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I want to diagonalize this matrix :

$$ A =\begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix} $$

the first step I did is calculates the eigenvalues​​ :

$$det(A-λ I_3)=0\iff \begin{vmatrix} -λ & -1 & 0 \\ -1 & -λ & 0 \\ 1 & 1 & 1-λ \\ \end{vmatrix} = 0$$

$$\iff-λ\begin{vmatrix} -λ & 0 \\ 1 & 1-λ \\ \end{vmatrix} + \begin{vmatrix} -1 & 0 \\ 1 & 1-λ \\ \end{vmatrix} = 0$$

$$\iff -λ³ + λ² + λ - 1 = 0$$

$$\iff -λ²(λ-1) + λ - 1 = 0$$ $$\iff (λ-1)(-λ²+1)= 0$$ $$\iff (1-\lambda^2)= 0 \quad\text{or}\quad (λ-1)=0$$ $$l_1 = l_2 = 1\quad; \quad l_3=-1$$

Then the second step is to calculate the eigenvectors :

but I don't know how to do this step , could you please tell me what is the method to do it ?

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  • $\begingroup$ related: math.stackexchange.com/questions/192801/… $\endgroup$ – vadim123 Apr 29 '13 at 18:49
  • $\begingroup$ @vadim123 I cant understand the step where he is reduce the matrix to row-eschelon $\endgroup$ – Aimad Majdou Apr 29 '13 at 18:59
  • $\begingroup$ How did you get these eigenvalues to be zero, your equation suggests 1,1,-1 $\endgroup$ – imranfat Apr 29 '13 at 19:03
  • $\begingroup$ @imranfat I cant understand you the eigenvalues I got are : l1, l2 and l3 <=> 1,1,-1 $\endgroup$ – Aimad Majdou Apr 29 '13 at 19:06
  • $\begingroup$ I see it, didn't read it right $\endgroup$ – imranfat Apr 29 '13 at 19:23
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We solve the system of equations $(A-\lambda I_3)X=0$ where $$ X =\begin{pmatrix} x\\y\\z \end{pmatrix} $$ is an eigenvector associated to the eigenvalue $\lambda$

For $\lambda=-1$ we find $$ v_1 =\begin{pmatrix} -1\\-1\\1 \end{pmatrix} $$ and for $\lambda=1$ we find $$v_2=\begin{pmatrix} 0\\0\\1 \end{pmatrix}\quad\text{and}\quad v_3 =\begin{pmatrix} -1\\1\\0 \end{pmatrix} $$

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  • $\begingroup$ for λ=−1, okey I understood it, but for the second case λ=1 wehn I solve the system $$(A-I_3)X = 0$$ I found x' + x'' = 0. but how did you find the v2 and the v3 ? $\endgroup$ – Aimad Majdou Apr 29 '13 at 19:23
  • $\begingroup$ @AimadMajdou You find $x+y=0$ so you can choose $z$ arbitrary and $x=-y$ and then you can find two linearly independant vectors $v_2$ and $v_3$ as in my answer that satisfy this condition. $\endgroup$ – user63181 Apr 29 '13 at 19:27
  • $\begingroup$ so as I think from to x +y= 0 => y =-x so v3=(1,-1,0) so why you choosed the other solution and not (1,-1,0) ? I didn't understand why z =1 in v2 from x+y=0 => z =0 not 1. $\endgroup$ – Aimad Majdou Apr 29 '13 at 20:07
  • $\begingroup$ @AimadMajdou There are many possibilies to choose $v_2$ and $v_3$ but the important thing is that these vectors must be linearly independant since $(v_1,v_2,v_3)$ must form a basis, hence the aim of the choice $z=0$ and $z=1$ in the two vectors is to ensure the independance of $v_2$ and $v_3$. $\endgroup$ – user63181 Apr 29 '13 at 20:17
  • $\begingroup$ what do you mean by linearly independant ? $\endgroup$ – Aimad Majdou Apr 29 '13 at 20:25

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