# Upper triangular matrix

• May 14th 2011, 03:26 PM
qwerty1234
Upper triangular matrix
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How do I form P and T? Is P just made up of 4 eigenvectors? If so, how do I find the 4th eigenvector? And how do I find it? Is upper triangular matrix just the eigenvalues along the main entry?

edit: the 4 eigenvalues are 2,2,-2,1
• May 14th 2011, 03:42 PM
TheEmptySet
Quote:

Originally Posted by qwerty1234
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How do I form P and T? Is P just made up of 4 eigenvectors? If so, how do I find the 4th eigenvector? And how do I find it? Is upper triangular matrix just the eigenvalues along the main entry?

edit: the 4 eigenvalues are 2,2,-2,1

Since two is a repeated eigenvalue that admits only one eigenvector you need to find a generalized eigenvalue vector

You need a vector

$\mathbf{P}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \end{pmatrix}$

$(A-2I)\mathbf{P}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$

Note that the right hand side is the eigenvector for lambda equals 2.

Note 2: Generalized eigenvectors are not unique you just need to pick one vector that works.

Once you have this vector p use the other 3 eigenvectors to make the transformation matrix.
• May 14th 2011, 03:45 PM
qwerty1234
Thanks, and how do I find T? Is upper triangular matrix just the eigenvalues along the main entry?

lambda 0 0 0
0 lambda 0 0
0 0 lambda 0
0 0 0 lambda
?

Thanks
• May 14th 2011, 03:53 PM
TheEmptySet
Quote:

Originally Posted by qwerty1234
Thanks, and how do I find T? Is upper triangular matrix just the eigenvalues along the main entry?

lambda 0 0 0
0 lambda 0 0
0 0 lambda 0
0 0 0 lambda
?

Thanks

No the matrix is not diagonalizable! You can get it into a Jordan canonical form which is an upper triangular matrix.

Since lambda = 2 is the repeated e-value its Jordan block will be a 2 by 2 block. The other blocks will be 1 by 1.

So you will get

$\begin{bmatrix} -2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}$
• May 14th 2011, 04:12 PM
qwerty1234
Quote:

Originally Posted by TheEmptySet
Since two is a repeated eigenvalue that admits only one eigenvector you need to find a generalized eigenvalue vector

You need a vector

$\mathbf{P}=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \end{pmatrix}$

$(A-2I)\mathbf{P}=\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$

Note that the right hand side is the eigenvector for lambda equals 2.

Thanks, i tried solving (A-2I)P, and I got:
1
0
-1
1

And i checked it with a online calculator.

BTW, do i just add the 4 eigenvectors to get P?
• May 14th 2011, 04:34 PM
TheEmptySet
Quote:

Originally Posted by qwerty1234
Thanks, i tried solving (A-2I)P, and I got:
1
0
-1
1

And i checked it with a online calculator.

BTW, do i just add the 4 eigenvectors to get P?

Yes but What I typed in the previous post is wrong the vector on the right hand side is not the eigenvector for lambda =2. I wrote down the wrong vector.

It should be

$(A-2I)\marhbf{P}=\begin{pmatrix}1 \\ 0 \\ -1 \\ 1 \end{pmatrix}$

and I get

$\mathbf{P} = \begin{pmatrix}-1 \\ -1 \\ 1 \\ 0 \end{pmatrix}$

Again sorry for the typo