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Chapter 2: Problem 16

Let \(V\) be a finite-dimensional vector space, and let \(T: V \rightarrow V\) be linear. (a) If \(\operatorname{rank}(T)=\operatorname{rank}\left(T^{2}\right)\), prove that \(R(T) \cap N(T)=\\{0\\}\). Deduce that $\mathrm{V}=\mathrm{R}(\mathrm{T}) \oplus \mathrm{N}(\mathrm{T})$ (see the exercises of Section 1.3). (b) Prove that $\mathrm{V}=\mathrm{R}\left(\mathrm{T}^{k}\right) \oplus \mathrm{N}\left(\mathrm{T}^{k}\right)\( for some positive integer \)k$.

Short Answer

Expert verified
In this exercise, we are supposed to prove two parts: (a) If we are given that rank(T) = rank(T^2), we can prove that R(T) 鈭� N(T) = {0} by restating the properties of rank, N(T), and R(T), and using the given condition. We first suppose that x 鈭� R(T) 鈭� N(T), then show that x must be the zero vector. This proves that R(T) 鈭� N(T) = {0}. We can then deduce that V = R(T) 鈯� N(T), using a result from Section 1.3. (b) To prove that V = R(T^k) 鈯� N(T^k) for some positive integer k, we first define a set of nested subspaces V_i = R(T^i) and find the smallest non-negative integer n such that dim(V_n) = dim(V_{n+1}). We can apply the result of part (a) to T^n instead of T, and deduce that V = R(T^n) 鈯� N(T^n). Let k = n, a positive integer, hence V = R(T^k) 鈯� N(T^k).

Step by step solution

01

Restate the properties of rank, N(T) and R(T)

Rank(T) is the dimension of the range space of T, denoted R(T). N(T) denotes the null space of T. We know that dim(V) = rank(T) + dim(N(T)).
02

Use the given condition

We are given that rank(T) = rank(T^2). Since dim(V) = rank(T) + dim(N(T)), we also have dim(V) = rank(T^2) + dim(N(T^2)).
03

Prove that R(T) 鈭� N(T) = {0}

Suppose x 鈭� R(T) 鈭� N(T). Then, x = Tv, where v 鈭� V, and T(x) = 0. Since x 鈭� N(T), we have T^2(v) = T(Tv) = T(x) = 0. This implies that v 鈭� N(T^2). On the other hand, since x 鈭� R(T), we know that v 鈭� R(T^2) as rank(T) = rank(T^2). Since R(T^2) and N(T^2) are disjoint subspaces, it must be true that v = 0. Thus, x = T(v) = T(0) = 0, proving that R(T) 鈭� N(T) = {0}.
04

Deduce that V = R(T) 鈯� N(T)

Since R(T) 鈭� N(T) = {0}, we can apply the result from Section 1.3 to deduce that V = R(T) 鈯� N(T). This concludes the proof for part (a). b) Proving that V = R(T^k) 鈯� N(T^k) for some positive integer k:
05

Define a set of nested subspaces

Define V_i as the range space of T^i, i.e., V_i = R(T^i). Since the dimensions of V_i are non-negative integers, there must be a smallest non-negative integer n such that dim(V_n) = dim(V_{n+1}).
06

Apply the result of part (a) to T^n

Since dim(V_n) = dim(V_{n+1}), we can apply the result of part (a) to T^n instead of T, and deduce that V = R(T^n) 鈯� N(T^n).
07

Use the result for positive integer k

Let k = n, a positive integer. Then, V = R(T^k) 鈯� N(T^k). This concludes the proof for part (b).

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Most popular questions from this chapter

For each of the following vector spaces \(V\) and bases \(\beta\), find explicit formulas for vectors of the dual basis \(\beta^{*}\) for \(\mathrm{V}^{*}\), as in Example 4 . (a) \(\mathrm{V}=\mathrm{R}^{3} ; \beta=\\{(1,0,1),(1,2,1),(0,0,1)\\}\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R) ; \beta=\left\\{1, x, x^{2}\right\\}\)

