?!? (need help as soon as possible, thx)

quasar987

Ok, what am I doing wrong with that? It's the second problem like this with which I arrive at an absurdity.. Essentially I must show that A is not diagonalizable. So I want to find the eigenvalues and eigenvectors and show that there are not enough eigenvectors to diagonalize A. To ease the calculations, I am told that -1 is an eigenvalue.

[tex]A=\left( \begin{array}{ccc} 2 & 1 & 0 \\
2 & 1 & -2 \\ -1 & 0 & -2 \end{array} \right)[/tex]

I must find the roots of the 3rd degree polynomial equation expressed by [itex]det(A-\lambda I)=0[/itex]. Let us develop that determinant according the 2nd column. We have

[tex]det(A-\lambda I) = \sum_{i=1}^3 (-1)^{i+2}a'_{i2}det((A-\lambda I)_{i2}) = - \left| \begin{array}{cc} 2 & -2 \\ -1 & -2-\lambda \end{array} \right| + (1-\lambda)\left| \begin{array}{cc} 2-\lambda & 0 \\ -1 & -2-\lambda \end{array} \right|[/tex]
[tex]= -[(2)(-2-\lambda)-(-2)(-1)]+(1-\lambda)[(2-\lambda)(-2-\lambda)] = (2\lambda + 6) + (1-\lambda)(\lambda^2 -4)[/tex]
[tex] = -\lambda^3 + \lambda^2 -2\lambda + 2 [/tex]

For which of course -1 is not a root!


The final exam is tomorrow, so thanks for your quick replies!

shmoe

Hi, it looks like you've made an error in finding the characteristic polynomial. In your second to last equality, you replaced [tex](2-\lambda)(-2-\lambda)[/tex] with [tex](-2\lambda -4)[/tex]. This still won't make -1 a root though.

Is it possible there's a typo in the question?

Muzza

According to Derive, A doesn't have an eigenvalue of -1...

quasar987

grr.. well alright, thanks guys.