1 votes 1 votes An arbitrary vector X is an eigen vector of the vector of the matrix A=[1 0 0 a 0 0 0 0 b], if (a, b)= (a) (0, 0) (b) (1, 1) (c) (0, 1) (d) (1, 2) Mathematical Logic linear-algebra matrix + – Prince Sindhiya asked Mar 2, 2018 • retagged May 6, 2021 by Shiva Sagar Rao Prince Sindhiya 1.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Given Matrix is A = [1 0 0 ; a 0 0 ; 0 0 b] Since it is a upper triangular matrix ; eigen values will be the diagonal elements (1, a , b); Now suppose 1 as the eigen value(c) of the arbitrary vector ; then we have the equation AX = cX => AX = 1X (1 0 0 ; a 0 0; 0 0 b) (X1; X2; X3) = 1(X1; X2; X3) Solving the above; u will get the following equations X1 = X1 ; aX1 = X1 ; bX3 = X3 So bX3 = X3 => b =1 -----> (1) aX1 = X1 => a=1 So (a,b) = (1,1) is the answer. Sai Sankalp answered Mar 5, 2018 Sai Sankalp comment Share Follow See all 0 reply Please log in or register to add a comment.