Hi!

Could you help me with the following problem:

Find an invertible matrix P such that

is upper triangular, where A is the matrix:

Thanks a lot for suggestions/help! (general algorithms welcome)

Best Wishes,

M

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- Jan 14th 2010, 09:52 AMMimi89Triangular Matrices
Hi!

Could you help me with the following problem:

Find an invertible matrix P such that

is upper triangular, where A is the matrix:

Thanks a lot for suggestions/help! (general algorithms welcome)

Best Wishes,

M

- Jan 14th 2010, 11:43 AMRoam
There is an invertible matrix P for which is a diagonal matrix. If P exists then A is diagonalizable. Do you know how to find P? Have you learnt about diagonalizability? Your 3x3 matrix A is diagonalizable if A has 3 linearly independent eigenvectors. So, the first step you have to find eigenvectors of A, say . Then form the matrix . the matrix will be diagonal and upper triangular and will have the eigenvalues corresponding to , respectively, as its successive diagonal entries. Let's see how you go.

- Jan 14th 2010, 06:50 PMtonio
- Jan 14th 2010, 06:51 PMtonio

Read http://www.millersville.edu/~rumble/...larization.pdf

There's a worked example in page 4.

Tonio - Jan 15th 2010, 12:01 AMRoamQuote:

This may be far from true: not every matrix is diagonalizable.

Tonio

- Jan 15th 2010, 02:20 AMtonio

Well, if A has 3 lin. ind. eigenvectors THEN, and only then, it is diagonalizable...but it could perfectly well be that it has no 3 lin. ind. eigenvectors, and STILL it'd be triangularizable!

The whole point of Schur's Triangularization Theorem is that ANY complex matrix is similar to a tringular matrix, even if it is not diagonalizable , and this is what the OP, imo, is trying to achieve.

Tonio

Ps. An nxn matrix doesn't have to have n different eigenvalues to be diagonalizable: this is a sufficient condition but not a necessary one.

An nxn matrix over a field F is diagonalizable iff it has n lin. ind. eigenvectors iff its minimal pol. in F[x] splits into different linear factors (again, NOT necessarily n different linear factors...just different linear factors) - Jan 15th 2010, 12:22 PMRoam
Gosh, I just might be forced to agree with you! Triangularization is probably what needs to be done, but the OP may not have yet done triangularization as it's often taught after diagonalization.

Hmm, Yes, the converse of what I said is false; it's possible for an nxn matrix to be diagonalizable without having n distinct eigenvalues. But if it has, you know it is diagonalizable. I guess the real key to diagonalizability is with the dimensions of the eigenspaces. - Jan 18th 2010, 11:35 AMMimi89
Thanks for the quick reply. The characteristic polynomial is , which has only one eigenvalue: 2. Its eigenspace is spanned by (1,-1,1). Hence, it isn't diagonalisable (for that don't we have to have the dimension of the eigenspace=dimension of the vector space?).

Thank you, I will do that now. Thanks for finding something with an example - those usually help me quite a lot.

Best,

M.