4 votes 4 votes The Eigen values of $A=\begin{bmatrix} a& 1& 0\\1 &a &1 \\0 &1 &a \end{bmatrix}$ are______ $a,a,a$ $0,a,2a$ $-a,2a,2a$ $a,a+\sqrt{2},a-\sqrt{2}$ Linear Algebra engineering-mathematics linear-algebra eigen-value + – Hirak asked Apr 25, 2019 • edited Apr 25, 2019 by Lakshman Bhaiya Hirak 1.3k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Lakshman Bhaiya commented Apr 25, 2019 i edited by Lakshman Bhaiya Apr 25, 2019 reply Follow Share Important properties of Eigen values:- $(1)$Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix. $(2)$ Product of all Eigen values$=Det(A)=|A|$ $(3)$ Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself. Example$:$$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$ Diagonal matrix Eigenvalues are $1,1,1$ $B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$ Upper triangular matrix Eigenvalues are $1,1,1$ $C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$ Lower triangular matrix Eigenvalues are $1,1,1$ ------------------------------------------------------------------ Apply the above properties to the your question then you will get answer $(d).$ Here $\text{Sum of all eigen values = a+a+a=3a}$ and $\text{Product of all eigen values =|A|=$a^{3}$-2a}$ 2 votes 2 votes Hirak commented Apr 30, 2019 reply Follow Share Thanks.. :) I made a BAD silly mistake while finding the determinant... my bad.. ! Thanks a ton.. :) 0 votes 0 votes Please log in or register to add a comment.
Best answer 6 votes 6 votes Plz see the picture below for detailed answer. 1. Using properties to eliminate options: 2. Using general procedure: SuvasishDutta answered Apr 27, 2019 • edited May 1, 2019 by SuvasishDutta SuvasishDutta comment Share Follow See all 6 Comments See all 6 6 Comments reply Hirak commented Apr 30, 2019 reply Follow Share Instead of doing all these i think if we find out the determinant of the given matrix and see which eigen values product given in the option matches then it will be much faster.. I made a silly mistake while calculating the determinant so my ans came wrong.. see that the determinant is a^3 -2a and only option d's product is a^3 - 2a so option d must be the ans.. This procedure will take less time.. :) 0 votes 0 votes SuvasishDutta commented Apr 30, 2019 reply Follow Share Here your method is possible which will save time. But you cannot apply this shortcut in all cases. For example, in the matrix given in the question, Determinant of A= a*(a-√2)*(a+√2) and eigen values are= a, a-√2, a+√2. But if there are options like (a,a+√2,a-√2) and (-a, a+√2, -a+√2) among the four options. Here the shortcut is not applicable for eliminating either of the option. In this situation we have to solve the characteristic equation. I have shown the general procedure for finding the eigen values. 0 votes 0 votes Hirak commented May 1, 2019 reply Follow Share -a, a+√2, -a+√2 Here the sum of the eigen values are not equal to the sum of the trace of the matrix. Hence we can cancel this option at the first glance.. :) 1 votes 1 votes SuvasishDutta commented May 1, 2019 reply Follow Share Thanks. You can apply properties which will save time. I have added it. 0 votes 0 votes Lakshman Bhaiya commented May 1, 2019 reply Follow Share @SuvasishDutta @Hirak Good explanation, and always believe in method rather than a shortcut. And property just saves time but for better understanding, the actual method is good enough. 0 votes 0 votes SuvasishDutta commented May 1, 2019 reply Follow Share Yes absolutely right @Lakshman Patel RJIT. 1 votes 1 votes Please log in or register to add a comment.