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Infer BA When AB Is the Identity

Goldman SachsData Scientist · Online Assessment
EasyData ScientistOnline AssessmentStatistics & Math

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This question tests the understanding of matrix invertibility, the properties of the identity matrix, and the fundamental differences between square and non-square matrix multiplication.

The Relationship Between Matrix Products

Let A be an m x n matrix and B be an n x m matrix. Suppose that their product AB is the identity matrix I of size m x m. That is, AB = I_m.

Task

  1. If m = n (i.e., A and B are square matrices), prove that BA = I_n.
  2. If m n (A is a "tall" matrix), explain the rank of A and determine if AB = I_m is possible.

Worked Example

Consider A = [[1, 0, 0], [0, 1, 0]] (a 2x3 matrix) and B = [[1, 0], [0, 1], [0, 0]] (a 3x2 matrix).

  • Calculate AB: [[1, 0, 0], [0, 1, 0]] * [[1, 0], [0, 1], [0, 0]] = [[1, 0], [0, 1]] = I_2.
  • Calculate BA: [[1, 0], [0, 1], [0, 0]] * [[1, 0, 0], [0, 1, 0]] = [[1, 0, 0], [0, 1, 0], [0, 0, 0]].
  • Result: In this case, BA is not the identity I_3.

Constraints

  • Assume all entries are real numbers.
  • Consider cases where the matrices are singular or full-rank.

Hints

Hint 1

Recall that for square matrices, if AB = I, then A and B must both be invertible, which implies B = A^-1.

Hint 2

For non-square matrices, consider the rank-nullity theorem: the rank of a product of matrices cannot exceed the rank of any of its factors.

Hint 3

Think about the dimensions of the null space for A and B when m does not equal n.

Follow-up questions

  • How does the concept of a Moore-Penrose pseudoinverse relate to the cases where BA is not equal to I?
  • If AB = I, what does this tell you about the linear independence of the columns of B and the rows of A?
  • Can you construct a scenario where BA is an idempotent matrix (i.e., (BA)^2 = BA) but not the identity?