# non-diagonalizable matrix help?

• Nov 28th 2009, 05:22 PM
drangelpsyche
non-diagonalizable matrix help?
Hello,

I've been asked to find a 3x3 matrix A such that

1) $A^3 - 4A^2 + 5A - 2I = 0$

and

2) The eigenvectors of A do not span $R^3$

I believe the second condition implies that A is not diagonalizable, which means it does not contain a complete set of eigenvectors (I am only aware that nilpotent matrices fall under this category, are there others?) and I know that 1) can be expressed as:
$(A - I)^2(A-2I)$

but I cannot figure out how to work this out. Can anyone help?
• Nov 28th 2009, 07:41 PM
tonio
Quote:

Originally Posted by drangelpsyche
Hello,

I've been asked to find a 3x3 matrix A such that

1) $A^3 - 4A^2 + 5A - 2I = 0$

and

2) The eigenvectors of A do not span $R^3$

I believe the second condition implies that A is not diagonalizable, which means it does not contain a complete set of eigenvectors (I am only aware that nilpotent matrices fall under this category, are there others?) and I know that 1) can be expressed as:
$(A - I)^2(A-2I)$

but I cannot figure out how to work this out. Can anyone help?

So the characteristic pol. of A is $p_A(x)=(x-1)^2(x-2)$ and its minimal pol. MUST be the same, otherwise, as the char. pol. and the min. pol. have the same irreducible factors, the min. pol would be $m_A(x)=(x-1)(x-2)$ , but then A would be diagonalizable as its min. pol. would be the product of different linear factors...
Now, check that the following matrix does the trick:

$A=\left(\begin{array}{rrr}0&1&0\\0&0&1\\2&-5&4\end{array}\right)$

This matrix is known as the companion matrix of the pol. $f(x)$ and it's used to prove that for any given monic polynomial there exists a square matrix whose char. pol. is that given monic pol.

Tonio
• Nov 28th 2009, 08:30 PM
drangelpsyche
Thank you very much for your help!