Eigenvalues

By Hal 2001 on Sunday, December 30, 2001 - 11:08 pm:

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.

By Kerwin Hui on Sunday, December 30, 2001 - 11:15 pm:

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

Kerwin

By Hal 2001 on Sunday, December 30, 2001 - 11:57 pm:

Hi Kerwin,

Another question,

Suppose that

A

is an

n \times n

square matrix such that

A^{- 1}

exists, and let

λ

be an eigenvalue of

A

. Show that

λ \neq 0

, and that

1 / λ

is an eigenvalue of

A^{- 1}

$A x = λ x$

$A^{- 1} A x = A^{- 1} λ x$

$x = λ A^{- 1} x$
What do I do next?

Thanks.
Hal.

By Kerwin Hui on Monday, December 31, 2001 - 09:20 am:

Hal,

A x = λ x

m

A^{m} = 0

(a - d)^{2} < - 4 b c

\geq 0

λ \neq 0

λ = 0

λ^{- 1}

A^{- 1}

λ

Kerwin

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

By Hal 2001 on Monday, December 31, 2001 - 10:34 am:

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)

If you have

A x = λ x

$A^{m} x = λ^{m} x$

If $x \neq 0$ 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 $λ \neq 0$ .
To clarify, what is our evector here?

Thanks.
Hal.

By Kerwin Hui on Monday, December 31, 2001 - 10:47 am:

Hal,

A^{m} x = λ^{m} x

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

Suppose $λ \neq 0$ , we have a non-zero eigenvector $x$ , and so RHS $\neq 0$ , but LHS $= 0$ . Contradiction.

(b) What you get is the condition $- 4 b c > 0$ , 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 $\pm i$ .
(c)Suppose $A$ has evect $x$ with eval $λ$ , then for any $μ$ , we have $A (μ x) = λ μ x = 0$ , i.e. $μ x$ is in the kernel of $A$ . However, the kernel must be trivial (Hit with $A^{- 1}$ to see why).

Kerwin

By Hal 2001 on Monday, December 31, 2001 - 11:01 am:

Thanks for your help Kerwin.

Have a nice new year.
Hal.