Inner product with (1,1) tensors: Diff. Geometry/ Lin algebra

[SNOOTCHIEBOOCHEE]

Homework Statement

Given g$$\equiv g_{ij} =$$
[-1 0;
0 1]

Show that A= $$A^{i}_{j}$$ =
[1 2
-2 1]

is symmetric wrt innter product g, has complex eigenvalues, but eigenvectros have zero length wrt the complex inner product.

The Attempt at a Solution

Im sure this is just a simple linear algebra problem but I am having trouble figuring out how to compute the dot procut with this 1,1 tensor.

My guess would be to break the matrix $$A^{i}_{j}$$ intro rows and compute { (row1)g(row2)^t} then to show this is symmetric calculate { (row2)g(row1)^t}. But that seems wrong.

I can calculate the eigenvalues (they come out to $$\pm i \sqrt{3}$$ )

Also i am lost on showing the eigenvecotrs are 0 wrt the this inner product. I would have no idea how to approach this even if i knew how to calculate inner product.

Thanks in advance

 

[eok20]

The metric g let's you lower the indices of A_i^j to get a tensor A_{ij}. Then the claim is that A_{ij} = A_{ji}. For the other part, I'm not quite sure what inner product they mean. I'm guessing g as a complex inner product?

 

[SNOOTCHIEBOOCHEE]

Ok but as written, how do you compute that complex inner product?