Determine this!
Let A = ( a i j ) be the 2 0 1 6 × 2 0 1 6 matrix with a i , i = 0 and a i , j = 1 when i = j , meaning that the diagonal entries are all 0 and the off-diagonal entries are all 1. Find det ( A ) .
The answer is -2015.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
Yes, that's a sensible way of doing it! (+!)
The matrix A + I 2 0 1 6 , whose entries are all 1, has the eigenvalues 2016 and 0, with multiplicity 2015. Thus A has eigenvalues 2015 and − 1 , with multiplicity 2015. The determinant is the product of the eigenvalues, det ( A ) = ( − 1 ) 2 0 1 5 2 0 1 5 = − 2 0 1 5
Sorry sir if I am asking a very trivial question,but I am not understanding how you obtained the eigenvalues of (A+I) and after that of A?
Log in to reply
The vector ( 1 , 1 , … , 1 ) T is an eigenvector of A of eigenvalue 2 0 1 5 , and any element of the 2 0 1 5 -dimensional subspace of vectors perpendicular to ( 1 , 1 , … , 1 ) T is an eigenvector with eigenvalue − 1 .
A + I has rank 1, meaning that the kernel is ( n − 1 ) -dimensional. This in turn means that 0 is an eigenvalue with multiplicity n − 1 . Now the last eigenvalue is n , the trace. The eigenvalues of A + k I are k more than those of A , with the same eigenvectors.
Hope this makes sense...
Log in to reply
Yes,I now understand the method by which you found the eigenvalues.Thank you very much for the explanation.
Awesome question!!
I did it by induction and concluded with the result
For n×n matrix det(A) = - (n - 1) if n is even and (n - 1) if n is odd.
Cool result @Otto Bretscher
Same here :P
Let R i denote the i th row of A and n = 2 0 1 6 . Perform the row operation R 1 ← i = 1 ∑ n R i . This leads to a matrix whose all elements in the first row is n − 1 . Then for each row R j , j > 1 , perform the row operation R j ← R j − n − 1 1 R 1 . This leads to the following matrix A ′ ′ = ( ( n − 1 ) O ( n − 1 ) × 1 ( n − 1 ) J 1 × ( n − 1 ) − I ( n − 1 ) × ( n − 1 ) ) where I , J , O respectively denotes the identity, all one and all zero matrices with indicated dimensions. Expanding the determinant of A ′ ′ along the first column, we readily have det ( A ′ ′ ) = ( − 1 ) n − 1 ( n − 1 ) . Since elementary row operations don't change the determinant, we have det ( A ) = det ( A ′ ′ ) = ( − 1 ) n − 1 ( n − 1 ) = − 2 0 1 5 .