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Question about eigenvector and identity matrix
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[QUOTE="Umayer, post: 4546903, member: 488400"] [h2]Homework Statement [/h2] Actually I figured it out, I forgot to put a minus when calculating the determinant. But I'll write down the matrix. This is the matrix given: [tex] \begin{pmatrix} 1 & 1 & 2\\ 4 & 0 & 2\\ -2 & 1 & 1 \end{pmatrix} [/tex] So when determining the eigenvalues the matrix will become: [tex] \begin{pmatrix} 1-λ & 1 & 2\\ 4 & -λ & 2\\ -2 & 1 & 1-λ \end{pmatrix} [/tex] [h2]Homework Equations[/h2] This is how I calculated the determinant: [itex]A11: (1-λ)([-λ(1-λ)]-2)=(1-λ)(λ^2-λ-2)[/itex] [itex]A12: -1([4(1-λ)]-[2*-2])=-8+4λ[/itex] (This is where I forgot the minus in front of the 1.) [itex]A13: 2([1*4]-[-2*-λ])=8-4λ[/itex] A12 and A13 cancels each other out so: [itex]detA = 0 → (1-λ)(λ^2-λ-2)=0[/itex] So the eigenvalues are: [itex]λ=1[/itex], [itex]λ=2[/itex] , [itex]λ=-1[/itex] I then had to calculate two eigenvectors, I chose 1 and 2 which are respectively:[tex]\frac{√6}{3} \begin{pmatrix} 0,5\\ 1\\ -0,5 \end{pmatrix} [/tex] [tex]\frac{1}{√35} \begin{pmatrix} -3\\ -5\\ 1 \end{pmatrix} [/tex] I believe this is the correct answer. [h2]The Attempt at a Solution[/h2] So now my question is when I was calculating the determinant, I forgot the minus so my answers became this: [itex]λ=2[/itex], [itex]λ^2=9[/itex] I inserted [itex]λ=3[/itex] into the eigenvalue matrix, it became this:[tex] \begin{pmatrix} -2 & 1 & 2\\ 4 & -3 & 2\\ -2 & 1 & -2 \end{pmatrix} [/tex] Performing row operations the matrix will become:[tex] \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} [/tex] Can a situation like this ever occur? I hope this clears a little bit up. [/QUOTE]
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Question about eigenvector and identity matrix
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