Help me about matrix in Matlab ?

Give a square matrix A. For k is positive integer. Find k that A^k = 0. (I'm Vietnamese, I don't know How to call k in English). Could you please help me write the code to find k.

Thanks you very much.

[Roger Stafford]

If A is a non-singular square matrix - that is, if it possesses an inverse - then no finite value of k can ever satisfy your condition. You seem to have posed an impossible problem. Are you sure this is what you meant to ask?

[Nguyen Trong Nhan]

yes, I meant that.please help me .

[Image Analyst]

You said "Give a square matrix A". Well how about if I give you an A that is all zeros?

[Walter Roberson]

Roger, it can happen in floating point arithmetic, though not algebraically. For example,

diag(rand(1,5))

raised to a large enough power will underflow to all 0's.

For example,

A = diag([0.757740130578333, 0.743132468124916, 0.392227019534168, 0.655477890177557, 0.171186687811562]);

is last non-zero at A^2685

Answer 1

[Walter Roberson]

There is in general no solution for this. If you do a singular value decomposition

[U,S,V] = svd(A);

then A = U * S * V' where S is a diagonal matrix, and A^k = U * S^k * V' . Then, A^k can only go to zero if S^k goes to 0. Algebraically that requires that the matrix be singular in the first place. In floating point arithmetic, it would require that the diagonal of the diagonal matrix S be all in (-1,+1) (exclusive on both ends) and then k would be the point at which the diagonal elements underflowed to 0. As the non-zero diagonal S entries of SVD are the square roots of the eigenvalues of A, this in turn requires that the eigenvalues are all strictly in the range (0,1) -- which is certainly not true for general matrices A.

[Roger Stafford]

No, that isn't true for 'svd' in general, Walter. It does hold true for 'eig' when it can obtain a complete set of orthogonal eigenvectors and eigenvalues.