Let \(V\) be a nonzero vector space, and let \(W\) be a proper subspace of \(V\) (i.e., \(W \neq V\) ). (a) Let \(\mathrm{g} \in \mathrm{W}^{*}\) and \(v \in \mathrm{V}\) with $v \notin \mathrm{W}\(. Prove that for any scalar \)a\( there exists a function \)\mathrm{f} \in \mathrm{V}^{*}\( such that \)\mathrm{f}(v)=a$ and \(\mathrm{f}(x)=\mathrm{g}(x)\) for all \(x\) in W. Hint: For the infinite- dimensional case, use Exercise 4 of Section \(1.7\) and Exercise 35 of Section 2.1. (b) Use (a) to prove there exists a nonzero linear functional $f \in \mathrm{V}^{*}\( such that \)\mathrm{f}(x)=0\( for all \)x \in \mathrm{W}$.

For each of the following pairs of ordered bases \(\beta\) and \(\beta^{\prime}\) for \(\mathrm{P}_{2}(R)\), find the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. (a) \(\beta=\left\\{x^{2}, x, 1\right\\}\) and $\beta^{\prime}=\left\\{a_{2} x^{2}+a_{1} x+a_{0}, b_{2} x^{2}+b_{1} x+b_{0}, c_{2} x^{2}+c_{1} x+c_{0}\right\\}$ (b) \(\beta=\left\\{1, x, x^{2}\right\\}\) and $\beta^{\prime}=\left\\{a_{2} x^{2}+a_{1} x+a_{0}, b_{2} x^{2}+b_{1} x+b_{0}, c_{2} x^{2}+c_{1} x+c_{0}\right\\}$ (c) \(\beta=\left\\{2 x^{2}-x, 3 x^{2}+1, x^{2}\right\\}\) and \(\beta^{\prime}=\left\\{1, x, x^{2}\right\\}\) (d) \(\beta=\left\\{x^{2}-x+1, x+1, x^{2}+1\right\\}\) and \(\beta^{\prime}=\left\\{x^{2}+x+4,4 x^{2}-3 x+2,2 x^{2}+3\right\\}\) (e) \(\beta=\left\\{x^{2}-x, x^{2}+1, x-1\right\\}\) and \(\beta^{\prime}=\left\\{5 x^{2}-2 x-3,-2 x^{2}+5 x+5,2 x^{2}-x-3\right\\}\) (f) \(\beta=\left\\{2 x^{2}-x+1, x^{2}+3 x-2,-x^{2}+2 x+1\right\\}\) and \(\beta^{\prime}=\left\\{9 x-9, x^{2}+21 x-2,3 x^{2}+5 x+2\right\\}\)

Let \(V\) and \(W\) be finite-dimensional vector spaces with ordered bases \(\beta=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) and \(\gamma=\left\\{w_{1}, w_{2}, \ldots, w_{m}\right\\}\), respectively. By Theorem \(2.6\) (p. 73), there exist linear transformations $\mathrm{T}_{i j}: \mathrm{V} \rightarrow \mathrm{W}$ such that $$ \mathrm{T}_{i j}\left(v_{k}\right)= \begin{cases}w_{i} & \text { if } k=j \\\ 0 & \text { if } k \neq j\end{cases} $$ First prove that $\left\\{\mathbf{T}_{i j}: 1 \leq i \leq m, 1 \leq j \leq n\right\\}\( is a basis for \)\mathcal{L}(\mathbf{V}, \mathbf{W})$. Then let \(M^{i j}\) be the \(m \times n\) matrix with 1 in the \(i\) th row and \(j\) th column and 0 elsewhere, and prove that $\left[\mathrm{T}_{i j}\right]_{\beta}^{\gamma}=M^{i j}$. Again by Theorem 2.6, there exists a linear transformation $\Phi_{\beta}^{\gamma}: \mathcal{L}(\mathrm{V}, \mathrm{W}) \rightarrow \mathrm{M}_{m \times n}(F)$ such that \(\Phi_{\beta}^{\gamma}\left(T_{i j}\right)=M^{i j}\). Prove that \(\Phi_{\beta}^{\gamma}\) is an isomorphism.

Suppose \(A\) is a square matrix. Show (a) \(A+A^{T}\) is symmetric, (b) \(A-A^{T}\) is skew-symmetric, (c) \(A=B+C,\) where \(B\) is symmetric and \(C\) is skew- symmetric.
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