# Eigenvalue problem

1. Jan 23, 2008

### neelakash

1. The problem statement, all variables and given/known data

Consider the following problem:

if $$\ A$$$$\psi=$$$$\lambda$$$$\psi$$,prove that

$$\ e ^ A$$$$\psi=$$$$\ e ^\lambda$$$$\psi$$

2. Relevant equations

3. The attempt at a solution

This is my attempt.Please check if I am correct.

If
$$\ e ^ A$$$$\psi=$$$$\ e ^\lambda$$$$\psi$$

is correct, we should have:

$$\ ln$$$$\ e^ A$$$$\psi=$$$$\ ln$$$$\ e ^\lambda$$$$\psi$$

or, $$\ ln \{e^A} +$$$$\ ln \psi=$$$$\ ln \{e^\lambda} +$$$$\ ln \psi$$

Now cancel $$\ ln \psi$$ from both sides and post-multiply the resulting equation by $$\psi$$

That is---

$$\ ln \{e^A} =$$$$\ ln \{e^\lambda}$$

or, $$\ [ln \{e ^ A}]$$$$\psi=$$$$\ [ln \{e ^\lambda}]$$$$\psi$$

Alternatively,

$$\ A$$$$\psi=$$$$\lambda$$$$\psi$$

So, we got the given equation from the equation to be proved.

Last edited: Jan 23, 2008
2. Jan 23, 2008

### neelakash

The LaTeX image is not coming good.But I think everyone will understand my procedure.

3. Jan 23, 2008

### Rainbow Child

Do you know the definition of the matrix $e^A$?

4. Jan 23, 2008

### neelakash

I know the definition of e^A

Let A,B and C be three matrices.

If I have A+C=B+C

what is wrong if I cancel C from both sides?

ln (psi) is a matrix if you consider psi as matrix.I am usimg psi as a complex function like psi=M+iN

5. Jan 23, 2008

### neelakash

I did according to your hint.That works well, but still I am having problem to see why the logaritm of a matrix/operator will not work.

6. Jan 23, 2008

### CompuChip

Also, your way of setting up the proof is flawed, as the following direct analogy shows

Theorem: If 1 = 1, prove that 0 = 1.

Proof:
If 0 = 1 is correct, then we obviously also have 1 = 0. Adding the two equations gives
Code (Text):

0 = 1
1 = 0  +
--------
1 = 1

which is the assumption given.