what is the largest gap between rank and approximate rank?

Having a nxn matrix A, prove rank(A)+rank(A^3)>=2xrank(A^2)...

• I have an exam question that I don't understand how to tackle having an nxn matrix, prove that rank(A)+rank(A^3)>=2xrank(A^2) I do understand that the rank of a square matrix is one of its dimensions and the rank of A and A^3 should be the same, still I am stuck on how to prove this, If anybody can help I'd appreciate it :D

• Answer:

  First consider the case where A is itself a single nxn Jordan block. If the block corresponds to a nonzero eigenvalue, then A is is invertible, and hence so are A^2 and A^3, and so rank(A) + rank(A^3) = n + n is the same thing as 2 * rank(A^2) = 2n. If the block corresponds to the eigenvalue 0, it is easy to check (with explicit computation) that rank(A) = n-1, rank(A^2) = max(n-2,0) [that is, n - 2 if it is positive, otherwise 0], and rank(A^3) = max(n-3,0) [that is, n - 3 if it is positive, otherwise 0]. So we have rank(A) + rank(A^3) = (n-1) + max(n-3,0) >= (n-1) + (n-3) >= 2n - 4 = 2(n - 2), and clearly also rank(A) + rank(A^3) >= 0 (since rank(A) and rank(A^3) are themselves nonnegative). So rank(A) + rank(A^3) >= max(2(n-2),0) = 2 max(n-2,0) = rank(A^2). Conclusion: for all n, the theorem is true if A is an nxn Jordan block. Now, a general matrix A is similar to a direct sum B_1 + B_2 + ... + B_k of some number of Jordan blocks B_1, ..., B_k, and a short calculation shows that for any positive integer p, the matrix A^p is similar to the direct sum B_1^p + B_2^p + ... + B_k^p. Since the rank is invariant under similarity, and adds over direct sums, we have rank(A) + rank(A^3) = rank(B_1 + ... + B_k) + rank(B_1^3 + B_2^2 + ... + B_k^3). = sum_{j=1}^k rank(B_j) + sum_{j=1}^k rank(B_j^3) = sum_{j=1}^k [rank(B_j) + rank(B_j^3)] >= sum_{j=1}^k [2 rank(B_j^2)] [since the result is known for Jordan blocks] = 2 sum_{j=1..k} rank(B_j^2) = 2 rank(B_1^2 + ... + B_k^2) = 2 rank(A^2).