/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Prove Theorem 10.4: Suppose \(W_... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 8

Prove Theorem 10.4: Suppose \(W_{1}, W_{2}, \ldots, W_{r}\) are subspaces of \(V\) with respective bases $$ B_{1}=\left\\{w_{11}, w_{12}, \ldots, w_{1 n_{1}}\right\\}, \quad \ldots, \quad B_{r}=\left\\{w_{r 1}, w_{r 2}, \ldots, w_{m_{r}}\right\\} $$ Then \(V\) is the direct sum of the \(W_{i}\) if and only if the union \(B=\bigcup_{i} B_{i}\) is a basis of \(V\). Suppose \(B\) is a basis of \(V\). Then, for any \(v \in V\), $$ v=a_{11} w_{11}+\cdots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{m_{r}} w_{m_{r}}=w_{1}+w_{2}+\cdots+w_{r} $$ where \(w_{i}=a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i}\). We next show that such a sum is unique. Suppose $$ v=w_{1}^{\prime}+w_{2}^{\prime}+\cdots+w_{r}^{\prime}, \quad \text { where } \quad w_{i}^{\prime} \in W_{i} $$ Because \(\left\\{w_{i 1}, \ldots, w_{i n_{i}}\right\\}\) is a basis of \(W_{i}, w_{i}^{\prime}=b_{i 1} w_{i 1}+\cdots+b_{i n_{i}} w_{i n_{i}}\), and so $$ v=b_{11} w_{11}+\cdots+b_{1 n_{1}} w_{1 n_{1}}+\cdots+b_{r 1} w_{r 1}+\cdots+b_{r n_{r}} w_{m_{r}} $$ Because \(B\) is a basis of \(V, a_{i j}=b_{i j}\), for each \(i\) and each \(j\). Hence, \(w_{i}=w_{i}^{\prime}\), and so the sum for \(v\) is unique. Accordingly, \(V\) is the direct sum of the \(W_{i}\). Conversely, suppose \(V\) is the direct sum of the \(W_{i}\). Then for any \(v \in V, v=w_{1}+\cdots+w_{r}\), where \(w_{i} \in W_{i}\). Because \(\left\\{w_{i j_{j}}\right\\}\) is a basis of \(W_{i}\), each \(w_{i}\) is a linear combination of the \(w_{i j_{j}}\), and so \(v\) is a linear combination of the elements of \(B\). Thus, \(B\) spans \(V\). We now show that \(B\) is linearly independent. Suppose $$ a_{11} w_{11}+\cdots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{r n_{r}} w_{m_{r}}=0 $$ Note that \(a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}} \in W_{i}\). We also have that \(0=0+0 \cdots 0 \in W_{i}\). Because such a sum for 0 is unique, $$ a_{i 1} w_{i 1}+\cdots+a_{i n_{i}} w_{i n_{i}}=0 \quad \text { for } i=1, \ldots, r $$ The independence of the bases \(\left\\{w_{i j_{i}}\right\\}\) implies that all the \(a\) 's are 0 . Thus, \(B\) is linearly independent and is a basis of \(V\).

Short Answer

Expert verified
In conclusion, Theorem 10.4 states that a vector space V is the direct sum of subspaces \(W_i\) if and only if the union of their respective bases B is a basis of V. We proved this theorem by showing that (1) if B is a basis for V, then V is the direct sum of the subspaces \(W_i\) and (2) if V is the direct sum of subspaces \(W_i\), then B is a basis for V. This establishes the equivalence between the direct sum decomposition and the union of subspaces' bases being a basis for the vector space V.

Step by step solution

01

Prove that if B is a basis for V, then V is the direct sum of the subspaces \(W_i\)

Since B is a basis for V, any vector v in V can be written as a unique linear combination of the elements of B: $$ v = a_{11}w_{11}+\cdots+a_{1n_{1}}w_{1n_{1}}+\cdots+a_{r1}w_{r1}+\cdots+a_{rn_{r}}w_{rn_{r}} $$ Since the elements of B belong to the subspaces \(W_i\), we can rewrite the above equation as: $$ v = w_1 + w_2 + \cdots + w_r $$ where \(w_i = a_{i1}w_{i1}+\cdots+a_{in_i}w_{in_i} \in W_i\). Now, let's prove the uniqueness of this decomposition. Suppose we have another representation of v: $$ v = w'_1 + w'_2 + \cdots + w'_r $$ where \(w'_i \in W_i\). Then, for each i, we have \(w_i = w'_i\), and thus the sum for v is unique. Hence, V is the direct sum of the subspaces \(W_i\).
02

Prove that if V is the direct sum of subspaces \(W_i\), then B is a basis for V

If V is the direct sum of subspaces \(W_i\), then any vector v in V can be uniquely written as the sum of vectors from the corresponding subspaces: $$ v = w_1 + w_2 + \cdots + w_r $$ Since the elements of B form bases for the subspaces \(W_i\), we can represent any vector v in V as a linear combination of the elements of B. Thus, B spans V. To prove that B is linearly independent, we consider the following equation: $$ a_{11}w_{11} + \cdots + a_{1n_{1}}w_{1n_{1}} + \cdots + a_{r1}w_{r1} + \cdots + a_{rn_{r}}w_{rn_{r}} = 0 $$ Because the direct sum decomposition is unique, we have: $$ a_{i1}w_{i1}+\cdots+a_{in_i}w_{in_i}=0 \quad \text { for } i=1,\ldots,r $$ Since each of the bases \(\{w_{ij}\}\) is linearly independent, all coefficients \(a_{ij}\) are zero, which implies that B is linearly independent as well. Thus, B is a basis of V. This completes the proof of the theorem.

