Let A be a square matrix such that for some positive integer m, A^m = 0. Show that every eigenvalue of A is zero.

Let A be a real 2 x 2 matrix which has non-zero non-real eigenvalues. Show that the non-diagonal elements of A are non-zero, but that the diagonal elements may be zero.

Thanks for your help in advance.
Hal.

Hint: The first part - proof by contradiction. The second part - what can you tell me about the trace and determinant of A?

Kerwin

Hi Kerwin,

Another question,

A x=lx

A^-1A x=A^-1lx

x=lA^-1 x
What do I do next?

Thanks.
Hal.

(a) Apply A x=lx m times and remember A^m=0
(b) You should get the condition (a-d)² < -4b c. Now LHS ³ 0.
(c) You have done most of the work. Now to see l ¹ 0, just choose an arbitrary multiple of our eigenvector and get a contradiction for l = 0. To see l^-1 is an eval of A^-1, divide both sides by l.

P.S. They look suspiciously like example sheet questions...

Hi Kerwin,

Yes, they are from example sheet A&G No.6 I was going through the questions again to make sure I understood the math. I am thankful for your help.

I think I understand (b). Since (a-d)² > 0. But if a and d are equal then condition that (a-d)² > = 0 exists - but does that contradict that eigenvalues are non-real?. How do we justify statement 'but that the diagonal elements may be zero'. Does that mean than both a, d could be zero or only one of them?

a)

A^m x=l^m x

If x ¹ 0 this implies that A^m=0.

Does this make sense? And x^m would be meaningless, right?
(c)

I understand that the eval for A^-1. But did not fully follow why l ¹ 0.
To clarify, what is our evector here?

Thanks.
Hal.

(a) A^m x=l^m x.

But A^m=0, so A^m x=0 for all x.

Suppose l ¹ 0, we have a non-zero eigenvector x, and so RHS ¹ 0, but LHS=0. Contradiction.


(b) What you get is the condition -4b c > 0, so b, c are non-zero. a and d can both be zero, e.g. the matrix with b=1, c=-1, a=d=0 has evals ±i.
(c) Suppose A has evect xwith eval l, then for any m, we have A(mx)=lmx=0, i.e. mx is in the kernel of A. However, the kernel must be trivial (Hit with A^-1 to see why).

Thanks for your help Kerwin.

Have a nice new year.
Hal.