Eigenpairs and Hermitian matrices

Somefantastik

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

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

2. Relevant equations



3. The attempt at a solution
[tex] \lambda_{1} = 2[/tex]
[tex]\lambda_{2} = -2[/tex]

[tex]v_{1} = \widehat{i}+2\widehat{j}[/tex]
[tex]v_{2} = \widehat{i}-2\widehat{j}[/tex]

[tex]u_{1} = \frac{1}{\sqrt{5}}\left(\widehat{i} + 2\widehat{j} \right)[/tex]
[tex]u_{2} = \frac{1}{\sqrt{5}}\left(\widehat{i} - 2\widehat{j}\right)[/tex]

[tex]v_{3} = u_{1} \times u_{2} = -\frac{4}{5}\widehat{k}[/tex]

[tex]u_{3} = -\widehat{k}[/tex]

[tex]Basis: \ U = \left\{ u_{1},u_{2},u_{3} \right}\}[/tex]

[tex]H = \left| \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{array} \right| [/tex]

How does this look?

lanedance

can you choose any values here?

why not just
[tex] \lambda_{1} = 1 [/tex]
[tex] \lambda_{2} = 2 [/tex]
[tex] \lambda_{3} = 3 [/tex]

and
[tex] u_{1} = (1,0,0)[/tex]
[tex] u_{2} = (0,1,0)[/tex]
[tex] u_{3} = (0,0,1)[/tex]

then
[tex]H = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2& 0 \\ 0 & 0 & 3 \end{array} \right| [/tex]

Somefantastik

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

lanedance

maybe it asking if you are given arbitrary
[tex] u_1, u_2, u_3 [/tex]
and corresponding eigenvalues
[tex] \lambda_1, \lambda_2, \lambda_3 [/tex]
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
[tex] D = P^{-1}.H.P[/tex]

where
[tex]D = \left| \begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2& 0 \\ 0 & 0 & \lambda_3 \end{array} \right| [/tex]

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

then H is
[tex] H = PDP^{-1}[/tex]

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


so running that through
[tex] PDP^{-1} = (PDP^{-1})^* = P^{-1}^*D^*P^* = P^{-1}^*DP^* [/tex]

hmmm... not totally sure where to go now

lanedance

maybe a little left & right muliplication
[tex] D = P^{-1}(PDP^{-1})P = (P^{-1}P^{-1}^*)D(P^*P) [/tex]

this will be satisfied if P is unitary
[tex]P^*P = PP^* = I[/tex]

lanedance

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

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

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

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

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

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

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!