# Eigenvectors of hermitian matrix are not coming out to be orthogonal

I have the following hermitian matrix before me: $$\begin{pmatrix}2 &1+i\\ 1-i& 3\end{pmatrix}$$ I calculated its characteristic polynomial as $$k^2-5k+4$$ which has $$1$$ and $$4$$ as its roots. Eigenvectors corresponding to these two eigenvalues are $$\begin{pmatrix}-(1+i) &1 \end{pmatrix}$$ and $$\begin{pmatrix}1 &1-i\end{pmatrix}$$ But these are not orthogonal. Why is it so? Am I doing something wrong?

• make sure you are considering in the complex inner product i.e. $<v,w>=v \bar{w}$ Mar 3 at 14:23
• @Moo The second eigenvector of yours can be obtained by multiplying mine one by $1+i$. Even then the vectors are not orthogonal. Mar 3 at 14:54
• @Brozovic Could you please elaborate on complex inner product? I have no knowledge about it. Mar 3 at 15:32

Check Eigen values are $$4,1$$ and eigenvectors are $$V_1=\begin{bmatrix} 1+i \\2 \end{bmatrix}$$ and $$V_2=\begin{bmatrix} -1-i \\ 1 \end{bmatrix}$$, then $$V_1^{\dagger} V_2=0$$, $$\dagger$$ means conjugate and transpose.