I am reviewing some more advanced linear algebra before I go to grad school, and I came across this question.
"Does there exist a Hermitian matrix with characteristic polynomial equal to $x^4- 1$? If there does, construct one and prove it is one. If there does not, prove there does not."
If we take our field to be the complex numbers, then we have 4 distinct eigenvalues $1,-1,i,-i$, each with algebraic/geometric multiplicity 1, so we have a diagonalizable matrix (similar to a diagonal matrix) with a basis of eigenvectors for our vector space. So I though about the matrix $$A= \begin{bmatrix} 1 & 0 &0 &0\\ 0& -1 &0&0 \\ 0 &0 &i &0\\ 0 &0&0&-i\\ \end{bmatrix}$$ However $$A^*= \begin{bmatrix} 1 & 0 &0 &0\\ 0& -1 &0&0 \\ 0 &0 &-i &0\\ 0 &0&0&i\\ \end{bmatrix}$$ So $A\neq A^*$. I know that a matrix is hermitian if and only if there exists an orthonormal basis consisting of eigenvectors of T. So why does my example not work out?
A square matrix is Hermitian if and only if it is unitarily diagonalizable with real eigenvalues. This proves that the answer is negative.