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Chapter 10: Problem 26

Let \(W\) be a subspace of a vector space \(V\). Suppose $\left\\{w_{1}, \ldots, w_{r}\right\\}\( is a basis of \)W$ and the set of \(\operatorname{cosets}\left\\{\bar{v}_{1}, \ldots, \bar{v}_{s}\right\\},\) where \(\bar{v}_{j}=v_{j}+W,\) is a basis of the quotient space. Show that the set of vectors \(B=\left\\{v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{r}\right\\}\) is a basis of \(V .\) Thus, $\operatorname{dim} V=\operatorname{dim} W+\operatorname{dim}(V / W)$.

Short Answer

Expert verified
Since B spans V and is linearly independent, it must be a basis for V. Furthermore, the dimension of V is the number of elements in B. Since B has \(s\) coset representatives from the basis of V/W and \(r\) basis vectors from W: \[\operatorname{dim} V = s + r = \operatorname{dim} W + \operatorname{dim}(V / W)\] This confirms the given statement in the exercise.

Step by step solution

01

Prove that the set B spans V

Let us consider an arbitrary vector v in V. By definition of the quotient space, we can write \(v = w + v_j\) with \(w \in W\) and \(v_j \in B\) (or equivalently, \(v+ W = v_j + W\)). Thus, since \(w\) is an element of W, it can be expressed as a linear combination of the basis vectors of W, i.e., \(w = a_1 w_1 + \dots + a_r w_r\). And since \(v_j\) already belongs to B, the vector \(v\) can be written as a linear combination of the elements of B. Therefore, the set B spans V.
02

Prove that the set B is linearly independent

Now we must show that the set B is linearly independent. Let's suppose we have a linear combination of the elements of B equal to the zero vector: \[a_1 v_{1} + \cdots + a_s v_{s} + b_1 w_{1}+ \cdots + b_r w_{r} = 0\] Since \(V\) is a vector space, we can subtract some of the terms from both sides of the equation: \[a_1 v_{1} + \cdots + a_s v_{s} = -b_1 w_{1} - \cdots - b_r w_{r}\] However, since the left side is a linear combination of the coset representatives \(\bar{v}_1,\ldots,\bar{v}_s\), and the right side is a linear combination of the basis vectors of W, their sum must belong to the same coset in V/W. That is: \[(a_1 v_{1} + \cdots + a_s v_{s}) + W = w + W\] where \(w \in W\). Therefore, by the uniqueness of coset representatives in the quotient space, we have that: \[w = a_1 v_{1} + \cdots + a_s v_{s}\] By the linear independence of \(\left\\{w_1, \ldots, w_r\right\\}\) and the fact that \(w\) belongs to W, we can deduce that \(a_1 = \cdots = a_s = 0\), and thus: \[0 = -b_1 w_{1} - \cdots - b_r w_{r}\] Now, because the basis vectors of W are linearly independent, we have that \(b_1 = \cdots = b_r = 0\). Therefore, we proved that the set B is linearly independent.
03

Conclude that B is a basis for V and compute the dimension of V

Since B spans V and is linearly independent, it must be a basis for V. Furthermore, the dimension of V is the number of elements in B. Since B has \(s\) coset representatives from the basis of V/W and \(r\) basis vectors from W: \[\operatorname{dim} V = s + r = \operatorname{dim} W + \operatorname{dim}(V / W)\] This confirms the given statement in the exercise.

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Most popular questions from this chapter

Suppose dim \(V=n .\) Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

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_{r n_{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 \(\left\\{w_{i 1}, \ldots, w_{i n}\right\\}\) 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}+\dots+b_{1 n_{1}} w_{1 n_{1}}+\cdots+b_{r 1} w_{r 1}+\cdots+b_{m_{r}} w_{r 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 \(\left\\{w_{i j_{i}}\right\\}\) is a basis of \(W_{i},\) each \(w_{i}\) is a linear combination of the \(w_{i j},\) 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}+\dots+a_{1 n_{1}} w_{1 n_{1}}+\cdots+a_{r 1} w_{r 1}+\cdots+a_{r n_{r}} w_{r n_{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 \(\left\\{w_{i j}\right\\}\) implies that all the \(a\) 's are \(0 .\) Thus, \(B\) is linearly independent and is a basis of \(V\)

Prove Theorem 10.11 on the Jordan canonical form for an operator \(T\) By the primary decomposition theorem, \(T\) is decomposable into operators \(T_{1}, \ldots, T_{r} ;\) that is, \(T=T_{1} \oplus \cdots \oplus T_{r},\) where \(\left(t-\lambda_{i}\right)^{m_{i}}\) is the minimal polynomial of \(T_{i} .\) Thus, in particular, \\[ \left(T_{1}-\lambda_{1} I\right)^{m_{1}}=\mathbf{0}, \ldots,\left(T_{r}-\lambda_{r} I\right)^{m_{r}}=\mathbf{0} \\] Set \(N_{i}=T_{i}-\lambda_{i} I .\) Then, for \(i=1, \ldots, r\) \\[ T_{i}=N_{i}+\lambda_{i} I, \quad \text { where } \quad N_{i}^{m^{i}}=\mathbf{0} \\] That is, \(T_{i}\) is the sum of the scalar operator \(\lambda_{i} I\) and a nilpotent operator \(N_{i},\) which is of index \(m_{i}\) because \(\left(t-\lambda_{i}\right)_{i}^{m}\) is the minimal polynomial of \(T_{i}\) Now, by Theorem 10.10 on nilpotent operators, we can choose a basis so that \(N_{i}\) is in canonical form. In this basis, \(T_{i}=N_{i}+\lambda_{i} I\) is represented by a block diagonal matrix \(M_{i}\) whose diagonal entries are the matrices \(J_{i j}\). The direct sum \(J\) of the matrices \(M_{i}\) is in Jordan canonical form and, by Theorem \(10.5,\) is a matrix representation of \(T\) Last, we must show that the blocks \(J_{i j}\) satisfy the required properties. Property (i) follows from the fact that \(N_{i}\) is of index \(m_{i} .\) Property (ii) is true because \(T\) and \(J\) have the same characteristic polynomial. Property (iii) is true because the nullity of \(N_{i}=T_{i}-\lambda_{i} I\) is equal to the geometric multiplicity of the eigenvalue \(\lambda_{i}\) Property (iv) follows from the fact that the \(T_{i}\) and hence the \(N_{i}\) are uniquely determined by \(T\)

Let \(V\) be a vector space of odd dimension (greater than 1 ) over the real field \(\mathbf{R}\). Show that any linear operator on \(V\) has an invariant subspace other than \(V\) or \(\\{0\\}\).

Let \(W=Z(v, T),\) and suppose the \(T\) -annihilator of \(v\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial of degree \(d .\) Show that \(f(T)^{s}(W)\) is a cyclic subspace generated by \(f(T)^{s}(v)\) and that it has dimension \(d(n-s)\) if \(n>s\) and dimension 0 if \(n \leq s\).
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