Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
is my linear algebra proof solution good?
here's the problem, suppose A is a square matrix and
A = kA^T where k≠±1
show that A =0.
my solution:
(kA^T)^T = kA
therefore;
kA = A
A(k-1)=0
since k cannot be ±1, A has to be zero so
A = 0
is this correct?
1 Answer
- 6 years agoFavorite Answer
Close - just made a simple mistake. You'll notice nothing stops k = -1 in your proof. You made a slight error.
We have
A = k*A^T and you computed (k*A^T)^T = kA. But this doesn't mean A=kA, we have to applied the transpose operation to both sides like so:
A = k*A^T
(A)^T = (k*A^T)^T
A^T = kA
But then substitute A^T=kA back in the original equation A = k*A^T to yield:
A = k*A^T = k*(kA) = k^2*A
Solving in the same way you did:
A - k^2*A = 0
(1-k^2)*A = 0
Now we have both k=±1 or A = 0. And so then we conclude A = 0.
So you were close - just be careful!