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Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.

$$A= \begin{bmatrix} +3 & +2\\ -2 & -3\\ \end{bmatrix} $$

$$A-\ell I = \begin{bmatrix} +3-\ell & +2\\ -2 & -3-\ell\\ \end{bmatrix} $$ Then Determinant should be zero : $$ \begin{vmatrix} +3-\ell & +2\\ -2 & -3-\ell\\ \end{vmatrix}=(3-\ell)(-3-\ell)+4=0 \to \ell^2-5=0 \to \ell_{1}=\sqrt 5,\ell_{2}=-\sqrt5 , $$ $$ \begin{bmatrix} 3-\ell_{1}& +2\\ -2 & -3-\ell{1}\\ \end{bmatrix}= \begin{bmatrix} 3-\sqrt 5=0.76& +2\\ -2 & -3-\sqrt 5=-5.24\\ \end{bmatrix}\to{R_1\Leftarrow\Rightarrow R2 \mapsto} \begin{bmatrix} -2 & -5.24\\ 0.76& +2\\ \end{bmatrix}\to{R_1=R1/{-2}\mapsto} \begin{bmatrix} 1 & 2.62\\ 0.76& +2\\ \end{bmatrix}\to{R_2=R2-0.76R1\mapsto} \begin{bmatrix} 1 & 2.62\\ 0 & 0.009\\ \end{bmatrix}\to{R_2=R2/0.009\mapsto} \begin{bmatrix} 1 & 2.62\\ 0 & 1\\ \end{bmatrix}\to{R_1=R1-2.62R1\mapsto} \begin{bmatrix} 1 & 0 | 0\\ 0 & 1 | 0\\ \end{bmatrix} $$ $then$ $$ V_1= \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} $$ I am stuck here , matrix of $0,0$ is the right answer Eigenvector 1 ?

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Careful there! You cant just simply approximate the square root and expect things to work out. These are delicate operations you are playing with. For the record, (0,0) is either not an eigenvector, or is an eigenvector of everything depending on your conventions. So that should be a red flag. –  Dhruv Ranganathan May 8 '13 at 21:31
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Don't do $3-\sqrt(5)=0.76$. Carry on with the $3-\sqrt(5)$ and operate with that. You're making numerical mistakes by aproximating that irrational number to a rational one. –  MyUserIsThis May 8 '13 at 21:32
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2 Answers

up vote 3 down vote accepted

The eigenvalues are correct. However you go wrong in the computation of the eigenvectors:

$$ \begin{bmatrix} 3-\sqrt{5} & 2\\ -2 & -3-\sqrt{5} \end{bmatrix} $$

Divide the first row by $3-\sqrt{5}$, which is the same as multiplying it by $(3+\sqrt{5})/2$, getting

$$ \begin{bmatrix} 1 & \frac{3+\sqrt{5}}{2}\\ -2 & -3-\sqrt{5} \end{bmatrix} $$

Now adding to the second row the first one multiplied by $2$ brings the matrix in the form

$$ \begin{bmatrix} 1 & \frac{3+\sqrt{5}}{2}\\ 0 & 0 \end{bmatrix} $$

so you know that one eigenvector is

$$ \begin{bmatrix} -\frac{3+\sqrt{5}}{2}\\ 1 \end{bmatrix} $$


The computations for the other eigenvalue are similar

$$ \begin{bmatrix} 3+\sqrt{5} & 2\\ -2 & -3+\sqrt{5} \end{bmatrix} $$ $$ \begin{bmatrix} 1 & \frac{3-\sqrt{5}}{2}\\ -2 & -3+\sqrt{5} \end{bmatrix} $$

$$ \begin{bmatrix} 1 & \frac{3-\sqrt{5}}{2}\\ 0 & 0 \end{bmatrix} $$

So the other eigenvector you're looking for is $$ \begin{bmatrix} -\frac{3-\sqrt{5}}{2}\\ 1 \end{bmatrix} $$


Of course a matrix that diagonalizes $A$ is

$$ P= \begin{bmatrix} -\frac{3+\sqrt{5}}{2} & -\frac{3-\sqrt{5}}{2}\\ 1&1 \end{bmatrix} $$

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The problem is that you rounded. The 0.009 in the second row is in fact a 0.

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