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Chapter 3: Problem 21

Let \(A\) and \(B\) be \(m \times n\) matrices. Show that \\[ \operatorname{rank}(A+B) \leq \operatorname{rank}(A)+\operatorname{rank}(B) \\]

Short Answer

Expert verified
To prove the inequality \(\operatorname{rank}(A+B) \leq \operatorname{rank}(A)+\operatorname{rank}(B)\), we consider the linearly independent rows of matrices A and B, calculate the sum of the matrices, and account for the possibility of linearly dependent rows being produced. Finally, we establish the desired inequality as \(\operatorname{rank}(C) = r_A + r_B - \text{overlap} \leq r_A + r_B\), which confirms the inequality holds.

Step by step solution

01

Set up and observe properties of matrices A and B

Let A have r_A linearly independent rows and B have r_B linearly independent rows. Then, we can write the matrices A and B in row echelon form or row-reduced echelon form. Without the loss of generality, let us assume r_A 鈮� r_B. Notice that the rank of each matrix is the number of linearly independent rows: \[ \operatorname{rank}(A) = r_A, \operatorname{rank}(B) = r_B \]
02

Calculate the sum of the matrices A and B

Now, we calculate the sum of the matrices A and B: \[ C = A + B \] By definition of matrix addition, each element in matrix C is the sum of corresponding elements in matrices A and B.
03

Observe the effect of addition on linearly independent rows

It is important to note that when we add the corresponding linearly independent rows from matrices A and B, it is possible that we may obtain a linearly dependent row. To account for this possibility, we will write C as a sum and think about its rank. We will have: \[ C = r_A + r_B - \text{overlap} \] where "overlap" is the number of linearly dependent rows that might have been produced when adding corresponding rows from A and B. This helps us to understand that the rank of the resulting matrix C should account for the possibility of linearly dependent rows being produced.
04

Establish the inequality for the rank of matrix C

Now, we are ready to show the inequality we want to prove: \[ \operatorname{rank}(C) = \operatorname{rank}(A+B) \leq \operatorname{rank}(A)+\operatorname{rank}(B) \] Since the rank of a matrix is the maximum number of linearly independent rows, we can rewrite the inequality using the variables r_A and r_B, \[ \operatorname{rank}(C) = r_A + r_B - \text{overlap} \leq r_A + r_B \] And finally, replace r_A and r_B with the ranks of matrices A and B: \[ \operatorname{rank}(A+B) \leq \operatorname{rank}(A)+\operatorname{rank}(B) \] Thus, the inequality has been proven.

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Most popular questions from this chapter

Prove that if \(S\) is a subspace of \(\mathbb{R}^{1},\) then either \(S=\\{\boldsymbol{0}\\}\) or \(S=\mathbb{R}^{1}\)

Which of the sets that follow are spanning sets for \(\mathbb{R}^{3} ?\) Justify your answers. (a) \(\left\\{(1,0,0)^{T},(0,1,1)^{T},(1,0,1)^{T}\right\\}\) (b) \(\left\\{(1,0,0)^{T},(0,1,1)^{T},(1,0,1)^{T},(1,2,3)^{T}\right\\}\) (c) \(\left\\{(2,1,-2)^{T},(3,2,-2)^{T},(2,2,0)^{T}\right\\}\) (d) \(\left\\{(2,1,-2)^{T},(-2,-1,2)^{T},(4,2,-4)^{T}\right\\}\) (e) \(\left\\{(1,1,3)^{T},(0,2,1)^{T}\right\\}\)

Let \(A\) be an \(m \times n\) matrix. Show that if \(A\) has linearly independent column vectors, then \(N(A)=\\{\mathbf{0}\\}\). \(\left[\text {Hint}: \text { For any } \mathbf{x} \in \mathbb{R}^{n}\right.\) \(\left.A \mathbf{x}=x_{1} \mathbf{a}_{1}+x_{2} \mathbf{a}_{2}+\cdots+x_{n} \mathbf{a}_{n} \cdot\right]\)

Let \(A\) be an \(n \times n\) matrix. Prove that the following statements are equivalent: (a) \(N(A)=\\{\mathbf{0}\\}\) (b) \(A\) is nonsingular. (c) For each \(\mathbf{b} \in \mathbb{R}^{n},\) the system \(A \mathbf{x}=\mathbf{b}\) has a unique solution.

Let \(S\) be the subspace of \(P_{3}\) consisting of all polynomials \(p(x)\) such that \(p(0)=0,\) and let \(T\) be the subspace of all polynomials \(q(x)\) such that \(q(1)=\) 0\. Find bases for (a) \(S\) (b) \(T\) (c) \(S \cap T\)
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