Triangular Matrices

**Mimi89** · January 14th 2010, 09:52 AM

Hi!

Could you help me with the following problem:

Find an invertible matrix P such that $P^-1 AP$

is upper triangular, where A is the matrix:

$\begin{pmatrix}3&2&1\\\!\!\!-1&2&1\\1&0&1\end{pmatrix}$

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

Best Wishes,

M

**Roam** · January 14th 2010, 11:43 AM

There is an invertible matrix P for which $P^{-1}AP$ 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 $P_1, P_2, P_3$ . Then form the matrix $P=[P_1,P_3,P_3]$ . the matrix $P^{-1}AP$ will be diagonal and upper triangular and will have the eigenvalues corresponding to $P_1, P_2, P_3$ , respectively, as its successive diagonal entries. Let's see how you go.

**tonio** · January 14th 2010, 06:50 PM

Originally Posted by Roam

There is an invertible matrix P for which $P^{-1}AP$ is a diagonal matrix.

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

Tonio

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 $P_1, P_2, P_3$ . Then form the matrix $P=[P_1,P_3,P_3]$ . the matrix $P^{-1}AP$ will be diagonal and upper triangular and will have the eigenvalues corresponding to $P_1, P_2, P_3$ , respectively, as its successive diagonal entries. Let's see how you go.

.

**tonio** · January 14th 2010, 06:51 PM

Originally Posted by Mimi89

Hi!

Could you help me with the following problem:

Find an invertible matrix P such that $P^-1 AP$

is upper triangular, where A is the matrix:

$\begin{pmatrix}3&2&1\\\!\!\!-1&2&1\\1&0&1\end{pmatrix}$

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

Best Wishes,

M

Read http://www.millersville.edu/~rumble/...larization.pdf
There's a worked example in page 4.

Tonio

**Roam** · January 15th 2010, 12:01 AM

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

Tonio

I see your point but I disagree. What if we found that A happens to have 3 linearly independent eigenvectors? It would satisfy the condition for diagonalizability. By the way, an nxn matrix which has n distinct real eigenvalues is diagonalizable; because we can make a set of n linearly independent eigenvectors by choosing one eigenvector from each eigenspace.

**tonio** · January 15th 2010, 02:20 AM

Originally Posted by Roam

I see your point but I disagree. What if we found that A happens to have 3 linearly independent eigenvectors? It would satisfy the condition for diagonalizability. By the way, an nxn matrix which has n distinct real eigenvalues is diagonalizable; because we can make a set of n linearly independent eigenvectors by choosing one eigenvector from each eigenspace.

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)

**Roam** · January 15th 2010, 12:22 PM

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.

**Mimi89** · January 18th 2010, 11:35 AM

Originally Posted by Roam

There is an invertible matrix P for which $P^{-1}AP$ 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 $P_1, P_2, P_3$ . Then form the matrix $P=[P_1,P_3,P_3]$ . the matrix $P^{-1}AP$ will be diagonal and upper triangular and will have the eigenvalues corresponding to $P_1, P_2, P_3$ , respectively, as its successive diagonal entries. Let's see how you go.

Thanks for the quick reply. The characteristic polynomial is $C(x)= (x-2)^3$ , 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?).

Originally Posted by tonio

Read http://www.millersville.edu/~rumble/Math.422/schur-triangularization.pdf
There's a worked example in page 4.

Tonio

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

Best,

M.

Thread: Triangular Matrices

LinkBack

Thread Tools

Display

Triangular Matrices

Similar Math Help Forum Discussions

[SOLVED] Product of atomic triangular matrices

Show that the product of two upper triangular matrices is upper triangular matrix

Subspace of 3x3 matrices containing symmetric and lower-triangular matices

Square Matrices- Upper Triangular

A Proof Involving Upper Triangular Matrices

Search Tags