4 votes 4 votes The condition for which the eigenvalues of the matrix $A=\begin{bmatrix} 2 & 1\\ 1 &k \end{bmatrix}$ are positive is $k > \frac{1}{2}$ $ k > −2$ $ k > 0$ $k< \frac{-1}{2}$ Linear Algebra eigen-value linear-algebra + – rahul sharma 5 asked Jun 26, 2017 • edited Apr 25, 2019 by Lakshman Bhaiya rahul sharma 5 4.2k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Prince Sindhiya commented Aug 17, 2018 i moved by Mk Utkarsh Sep 30, 2018 reply Follow Share https://gateoverflow.in/134860/gate-me-2016-eigen-value Duplicate 1 votes 1 votes Lakshman Bhaiya commented Nov 6, 2018 reply Follow Share $A)K>\frac{1}{2}$ 0 votes 0 votes aditi19 commented Nov 6, 2018 reply Follow Share how? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes By the properties of eigen values, if all the principal minors of A are possitive then all the eigen values of A are also possitive. so |A2*2| > 0 ie 2k - 1 > 0 k>1/2 pawan kumarln answered Jun 27, 2017 pawan kumarln comment Share Follow See all 2 Comments See all 2 2 Comments reply joshi_nitish commented Jun 29, 2017 reply Follow Share you are taking det(A)>0.. but what if both eigen values are negative then also det(A)>0 will follow.. 0 votes 0 votes Abhisek Das commented Feb 24, 2018 reply Follow Share We know that Det(A) = product of eigen values and Tr(A) = sum of eigen values. Therefore, to have both eigen values as positive , det(A) should be >0 & Tr(A) >0.So, we can check for both the conditions on the given matrix: Det(A) > 0 => 2k-1>0 => k>1/2 and Tr(A) > 0 => k+2 > 0 => k>-2 Hence, the first solution satisfies both the condition. So, the answer is A) .. I think this is a better approach than the Best Answer here @Joshi_nitish 10 votes 10 votes Please log in or register to add a comment.