Suppose $A$ is a symmetric matrix with complex entries. Let $c$ and $k$ be two distinct eigenvalues and $X$ and $Y$ be two eigenvectors belonging to $c$ and $k$. $AX=cX => (AX)^T=(cX)^T => X^TA^T=cX^T => X^TA=cX^T => X^TAY=cX^TY => X^TkY=cX^TY => (k-c)X^TY=0 => X^TY=0$,since $c≠k$.

My teacher gave the result for real symmetric matrices. I don't see why it won't be true for complex matrices. Isn't there a notion of orthogonality in $\Bbb C^n$?


Yes, there is. But the usual concept of orthogonal vectors in $\mathbb{C}^n$ is: $(z_1,\ldots,z_n)$ and $(w_1,\ldots,w_n)$ are orthogonal when $\sum_{j=1}^nz_j\overline{w_j}=0$. And, for this concept of orthogonal vectors, it is not true in general that eigenvectors belonging to distinct eigenvalues of a complex symmetric matrix are orthogonal.

However, it is true for Hermitian matrices (and the proof is similar).

  • $\begingroup$ Oh I see. So even if $X^TY=0$, $X$ and $Y$ in general cannot be called orthogonal? We need $X^T\overline Y=0$ for that? $\endgroup$ – Hrit Roy Mar 30 '18 at 15:47
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    $\begingroup$ @HritRoy In fact. If $v=(1,i)$, then $v$ is not orthogonal to itself, but $v^Tv=0$. $\endgroup$ – José Carlos Santos Mar 30 '18 at 15:48
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    $\begingroup$ In fact, a matrix that is both symmetric and real is Hermitian when viewed as a member of $\mathbb{C}^{n \times n}$. In this sense, the result for real matrices is a special case of the result for complex matrices; it's just that the real result doesn't generalize in the most obvious way. $\endgroup$ – Michael Seifert Mar 30 '18 at 15:57

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