Find the eigenvalues and their corresponding eigenspaces for the matrix,
The characteristic polynomial of is , so the eigenvalues are .
The eigenspace for eigenvalue 2 is two-dimensional and is spanned by the vectors found by solving the equation ; it is,
The eigenspace for eigenvalue 3 is one-dimensional and is spanned by the solution to the equation ; it is,
Remember that the eigenvalues of the matrix,
are , with eigenvector for ,
and eigenvectors for ,
Find an invertible matrix and a diagonal matrix so that .
Based on the known eigenvalues, one choice for is,
Based on this choice of , one valid choice for is,
Can the matrix
be diagonalized? Explain why or why not.
The characteristic polynomial of is, , so the eigenvalues are . However, the only eigenvector of is
As there is only one eigenvector for the matrix , it is not diagonalizable.
Can the matrix
be diagonalized? Explain why or why not.
The characteristic polynomial of is, . This polynomial has discriminant “” of . As the discriminant is greater than zero there are two unique real roots to the polynomial, and hence two different real eigenvalues for . As each eigenvalue has at least one eigenvector, has at least two eigenvectors. So, as is a matrix, this means that yes, is diagonalizable.
For a positive integer and the matrix,
find a formula for .
diagonalizes as the matrix product,
Since , a formula for is,