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Chapter 5: Problem 85

Suppose \(F\) and \(G\) are linear operators on \(V\) and that \(F\) is nonsingular. Assume that \(V\) has finite dimension. Show that \(\operatorname{rank}(F G)=\operatorname{rank}(G F)=\operatorname{rank}(G)\)

Short Answer

Expert verified
In conclusion, for linear operators \(F\) and \(G\) on a finite dimensional vector space \(V\), with \(F\) being nonsingular, we have shown that \(\operatorname{rank}(FG) = \operatorname{rank}(GF) = \operatorname{rank}(G)\).

Step by step solution

01

Recall definitions and properties

Recall the following definitions and properties related to linear operators: 1. rank(T) = dim(Im(T)), where Im(T) is the image of a linear operator T. 2. nullity(T) = dim(Ker(T)), where Ker(T) is the kernel (null space) of a linear operator T. 3. Dim(V) = rank(T) + nullity(T) for a linear operator T on V, also known as the Rank-Nullity theorem. 4. If T is a nonsingular linear operator, then rank(T) = dim(V) and nullity(T) = 0.
02

Determine the rank of FG

We want to show that rank(FG) = rank(G). Let T = FG. Then, rank(T) = dim(Im(T)) Since T is the composition of two linear operators, then Im(T) = Im(F)鈭�(G(Im(G))) Note that G(Im(G)) = Im(G) since G is a linear operator. Now, applying the Rank-Nullity theorem to G, we have dim(V) = rank(G) + nullity(G) Since F is nonsingular, rank(F) = dim(V). Then, dim(Im(F)) = dim(V) Also, Im(F)鈭�(Im(G)) 鈯� Im(G) Thus, we can conclude that rank(T) = rank(FG) = rank(G).
03

Determine the rank of GF

We want to show that rank(GF) = rank(G). Let T = GF. Then, rank(T) = dim(Im(T)) Since T is the composition of two linear operators, then Im(T) = G(Im(F)) Now, applying the Rank-Nullity theorem to G, we have dim(V) = rank(G) + nullity(G) Since F is nonsingular, rank(F) = dim(V). Then, dim(Im(F)) = dim(V) Since G is a linear operator, G(Im(F)) 鈯� Im(G) Thus, we can conclude that rank(T) = rank(GF) = rank(G).
04

Conclusion

In conclusion, we have shown that for linear operators F and G on a finite dimensional vector space V, with F being nonsingular, rank(FG) = rank(GF) = rank(G).

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Most popular questions from this chapter

Suppose \(F_{1}, F_{2}, \ldots, F_{n}\) are linear maps from \(V\) into \(U .\) Show that, for any scalars \(a_{1}, a_{2}, \ldots, a_{n}\) and for any \(v \in V\) \\[ \left(a_{1} F_{1}+a_{2} F_{2}+\cdots+a_{n} F_{n}\right)(v)=a_{1} F_{1}(v)+a_{2} F_{2}(v)+\cdots+a_{n} F_{n}(v) \\]

Let \(A\) be an \(n \times n\) matrix that is similar to an upper triangular matrix and has the distinct eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\( with corresponding multiplicities \)m_{1}, m_{2}, \ldots, m_{k}$. Prove the following statements. (a) \(\operatorname{tr}(A)=\sum^{k} m_{i} \lambda_{i}\) (b) $\operatorname{det}(A)=\left(\lambda_{1}\right)^{m_{1}}\left(\lambda_{2}\right)^{m_{2}} \ldots\left(\lambda_{k}\right)^{m_{k}}$.

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) be an ordered basis for \(\mathrm{V}\). Prove that \(\lambda\) is an eigenvalue of \(\mathrm{T}\) if and only if \(\lambda\) is an eigenvalue of \([\mathrm{T}]_{\beta}\).

Let \(F\) and \(G\) be the linear operators on \(\mathbf{R}^{2}\) defined by \(F(x, y)=(x+y, 0)\) and \(G(x, y)=(-y, x) .\) Find formulas defining the linear operators: (a) \(F+G,\) (b) \(5 F-3 G,\) (c) \(F G,(d) G F,(e) F^{2},(f) G^{2}\)

Determine whether \(\lim _{m \rightarrow \infty} A^{m}\) exists for each of the following matrices \(A\), and compute the limit if it exists. (a) \(\left(\begin{array}{ll}0.1 & 0.7 \\ 0.7 & 0.1\end{array}\right)\) (b) \(\left(\begin{array}{ll}-1.4 & 0.8 \\ -2.4 & 1.8\end{array}\right)\) (c) \(\left(\begin{array}{ll}0.4 & 0.7 \\ 0.6 & 0.3\end{array}\right)\) (d) \(\left(\begin{array}{ll}-1.8 & 4.8 \\ -0.8 & 2.2\end{array}\right)\) (e) \(\left(\begin{array}{rr}-2 & -1 \\ 4 & 3\end{array}\right)\) (f) \(\left(\begin{array}{ll}2.0 & -0.5 \\ 3.0 & -0.5\end{array}\right)\) (g) $\left(\begin{array}{rrr}-1.8 & 0 & -1.4 \\ -5.6 & 1 & -2.8 \\ 2.8 & 0 & 2.4\end{array}\right)$ (h) $\left(\begin{array}{rrr}3.4 & -0.2 & 0.8 \\ 3.9 & 1.8 & 1.3 \\ -16.5 & -2.0 & -4.5\end{array}\right)$ (i) $\left(\begin{array}{rrr}-\frac{1}{2}-2 i & 4 i & \frac{1}{2}+5 i \\ 1+2 i & -3 i & -1-4 i \\ -1-2 i & 4 i & 1+5 i\end{array}\right)$ (j) $\left(\begin{array}{ccc}\frac{-26+i}{3} & \frac{-28-4 i}{3} & 28 \\\ \frac{-7+2 i}{3} & \frac{-5+i}{3} & 7-2 i \\ \frac{-13+6 i}{6} & \frac{-5+6 i}{6} & \frac{35-20 i}{6}\end{array}\right)$
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