Hi, I am not 100% sure my working is correct for this question. Could someone be kind enough to check my working please.
The characteristic polynomial of A:
So the eigenvalues are and .
To find the eigenvectors we sub in and into the equations:
to solve for x and y for both eigenvalues.
So for example, for I get and .
Am I correct in my working out so far?
Thanks
There is a well known theorem: if is a simple eigenvalue then, the corresponding eigenspace has dimension 1. This means that a non null vector of form a basis of (because is linearly independent). In our case, and choosing (for example) we obtain from the first equation that . You needn't verify the second equation because necessarily the rank of the system matrix is 1.
or you could just solve the system Av = λv for each eigenvalue λ.
for example, with λ = i, and writing a+bi = z, c+di = w this leads to:
that is:
so:
so any eigenvector corresponding to i is of the form (z,(1+i)z)...z, of course, must be non-zero. it is convenient to choose z = 1+0i = 1.