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Most popular questions from this chapter

Let \(V\) be a seven-dimensional vector space over \(\mathbf{R}\), and let \(T: V \rightarrow V\) be a linear operator with minimal polynomial \(m(t)=\left(t^{2}-2 t+5\right)(t-3)^{3}\). Find all possible rational canonical forms \(M\) of \(T\).

Let \(V\) be a seven-dimensional vector space over \(\mathbf{R}\), and let \(T: V \rightarrow V\) be a linear operator with minimal polynomial \(m(t)=\left(t^{2}-2 t+5\right)(t-3)^{3} .\) Find all possible rational canonical forms \(M\) of \(T\) Because \(\operatorname{dim} V=7\), there are only two possible characteristic polynomials, \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)^{2}\) \((t-3)^{3}\) or \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)(t-3)^{5} .\) Moreover, the sum of the orders of the companion matrices must add up to 7. Also, one companion matrix must be \(C\left(t^{2}-2 t+5\right)\) and one must be \(C\left((t-3)^{3}\right)=\) \(C\left(t^{3}-9 t^{2}+27 t-27\right)\). Thus, \(M\) must be one of the following block diagonal matrices: (a) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],\left[\begin{array}{rr}0 & -9 \\ 1 & 6\end{array}\right]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],[3],[3]\right)\)(b) We have that $$ w_{k}=0+\cdots+0+w_{k}+0+\cdots+0 $$ is the unique sum corresponding to \(w_{k} \in W_{k}\); hence, \(E\left(w_{k}\right)=w_{k} .\) Then, for any \(v \in V\), $$ E^{2}(v)=E(E(v))=E\left(w_{k}\right)=w_{k}=E(v) $$ Thus, \(E^{2}=E\), as required.

Prove Theorem 10.8: In Theorem \(10.7\) (Problem 10.9), if \(f(t)\) is the minimal polynomial of \(T\) (and \(g(t)\) and \(h(t)\) are monic), then \(g(t)\) is the minimal polynomial of the restriction \(T_{1}\) of \(T\) to \(U\) and \(h(t)\) is the minimal polynomial of the restriction \(T_{2}\) of \(T\) to \(W\). Let \(m_{1}(t)\) and \(m_{2}(t)\) be the minimal polynomials of \(T_{1}\) and \(T_{2}\), respectively. Note that \(g\left(T_{1}\right)=0\) and \(h\left(T_{2}\right)=0\) because \(U=\) Ker \(g(T)\) and \(W=\) Ker \(h(T)\). Thus, $$ m_{1}(t) \text { divides } g(t) \quad \text { and } \quad m_{2}(t) \text { divides } h(t) $$ By Problem 10.9, \(f(t)\) is the least common multiple of \(m_{1}(t)\) and \(m_{2}(t) .\) But \(m_{1}(t)\) and \(m_{2}(t)\) are relatively prime because \(g(t)\) and \(h(t)\) are relatively prime. Accordingly, \(f(t)=m_{1}(t) m_{2}(t)\). We also have that \(f(t)=g(t) h(t)\). These two equations together with (1) and the fact that all the polynomials are monic imply that \(g(t)=m_{1}(t)\) and \(h(t)=m_{2}(t)\), as required.

Let \(W\) be a subspace of a vector space \(V\). Show that the following are equivalent: (i) \(\quad u \in v+W\) (ii) \(\quad u-v \in W\) (iii) \(\quad v \in u+W\)

Suppose \(W\) is a subspace of a vector space \(V\). Show that the operations in Theorem \(10.15\) are well defined; namely, show that if \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W\), then (a) \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W \quad\) and (b) \(k u+W=k u^{\prime}+W \quad\) for any \(k \in K\) (a) Because \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W\), both \(u-u^{\prime}\) and \(v-v^{\prime}\) belong to \(W\). But then \((u+v)-\left(u^{\prime}+v^{\prime}\right)=\left(u-u^{\prime}\right)+\left(v-v^{\prime}\right) \in W .\) Hence, \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W .\) (b) Also, because \(u-u^{\prime} \in W\) implies \(k\left(u-u^{\prime}\right) \in W\), then \(k u-k u^{\prime}=k\left(u-u^{\prime}\right) \in W\); accordingly, \(k u+W=k u^{\prime}+W\)
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