# Linear Algebra - Eigenvalue Problem

chill_factor

## Homework Statement

Let there be 3 vectors that span a space: { |a>, |b>, |c> } and let n be a complex number.

If the operator A has the properties:

A|a> = n|b>
A|b> = 3|a>
A|c> = (4i+7)|c>

What is A in terms of a square matrix?

det(A-Iλ)=0

## The Attempt at a Solution

I don't even know how to start. Can someone give me a starting hint?

susskind_leon
Imagine |a>=(1,0,0), |b>=(0,1,0), |c>=(0,0,1)
What would the square matrix be then?
Once you've got that, how can you relate your "hypothetical" square matrix to the correct one?

Homework Helper
Since A maps a three dimensional vector space to itself, it can be represented as a 3 by 3 matrix and so can be written
$$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$$

Since you have basis vectors |a>, |b>, and |c>, they can be written as <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>, respectively. What do you get when you multiply each of those by the matrix above?

You are told that "A|a> = n|b>", that is the what you got by multiplying A by <1, 0, 0> must be equal to <0, n, 0> for this complex number n. Compare the two.

Similarly, you are told that "A|b> = 3|a>". In other words, what you got by mutiplying A by <0, 1, 0> must be equal to <3, 0, 0>.

Finally, you are told that "A|c> = (4i+7)|c>" so that what you got by multiplying A by <0, 0, 1> must be equal to <0, 0, 4i+7>.

chill_factor
Ok, I didn't think of that. Writing them out in terms of specific numbers is helpful. However, why can we do that?

I think I understand why they were written that way a little. Is it because they are orthogonal elements of a Hilbert Space and since they are orthogonal, we can define a coordinate system such that one is <1,0,0> and the other 2 are necessarily orthogonal to it and others, following the right hand rule?