Matrix Inverses and Systems of Linear Equations

Problem Statement

The problem T25, selected from page 216 of “A First Course in Linear Algebra”, by Robert A. Bleezer.

T25 The notation A^k means a repeated matrix product between k copies of the square matrix A.

  1. Assume A is an n x n matrix where A^2 = O (which does not imply that A = O.)
    Prove that In - A is invertible by showing that In + A is an inverse of In - A.

  2. Assume that A is an n x n matrix where A^3 = O. Prove that In - A is invertible.

  3. Form a general theorem based on your observations from parts (1) and (2) and provide a proof.

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Here are the solutions: For ease of writing, I’ll write I for In.

  1. Given A^2 = O. It’s required to prove that (In - A) is invertible.

Proof: We know from algebra that (I - A).(I + A) = I^2 - A^2. The rule applies for Matrix multiplication too.
Now, A^2 = 0 (given), and I^2 = I.
Therefore, (I - A).(I + A) = I, i.e. (I - A) is invertible, and (I + A) is the inverse.

  1. Given A^3 = O. It’s required to prove that (In - A) is invertible.

Proof: We also know from algebra that (I - A).(I^2 + I.A + A^2) = I^3 - A^3. The rule applies for Matrix multiplication too.
Now, A^3 = 0 (given), I^3 = I.
Therefore, (I - A).(I^2 + I.A + A^2) = I, i.e. (I - A) is invertible, and (I^2 + I.A + A^2) is the inverse.

  1. We can generalize that if A^k = 0, then I - A is invertible.

Proof: We know factorization rule in algebra that (a^n - b^n) = (a - b).(a^(n-1) + a^(n-2).b + … + a.b^(n-2) + b^(n-1)). The rule applies for Matrix multiplication too.
Therefore, (I - A).(I^(k-1) + I^(k-2).A + … + I.A^(k-2) + A^(k-1)) = (I^k - A^k)
Now, A^k = 0 (given), and I^k = I.
Therefore, (I - A).(I^(k-1) + I^(k-2).A + … + I.A^(k-2) + A^(k-1)) = I, i.e. (I - A) is invertible, and (I^(k-1) + I^(k-2).A + … + I.A^(k-2) + A^(k-1)) is the inverse.

   Furthermore, this can be generalized for I + A. (Based on factorization rule for (a^n + b^n).

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