2 votes 2 votes This matrix is singular with rank one. Find three $λ$’s and three eigenvectors. $\begin{bmatrix}1\\2 \\1 \end{bmatrix}$ $\begin{bmatrix}2&1 &2 \end{bmatrix}$ = $\begin{bmatrix}2 & 1 &2\\4 & 2 & 4\\2 & 1 &2\end{bmatrix}$ Linear Algebra engineering-mathematics linear-algebra eigen-value + – Shubhanshu asked Jul 24, 2017 • edited Mar 16, 2018 by Sukanya Das Shubhanshu 1.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes λ= 0,0,6 and respective eigen_vectors are, [k1, k2, -k1-k2/2], [k1, k2, -k1-k2/2], k[1, 2, 1]......... where k1, k2 and k are arbitrary constants.. joshi_nitish answered Jul 24, 2017 joshi_nitish comment Share Follow See all 5 Comments See all 5 5 Comments reply Shubhanshu commented Jul 24, 2017 reply Follow Share I am getting. for lamnda = 0 eigen vectors are = [-1, 2, 0] and [-1,0, 1] and for lambda = 6 E.vector is [1, 2, 1] 0 votes 0 votes joshi_nitish commented Jul 24, 2017 reply Follow Share yes, your answer is correct, but you are taking particular solution.. i have taken general solution in term of arbitrary constants... 0 votes 0 votes ankitgupta.1729 commented Mar 17, 2018 reply Follow Share @Joshi_Nitish ,I think for λ= 0 , Eigen vector should be [k1, k2, (-2k1-k2)/2] .. 0 votes 0 votes joshi_nitish commented Mar 17, 2018 reply Follow Share (-2k1-k2)/2 = -k1-k2/2 isn't it? 1 votes 1 votes ankitgupta.1729 commented Mar 17, 2018 reply Follow Share ohh sorry.. I read it wrong.. 0 votes 0 votes Please log in or register to add a comment.