Question

: Linear Algebra: (Diagonalization) Let matrix A = {{2,1},{-2,1}}. Find a matrix B such that B^2 = A.

Answer 1

For clarification: A = \[\left[\begin{matrix}2 & 1 \\ -2 & -1\end{matrix}\right]\]

Answer 2

you could rewrite A= S L S^-1, where L is the diagonal matrix of eigenvalues and S is the matrix of eigenvectors. Here, we see that A is singular (row 2 = - row 1), so one lambda =0. The trace 2-1= 1 is the sum of the lambdas, so the other is 1
the square root of A is S L^(1/2) S^-1, but L is has just 1 and 0 so it doesn't change. We conclude A*A= A. multiply out to show it works.

Answer 3

I'm sorry, just to be sure, did you use the matrix I had in the question or in the comment? I forgot the last -1 on the matrix A in the question.

Answer 4

I used
[ 2 1]
[-2 -1]

Answer 5

thanks, just wanted to be sure :D

Answer 6

would there be another matrix B though that if you square you get A?

Answer 7

You know that -1 is also a square root of 1. so L^(1/2) could also be
[-1 0]
[ 0 0]
= -L
so you could deduce that B= S* -L*inv(S) = -S*L*inv(S)= -A is another solution
That is, for
B= [-2 -1]
[ 2 1]
B*B= A