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Chapter 10: Problem 45

Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if \(\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right]=\operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\)

Short Answer

Expert verified
In summary, we showed that the span of the \(W_i\)s is equal to the direct sum of the \(W_i\)s (i.e., \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\)) if and only if the dimension of the span of the \(W_i\)s is equal to the sum of the dimensions of the \(W_i\)s (i.e., \(\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right]=\operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\)). We demonstrated this by proving the statement in both directions (Part 1 and Part 2).

Step by step solution

01

Part 1: Span of \(W_i\)s equals the direct sum of \(W_i\)s implies equal dimensions

First, assume the span of the \(W_i\)s is equal to the direct sum of the \(W_i\)s (i.e., \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\)). Let's recall that the dimension of a direct sum of subspaces is equal to the sum of the dimensions of the subspaces. Then, we have: \[\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right] = \operatorname{dim}(W_{1} \oplus \cdots \oplus W_{r}) \\ = \operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\] Hence, if the span of the \(W_i\)s is equal to the direct sum of the \(W_i\)s, then the dimension of the span of the \(W_i\)s is equal to the sum of the dimensions of the \(W_i\)s.
02

Part 2: Equal dimensions implies span of \(W_i\)s equals the direct sum of \(W_i\)s

Now, assume that the dimension of the span of the \(W_i\)s is equal to the sum of the dimensions of the \(W_i\)s (i.e., \(\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right]=\operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\)). Since the span of the \(W_i\)s contains the set \(\bigcup_{i=1}^r W_i\), it must also include the direct sum of the \(W_i\)s, denoted as \(W_{1} \oplus \cdots \oplus W_{r}\). Therefore, we have: \[\operatorname{span}\left(W_{i}\right) \supseteq W_{1} \oplus \cdots \oplus W_{r}\] Now notice that: \[\operatorname{dim} \left[\operatorname{span}\left(W_{i}\right)\right] = \operatorname{dim}(W_{1} \oplus \cdots \oplus W_{r})\] As the dimensions are equal and since \(\operatorname{span}\left(W_{i}\right) \supseteq W_{1} \oplus \cdots \oplus W_{r}\), we can conclude that \(\operatorname{span}\left(W_{i}\right) = W_{1} \oplus \cdots \oplus W_{r}\). Hence, if the dimension of the span of the \(W_i\)s is equal to the sum of the dimensions of the \(W_i\)s, then the span of the \(W_i\)s is equal to the direct sum of the \(W_i\)s. In conclusion, we have shown that the span of the \(W_i\)s is equal to the direct sum of the \(W_i\)s if and only if the dimension of the span of the \(W_i\)s is equal to the sum of the dimensions of the \(W_i\)s.

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Most popular questions from this chapter

Suppose \(V=U \oplus W\) and that \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) is a basis of \(U .\) Show that \(\left\\{u_{1}+W, \ldots, u_{n}+W\right\\}\) is a basis of the quotient spaces \(V / W\). (Observe that no condition is placed on the dimensionality of \(V\) or \(W\).)

Let \(A\) be a \(4 \times 4\) matrix with minimal polynomial \(m(t)=\left(t^{2}+1\right)\left(t^{2}-3\right) .\) Find the rational canonical form for \(A\) if \(A\) is a matrix over (a) the rational field \(\mathbf{Q},(\mathbf{b})\) the real field \(\mathbf{R},(\mathrm{c})\) the complex field \(\mathbf{C}\).

Suppose the characteristic polynomial of \(T: V \rightarrow V\) is \(\Delta(t)=f_{1}(t)^{n_{1}} f_{2}(t)^{n_{2}} \cdots f_{r}(t)^{n_{r}},\) where the \(f_{i}(t)\) are distinct monic irreducible polynomials. Let \(V=W_{1} \oplus \cdots \oplus W_{r}\) be the primary decomposition of \(V\) into \(T\) invariant subspaces. Show that \(f_{i}(t)^{n_{i}}\) is the characteristic polynomial of the restriction of \(T\) to \(W_{i}\).

Let \(T: V \rightarrow V\) be linear. Let \(W\) be a \(T\) -invariant subspace of \(V\) and \(\bar{T}\) the induced operator on \(V / W\). Prove (a) The T-annihilator of \(v \in V\) divides the minimal polynomial of \(T\) (b) The \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) divides the minimal polynomial of \(T\) (a) The \(T\) -annihilator of \(v \in V\) is the minimal polynomial of the restriction of \(T\) to \(Z(v, T) ;\) therefore, by Problem \(10.6,\) it divides the minimal polynomial of \(T\) (b) The \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) divides the minimal polynomial of \(\bar{T},\) which divides the minimal polynomial of \(T\) by Theorem 10.16 Remark: In the case where the minimum polynomial of \(T\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial, then the \(T\) -annihilator of \(v \in V\) and the \(\bar{T}\) -annihilator of \(\bar{v} \in V / W\) are of the form \(f(t)^{m},\) where \(m \leq n\).

Let \(A=\left[\begin{array}{ccccc}0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ccccc}0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right] .\) The reader can verify that \(A\) and \(B\) are both nilpotent of index \(3 ;\) that is, \(A^{3}=0\) but \(A^{2} \neq 0,\) and \(B^{3}=0\) but \(B^{2} \neq 0 .\) Find the nilpotent matrices \(M_{A}\) and \(M_{B}\) in canonical form that are similar to \(A\) and \(B,\) respectively. Because \(A\) and \(B\) are nilpotent of index \(3, M_{A}\) and \(M_{B}\) must each contain a Jordan nilpotent block of order \(3,\) and none greater then \(3 .\) Note that \(\operatorname{rank}(A)=2\) and \(\operatorname{rank}(B)=3,\) so nullity \((A)=5-2=3\) and nullity \((B)=5-3=2 .\) Thus, \(M_{A}\) must contain three diagonal blocks, which must be one of order 3 and two of order \(1 ;\) and \(M_{B}\) must contain two diagonal blocks, which must be one of order 3 and one of order 2 Namely, \\[ M_{A}=\left[\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \quad \text { and } \quad M_{B}=\left[\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \\]
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