5 votes 5 votes Consider the $5\times 5$ matrix below : $\begin{bmatrix} 1&0 &0 &0 &1 \\ 0& 1 & 1 & 1 & 0\\ 0& 1 &1 &1 &0 \\ 0& 1 &1 &1 &0 \\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}$ The product of the non-zero eigenvalues of the matrix is _____. Linear Algebra tbb-mockgate-4 numerical-answers engineering-mathematics linear-algebra matrix eigen-value + – Bikram asked May 14, 2017 retagged Sep 11, 2020 by ajaysoni1924 Bikram 794 views answer comment Share Follow See 1 comment See all 1 1 comment reply JashanArora commented Jan 17, 2020 reply Follow Share Trace = 5. (Sum of diagonal elements) There will be 5 (not necessarily distinct) eigenvalues, because this is a $5\times 5$ matrix. Let the eigen values be $a, b,c,d,e$ $a+b+c+d+e=5$ The rank of this matrix is 2. How? $R_5 \leftarrow R_5-R_1$. Last row becomes 0. So, rank isn't 5. For any $4\times 4$ matrix, we'll still not get non-zero determinant. Same for $3\times 3$ matrix. But for this $2\times 2$ matrix $\begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}$ Determinant is non-zero. So, rank =2. Since rank = 2, we only have two non-zero eigenvalues. So, the equation reduces to $a+b=5$ Now, eigenvalues can be $(1,4)$ or $(2,3)$ => The answer could be 4 or 6. 2 votes 2 votes Please log in or register to add a comment.
Best answer 22 votes 22 votes 6 will be answer... akash.dinkar12 answered Aug 16, 2017 selected Aug 16, 2017 by Bikram akash.dinkar12 comment Share Follow See 1 comment See all 1 1 comment reply mehul vaidya commented Jun 6, 2019 reply Follow Share @Bikram @akash.dinkar12 @Arjun I am little doubtful about this solution. From first look it looks correct. But I have below doubt All Five equations x1 + x5 = lambda * x1 x2 + x3 + x4 = lambda * x2 x2 + x3 + x4 = lambda * x3 x2 + x3 + x4 = lambda * x4 x1 + x5 = lambda * x5 must hold true for lambda =2 and also for lambda = 3 but according to this explanation x1 + x5 = lambda * x1 & x1 + x5 = lambda * x5 hold true for lambda =2 and x2 + x3 + x4 = lambda * x2 x2 + x3 + x4 = lambda * x3 x2 + x3 + x4 = lambda * x4 holds true for lambda = 3 To solve system of five equation , we should substitute one equation in another to find relation between all five variable. Now if we use lambda =2 and try to find eigen vector , we can see that x1 and x5 are related (same) x2 , x3 , x4 are related (same). but there is no correlation between this two group. hence we fail to find eigen vector I am 70% sure about my explanation and not 100% . Please correct me If my explanation is wrong 0 votes 0 votes Please log in or register to add a comment.