How do I show that A is not diagonalizable?

[quasar987]

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

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.

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

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

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

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 $$(2-\lambda)(-2-\lambda)$$ with $$(-2\lambda -4)$$. 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.