Jul 8, 2015 · In short, diagonalizable of a matrix means that the matrix has independent set of eigenvectors that can span the entire space. There are several ways to decide whether matrix has independent set of eigenvectors that can span the entire space. One is that if matrix has distinct eigenvalues, it will be diagonalizable.

Nov 6, 2016 · A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue. Meaning, if you find matrices with distinct eigenvalues (multiplicity = 1) you should quickly identify those as diagonizable.

Aug 17, 2014 · The rank of a diagonalizable matrix is the same as the rank of its diagonalization. The latter is easy to compute by looking at its entries, since the rank of a diagonalized matrix is simply the number of nonzero entries.

But, there are non-diagonalizable matrices that aren't rotations - all non-zero nilpotent matrices. My intuitive view of nilpotent matrices is that they ''gradually collapse all dimensions/gradually lose all the information'' (if we use them over and over again), so it's clear to …

Dec 21, 2017 · I'll deal with complex matrices. A normal matrix is diagonalizable with a unitary matrix (this is a rather deep result).

Sep 2, 2024 · I'm trying to go about the proof that any matrix that is nilpotent (i.e. $\exists N \in\Bbb N. A^N = \mathbf{0}$) cannot be diagonalizable. I believe that the best way to go about this is by showing that a given eigenvalue's geometric multiplicity is not the same as …

Apr 11, 2015 · So now repeat the procedure with the matrix $\mathbf{A}_1$ and iterate, each iteration leaving a lower right square matrix of dimension one fewer than the one from the former iteration. When we reach the lower right corner, we have an upper triangular matrix, similar to the original by the product of all the unitary transformations at each step.

The matrix is diagonalizable if and only if all the blocks have size $1$, so if the minimal polynomial is ...

$\begingroup$ Ok, I see now. I just noticed a small other thing: Now, if we have two commuting, diagonalizable matrices Then the proof follows.

Nov 16, 2015 · We arrive at the well-known fact that orthogonal diagonalizable real matrices are symmetric (and vise-versa, real symmetric matrix has only real eigenvalues and is orthogonal diagonalizable.). So only non-symmetric matrices could be diagonalizable but not orthogonal diagonalizable .