a.) How many orthonormal eigenvector bases does a symmetric $n$ x $n$ matrix have? Now let $A=\pmatrix{a&b\\c&d}$, write down necessary and sufficient conditions on the entries a, b, c, d that ensures that A has only real eigenvalues.
b.) Let $A^T =-A$ be a real, skew-symmetric $n$ x $n$ matrix. Prove that the only possible real eigenvalue of A is $\lambda = 0$?
Answer for a:
If all eigenvalues are distinct there are $2^n$ different bases. If the eigen values are repeated there are infinitely many.
How did they get that? Lets say I have a $2$ x $2$ matrix and it has distinct eigenvalues (lets say 1 and 2 are the eigenvalues) wouldn't the eigenvectors be equal to the amount of eigenvalues, so in this case it will equal 2? But the answer says it equals 4?