Eigenpairs and Hermitian matrices

[Somefantastik]

Homework Statement

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})

Homework Equations

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?

 

[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|

 

[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
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

 

[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

 

[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

 

[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!