} then there exists a subspace W' of V so that the enlarged collection W_1,W_2,\dotsc,W_m,W' is a complete set of independent subspaces.
Proof idea.
Apply Proposition 28.6.6, taking \(W\) to be the span of the initial collection of subspaces \(W_1,W_2,\dotsc,W_m\) all together.
Proposition 28.6.8.
Suppose W_1,W_2,\dotsc,W_m form a complete set of independent subspaces of a vector space V\text{.} Then for every vector \uvec{w} in V there exists a unique collection of vectors \uvec{w}_1,\uvec{w}_2,\dotsc,\uvec{w}_m\text{,} with each \uvec{w}_j in the corresponding space W_j\text{,} so that
Proof idea.
Choose a collection of bases \(\basisfont{B}_1,\basisfont{B}_2,\dotsc,\basisfont{B}_m\) for the subspaces \(W_1,W_2,\dotsc,W_m\text{,}\) and express \(\uvec{w}\) as a linear combination of these basis vectors taken all together.
Proposition 28.6.9.
A collection of vectors \uvec{w}_1,\uvec{w}_2,\dotsc,\uvec{w}_m is linearly independent if and only if the collection of subspaces
is independent.
Proof.
We leave this proof to you, the reader.
Theorem 28.6.10.
The various eigenspaces of an n \times n matrix always form an independent set of subspaces of \R^n\text{.}
Proof idea.
Combine Proposition 28.6.9 with Proposition 25.6.6.