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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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Let \(A\) and \(B\) be \(n \times n\) matrices such that \(A B\) is invertible. (a) Prove that \(A\) and \(B\) are invertible. Hint: See Exercise 12 of Section 2.3. (b) Give an example to show that a product of nonsquare matrices can be invertible even though the factors, by definition, are not.

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