Question

Show that if the columns of B are linearly dependent, then so are the columns of AB. b. Suppose CA = In. Show that the equation Ax = 0 has only the trivial solution. Explain why A cannot have more columns than rows. c. Suppose AD = Im. Show that for any b in ℝm, the equation Ax = b has a solution. Explain why A cannot have more rows than columns. d. Let A be an invertible n×n matrix, and let B be an n×p matrix. Show that the equation AX = B has a unique solution A-1B.

Answer 1

Final Answers:

a. If the columns of B are linearly dependent, then the columns of AB are also linearly dependent.

b. If CA = Iₙ, then the equation Ax = 0 has only the trivial solution, and A cannot have more columns than rows.

c. If AD = Iₘ, for any b in ℝₘ, the equation Ax = b has a solution, and A cannot have more rows than columns.

d. For an invertible n×n matrix A and an n×p matrix B, the equation AX = B has a unique solution A⁻¹B.

Step-by-step explanation:

a. Consider the product AB. Let's say the columns of B are linearly dependent. This means there exist constants c₁, c₂, ..., cₖ (not all zero) such that c₁B₁ + c₂B₂ + ... + cₖBₖ = 0. Now, when we multiply both sides by A, we get c₁AB₁ + c₂AB₂ + ... + cₖABₖ = 0. Since c₁B₁ + c₂B₂ + ... + cₖBₖ = 0, the columns of AB are also linearly dependent.

b. If CA = Iₙ, then A⁻¹ exists. Suppose Ax = 0 has a non-trivial solution x₀. Multiply both sides by A⁻¹ to get A⁻¹Ax₀ = A⁻¹0. This simplifies to x₀ = 0, contradicting the assumption. Hence, the only solution to Ax = 0 is the trivial solution. Additionally, A cannot have more columns than rows because the inverse A⁻¹ would not exist.

c. If AD = Iₘ, then A⁻¹ exists. For any vector b in ℝₘ, the equation Ax = b has a solution x = A⁻¹b. This is because A⁻¹ADx = A⁻¹b, leading to x = A⁻¹b. A cannot have more rows than columns because the solution space would be underdetermined.

d. Given A is invertible, we can multiply both sides of AX = B by A⁻¹ to obtain X = A⁻¹B. This unique solution exists because the product of an invertible matrix and any matrix produces a unique result, ensuring a unique solution for X.

Answer 2

The question involves proving linear dependence in matrix multiplication, existence and uniqueness of solutions to linear systems, and constraints relating to the dimensionality of matrices. The properties of invertible matrices and the implications of matrix multiplication in terms of solutions to linear equations are discussed.

Step-by-step explanation:

Let's consider matrix B with its columns being linearly dependent. This means there exists a nontrivial solution to Bx = 0. For matrix AB, if we multiply both sides of the equation Bx = 0 by A, we get ABx = A0 which simplifies to ABx = 0. Since x != 0 (nontrivial), the columns of AB must also be linearly dependent.

For part b, if CA = In, where In is the identity matrix, then Ax = 0 implies that CAx = C0 which simplifies to Inx = 0. The only solution to this is x = 0, proving that Ax = 0 has only the trivial solution. Additionally, matrix A cannot have more columns than rows, as that would imply linear dependence, violating the invertibility implicit in CA = In.

In part c, if AD = Im, the presence of b in ℝm guarantees that the equation Ax = b has a solution for all b because D ensures a mapping from ℝm to the column space of A. Matrix A cannot have more rows than columns because it would not span ℝm, preventing Ax = b from having a solution for every b.

Finally, given an invertible matrix A and B, the equation AX = B has a unique solution X = A-1B, due to the properties of invertible matrices which include a unique inverse and one-to-one correspondence between the matrix elements and their solutions.