Jordan normal form of 4x4 matrix

[Shadow]

Given the matrix [math] A =

\left( \begin{array}{cccc}

2 & 1 & 0 & -1\\

0 & 2 & -1 & -1\\

0 & 0 & 3 & 1\\

0 & 0 & -1 & 1

\end{array}\right) [/math], find its Jordan matrix and the matrix T, for which [math]A=TJT^{-1}[/math].

 

All four of the eigenvalues are 2, and the Jordan matrix will consist of two Jordan blocks; [math] J =

\left( \begin{array}{cccc}

2 & 1 & 0 & 0\\

0 & 2 & 0 & 0\\

0 & 0 & 2 & 1\\

0 & 0 & 0 & 2

\end{array}\right) [/math]. Thus, if we label the eigenvectors [math]h_1[/math], [math]h_2[/math], [math]h_3[/math], [math]h_4[/math], we have

 

[A].[[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ] = [[math]h_1[/math] | [math]h_2[/math] | [math]h_3[/math] | [math]h_4[/math] ].[J],

 

and from this

 

A[math]h_1[/math] = 2[math]h_1[/math]

A[math]h_2[/math] = [math]h_1[/math] + 2[math]h_2[/math]

A[math]h_3[/math] = 2[math]h_3[/math]

A[math]h_4[/math] = [math]h_3[/math] + 2[math]h_4[/math]

 

But this would mean that [math]h_1[/math] = [math]h_3[/math] and thus [math]h_2[/math] = [math]h_4[/math], which is nonsense. And I'm stumped as to what to do about it. Any ideas?

[Schrödinger's hat]

Degenerate nxn matrices don't always have n linearly independant eigenvectors.

[Shadow]

I did not know that, but nevertheless this one does; the solution is [math] T = \left( \begin{array}{cccc} -1 & 0 & 0 & 0\\ -1 & 0 & -1 & 0\\ 1 & 0 & 1 & 1\\ -1 & 1 & -1 & 0 \end{array}\right) [/math]

Edited December 11, 2011 by Shadow

[DrRocket]

Degenerate nxn matrices don't always have n linearly independant eigenvectors.

 

If you had n linearly independent eigenvectors then the matrix would be diagonalizable and you would not need Jordan normal form.