A matrix is diagonalizable if there exists a nonsingular matrix such that is diagonal. In different phrases, a diagonalizable matrix is one that’s much like a diagonal matrix.
The situation is equal to with nonsingular, that’s, , , the place . Therefore is diagonalizable if and provided that it has an entire set of linearly impartial eigenvectors.
A Hermitian matrix is diagonalizable as a result of the eigenvectors could be taken to be mutually orthogonal. The identical is true for a traditional matrix (one for which ). A matrix with distinct eigenvalues can be diagonalizable.
Theorem 1.
If has distinct eigenvalues then it’s diagonalizable.
Proof. Let have eigenvalues with corresponding eigenvectors . Suppose that for some . Then
which suggests since for and . Premultiplying by exhibits, in the identical means, that . Persevering with on this means we discover that . Subsequently the are linearly impartial and therefore is diagonalizable.
A matrix can have repeated eigenvalues and be diagonalizable, as diagonal matrices with repeated diagonal entries present. What is required for diagonalizability is that each -times repeated eigenvalue has linearly impartial eigenvectors related to it. Equivalently, the algebraic and geometric multiplicities of each eigenvalue should be equal, that’s, the eigenvalues should all be semisimple. One other equal situation is that the diploma of the minimal polynomial is the same as the variety of distinct eigenvalues.
The best instance of a matrix that isn’t diagonalizable is . This matrix is a Jordan block with the eigenvalue . Diagonalizability is definitely understood by way of the Jordan canonical type: is diagonalizable if and provided that all of the Jordan blocks in its Jordan type are .
Most matrices are diagonalizable, within the sense that the diagonalizable matrices are dense in , that’s, any matrix in is arbitrarily near a diagonalizable matrix. This property is beneficial as a result of it may be handy to show a outcome by first proving it for diagonalizable matrices after which arguing that by continuity the outcome holds for a common matrix.
Is a rank- matrix diagonalizable, the place are nonzero? There are zero eigenvalues with eigenvectors any set of linearly impartial vectors orthogonal to . If then is the remaining eigenvalue, with eigenvector , which is linearly impartial of the eigenvectors for , and is diagonalizable. If then all of the eigenvalues of are zero and so can’t be diagonalizable, as the one diagonalizable matrix whose eigenvalues are all zero is the zero matrix. For the matrix talked about above, and , so , confirming that this matrix just isn’t diagonalizable.