1. The problem statement, all variables and given/known data I think my teacher made a mistake in his homework answer. I need to verify this for practice. The answer I got is below. The answer the teacher has is in the pdf. 2. Relevant equations Please refer to attached pdf 3. The attempt at a solution So there is two eigenvalues= 4 and 2 but the eigenvalue 2 has 2 eigenvectors [-1 1 0]^{T} and [0 0 1]^{T} but my teacher has only one [-1 1 0]^{T}. That's why he says A is not diagonalizable. Do you think it's correct?
Re: Diagonalization I get the same result as your teacher for [tex] \lambda = 2 [/tex] [tex] A= \begin{bmatrix} 3 & 1 & 0\\ 0 & 2 & 1\\ 1 & 1 & 3 \end{bmatrix} [/tex] so for lamda = 2, [tex] (A-2I)\vec{v}=\vec{0} [/tex] [tex] \begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix} [/tex] I've personally always found it easier to not do row operations here and just jump straight in. From that you get [tex] v_3 = 0 [/tex] and [tex] v_1 = -v_2 [/tex] so therefore [tex] \vec{v} = \begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix} [/tex] Since it is a 3x3 matrix, it needs 3 eigenvectors to be diagonalizable. Hope this helps.
Re: Diagonalization Hi, thanks for replying. Attached is how I got my vectors, do you think my steps are correct.
Re: Diagonalization No. Why do you think (0, 0, 1) is an eigenvector? You seem to just pull that out of thin air.
Re: Diagonalization You seem to have made a mistake in the step [tex] (A-2I)\vec{v}=\vec{0} [/tex] Why is your second row 1 0 1 instead of 0 0 1 ? Have you said [tex] v_2= \delta [/tex] ? I'm not sure if I have read that correctly. If it is a delta, you can't say that unless the whole row equals zero. i.e. [tex] \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix} [/tex]