# Eigenpairs and Hermitian matrices

1. Oct 22, 2009

### Somefantastik

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

Choose $$\lambda_{1}, \lambda_{2}, \lambda_{3}$$ along with a set of vectors $${v_{1},v_{2},v_{3}}$$ and construct an Hermitian matrix H with the eigenpairs $$(\lambda_{1},v_{1}),(\lambda_{2},v_{2}),(\lambda_{3},v_{3})$$

2. Relevant equations

3. The attempt at a solution
$$\lambda_{1} = 2$$
$$\lambda_{2} = -2$$

$$v_{1} = \widehat{i}+2\widehat{j}$$
$$v_{2} = \widehat{i}-2\widehat{j}$$

$$u_{1} = \frac{1}{\sqrt{5}}\left(\widehat{i} + 2\widehat{j} \right)$$
$$u_{2} = \frac{1}{\sqrt{5}}\left(\widehat{i} - 2\widehat{j}\right)$$

$$v_{3} = u_{1} \times u_{2} = -\frac{4}{5}\widehat{k}$$

$$u_{3} = -\widehat{k}$$

$$Basis: \ U = \left\{ u_{1},u_{2},u_{3} \right}\}$$

$$H = \left| \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{array} \right|$$

How does this look?

2. Oct 23, 2009

### lanedance

can you choose any values here?

why not just
$$\lambda_{1} = 1$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 3$$

and
$$u_{1} = (1,0,0)$$
$$u_{2} = (0,1,0)$$
$$u_{3} = (0,0,1)$$

then
$$H = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2& 0 \\ 0 & 0 & 3 \end{array} \right|$$

3. Oct 23, 2009

### Somefantastik

If that's all this is, am I misunderstanding or is this a really trivial question?

4. Oct 23, 2009

### lanedance

maybe it asking if you are given arbitrary
$$u_1, u_2, u_3$$
and corresponding eigenvalues
$$\lambda_1, \lambda_2, \lambda_3$$
can you find a hermitian matrix H, with the given eigenpairs?

one way you could do is assume you have found H and it can be diagonalised (which you know it can, from the eigenvectors), then the diagonal form of H is
$$D = P^{-1}.H.P$$

where
$$D = \left| \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2& 0 \\ 0 & 0 & \lambda_3 \end{array} \right|$$

now the matrix P is simply given by the eigenvectors
where
$$P = \left| \begin{array}{ccc} u_1, u_2, u_3 \end{array} \right|$$

then H is
$$H = PDP^{-1}$$

that gives the eigenpairs, however I doubt it would automatically be hermitian... for that you need, where the star means conjugate transpose
$$H = H^*$$

so running that through
$$PDP^{-1} = (PDP^{-1})^* = P^{-1}^*D^*P^* = P^{-1}^*DP^*$$

hmmm... not totally sure where to go now

Last edited: Oct 23, 2009
5. Oct 23, 2009

### lanedance

maybe a little left & right muliplication
$$D = P^{-1}(PDP^{-1})P = (P^{-1}P^{-1}^*)D(P^*P)$$

this will be satisfied if P is unitary
$$P^*P = PP^* = I$$

6. Oct 23, 2009

### lanedance

from before P is given by the eigenvectors (choose up to some multiplicative constant)
$$P = \left| \begin{array}{ccc} u_1, u_2, u_3 \end{array} \right|$$

so in subscript form if j is the eigenvector and i is it's component, write
$$P = P_{ij} = u_{j}^{(i)}$$

now for the transpose conjugate of a matrix
$$P^* = (P^*)_{ij} = \bar{P}_{ji} = \bar{u}_{i}^{(j)}$$

matrix multipication is given by (with summation over the repeated indice)
$$P^*P = (P^*)_{ij}P_{jk} = \bar{u}_{i}^{(j)} u_{k}^{(j)}$$
which looks just like a complex inner product between eigenvectors

so to satisfy the unitary condition we were looking for
$$PP^* = I \rightarrow \bar{u}_{i}^{(j)} u_{k}^{(j)} = \delta_{ik}$$

ie. all the columns of P must form an orthonormal basis, which is a property of unitary matricies in general. So the condition will be satisfied if you just normalise your eigenvectors to length 1.

got a bit carried away there... but from this you should be able to write the general form of H no worries

7. Oct 24, 2009

### Somefantastik

Ok that helps. I guess I got confused and thought he was looking for D, but I think you're right; I think he's looking for PDP-1

Thanks!