something MATH.......

[mitsun]

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.

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A = ( a         b )
    ( -a^2/b   -a ) 

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.

[lovelove]

A <> 0, A^2 = 0 , where A is a 2x2 matrix.

Write A using 2 real variables.

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Given: A is a 2x2 matrix.
Prove: A^4 = 0 -> A^2 = 0.

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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 

[perceval]

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.

Code: Select all

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.

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