(adsbygoogle = window.adsbygoogle || []).push({}); ?!? (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.

[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!

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

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