Is Matrix A Triangulable but Not Diagonalizable?

[kmjonec]

Homework Statement

Show that A = 3 4 3
-1 0 -1
1 2 3

is not diagonalizable but is triangulable and carry out triangulation (A has rational entries)

I found that the only eigenvalue is 2, and that the characteristic equation is (x-2)³, but I'm not sure how to prove that a matrix is not diagonalizable. I know that if it has 3 distinct eigenvalues then it will be diagonalizable (for a 3x3 matrix), but if it has less than 3 that doesn't necessarily imply it isn't.

As for triangulating the matrix, I found that (1,-1, 1) is an eigenvector and {(1,-1,1), (0,1,0),(0,0,1)} is a basis of Q³. But I'm not sure how to find an S so that S^-1AS is upper triangualar. Any advice?

 

[jbunniii]

A matrix M is diagonalizable if and only if it is similar to a diagonal matrix D, which means there is an invertible matrix S such that

S^{-1}MS = D

But this is equivalent to

MS = SD

and you can see from this that the columns of S must be eigenvectors of M.

That means that the condition that S must be invertible is equivalent to requiring that the eigenvectors must be linearly independent.

 

[HallsofIvy]

An n by n matrix is "diagonalizable" if and only if it has n independent eigenvectors (eigenvectors corresponding to distinct eigenvalues are always independent so if there are n distinct eigenvalues, the matrix is diagonalizble).

Here, you say the only eigenvalue is 2 so you need to look for eigenvectors:
\begin{bmatrix}3 & 4 & 3 \\ -1 & 0 & -1 \\ 1 & 2 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}
or
\begin{bmatrix}1 & 4 & 3 \\ -1 & -2 & -1 \\ 1 & 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}


That is equivalent to the three equations 3x+ 4y+ 3z= 2x, -x- z= 2y, and x+ 2y+ 3z= 2z or x+ 4y+ 3z= 0, x+ 2y+ z= 0, and x+ 2y+ z= 0. The second and third equations are exactly the same and are equivalent to z= -x- 2y. putting that into the first equation, x+ 4y- 3x- 6y= -2x- 2y= 0 or y= -x. With y= -x, z= -x+ 2x= x. Any eigenvector must be of the form <x, -x, x>= x<1, -1, 1>. We get only one "independent" eigenvector" so this is NOT diagonalizable. I have never seen the word "triangulable" before. Any matrix that is not diagonalizable can be put in "Jordan normal form" which has "1" directly above some of the diagonal elements and "0"s else where. For this matrix, the Jordan normal form is
\begin{bmatrix}2 &amp; 1 &amp; 0 \\ 0 &amp; 2 &amp; 1 \\ 0 &amp; 0 &amp; 2\end{bmatrix}.

The matrix, S, that jbuniii refers to (that gives the Jordan normal form, not a diagonal matrix) would be a matrix having the eigenvector <1, -1, 1> as first column, the "generalized eigenvector" <x, y, z> such that
\begin{bmatrix}1 &amp; 4 &amp; 3 \\ -1 &amp; -2 &amp; -1 \\ 1 &amp; 2 &amp; 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}
as second column and the "generalized eigenvector" <u, v, w> such that
\begin{bmatrix}1 &amp; 4 &amp; 3 \\ -1 &amp; -2 &amp; -1 \\ 1 &amp; 2 &amp; 2\end{bmatrix}\begin{bmatrix}u \\ v \\ w\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}
for the x, y, z just found, as third column.