Recall that the column vector Ax is the linear combination of columns of A with coefficients from x. Therefore, if Ax = 0 for some x, then columns of A are linearly dependent, and so A is degenerate.
can a matrix be diagonalized if it has an eigenvalue of 0? I know that the diagonal must be invertible for this to happen but (yesterday I was solving a problem, a 3x3 matrix and I got 3 eigenvalues, 2 of which were 0, but if the two are 0, then the diagonal would look something like x 0 0/ 000/ 000 which I dont think is invertible... Am I going wrong somewhere? Also, if we were to find A^k=P^-1D^KP and we have such a matrix wouldnt we get an incorrect answer for A^K? I have a feeling I am missing something...
Well, I'm not certain what emakarov means by "degenerate". Certainly, if A has 0 as an eigenvalue, then there exist non-zero v such that Av= 0v= 0 so A is not invertible. But as far as "diagonalizable" is concerned, that depends entirely upon how many independent eigenvectors A has, not on what the eigenvalues are. An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. For example, the matrix has eigenvalue 1 and 0 and is diagonal while has 0 as a double eigenvalue and is NOT Diagonalizable.
(eigenvectors corresponding to distinct eigenvalues are necessarily independent so if an n by n matrix has n distinct eigenvalues, it has n independent eigenvectors and is diagonalizable. In order to have a non-diagonalizable matrix, we must have at least one eigenvalue with algebraic multiplicity greater than 1.)