The three original questions are all essentially the same (as I think Bill was attempting to lead up to). You can Google "matrix square root" to get more information than you want.
The solution to A*A = 0 can be derived with a little algebra (exercise for the reader). Verifying the solution is easy by direct matrix multiplication.
For amusement, this is the only matrix with determinant = trace = 0. The determinant property is necessary, but the trace property is not obvious, at least to me.
Re: something MATH.......
Posted: Tue Apr 02, 2013 5:47 am
by lovelove
Bill Spight wrote:A <> 0, A^2 = 0 , where A is a 2x2 matrix.
Write A using 2 real variables.
----
Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.
I think I get the first question.
A = ( a b ); 'a, b, c, d' are real numbers; 'n' is a natural number
( c d )
If A^n = O, 'A' cannot have an inverse matrix, so the determinant is zero.
using the Caily-Hamilton thing, A^2 = (a+d)A
when n >= 3, A^n = (a+d)^(n-1)A = O, a+d = 0, A^2 = O
when n = 1 or n = 2, A^2 = O
Therefore A^2 = O
Re: something MATH.......
Posted: Tue Apr 02, 2013 3:10 pm
by perceval
lovelove wrote:
Bill Spight wrote:A <> 0, A^2 = 0 , where A is a 2x2 matrix.
Write A using 2 real variables.
----
Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.
I think I get the first question.
A = ( a b ); 'a, b, c, d' are real numbers; 'n' is a natural number
( c d )
If A^n = O, 'A' cannot have an inverse matrix, so the determinant is zero.
using the Caily-Hamilton thing, A^2 = (a+d)A
when n >= 3, A^n = (a+d)^(n-1)A = O, a+d = 0, A^2 = O
when n = 1 or n = 2, A^2 = O
Therefore A^2 = O
OK for Q1.
Now for Q2:
you have to either prove the proposed assertions, or find a counter example.
For 2 i suggest finding a counter example .
bigger hint:
use the result from Q1
more hint:
For example take a matrix A different from 0 such that A^2=0.
.