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I have the following problem:

Suppose that $A$ is a symmetric matrix, with $A$ = $A^{T}$ . Suppose $\vec{v}$ and $\vec{w}$ are eigenvectors of $A$ associated with distinct eigenvalues. Show that $\vec{v}$ and $\vec{w}$ must be orthogonal. (Hint: Show that a$\vec{v}$ $\cdot$ $\vec{w}$ = $\vec{v}$ $\cdot$ b$\vec{w}$.)

I am unsure how to approach this, even with the hint taken into account. I tried to use the fact that orthogonal complement of Im(A) is in Ker of A transpose, and since they are equal it is also in ker A, but that didn't get me anywhere (I am probably thinking in the wrong direction). Thanks in advance for your hints!

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  • $\begingroup$ the main property is $\langle Ax,y\rangle=\langle x,A^Ty\rangle$. that is the property of the transpose (assuming everything is over the reals). $\endgroup$
    – yoyo
    Commented Mar 9, 2011 at 16:11
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    $\begingroup$ Does this answer your question? Eigenvectors of real symmetric matrices are orthogonal $\endgroup$
    – user53259
    Commented Jul 25, 2020 at 22:50

2 Answers 2

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Let $\vec{v}$ be the eigenvector corresponding to $\lambda$ and $\vec{w}$ be the eigenvector corresponding to $\mu$, then we have $Av = \lambda v$ and $Aw = \mu w$.

$v^T(Aw) = (Aw)^Tv$ since it is a scalar. And hence $v^TAw = w^TA^Tv$. Further since $A=A^T$, we get $v^TAw = w^TAv$.

Now use the fact that $Aw = \mu w$ and $Av = \lambda v$.

And use the fact that $\lambda \neq \mu$, to get that the eigenvectors are orthogonal.

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Just writing Sivaram's solution in another way.

For any matrix $A \in \mathfrak{gl}(\mathbb K,n)$, considering the inner product $\langle v,w\rangle = \sum_{i=1}^nv_iw_i$, we have that $$\langle Av,w\rangle = \langle v,A^Tw\rangle,$$ for all $v$ and $w \in \mathbb K^n$.

So, if $Av=\lambda v$ and $Aw=\mu w$ and $A=A^T$, then $$\lambda\langle v,w\rangle = \langle Av,w\rangle = \langle v,Aw\rangle = \mu\langle v,w\rangle.$$ From where we conlude that $\lambda = \mu$ or $\langle v,w\rangle=0$.

Observe that, for $\mathbb K = \mathbb C$ and the inner product given by $\langle v,w\rangle = \sum_{i=1}^nv_i\overline{w_i}$, we would need the condition $A=\overline{A}^T$ in order to get $$\langle Av,w\rangle = \langle v,Aw\rangle,$$ for all $v$ and $w \in \mathbb K^n$.

In fact, for any matrix $A \in \mathfrak{gl}(\mathbb K,n)$ and inner product $\langle,\rangle$ there is a matrix $A^* \in \mathfrak{gl}(\mathbb K,n)$ such that $$\langle Av,w\rangle = \langle v,A^*w\rangle,$$ for all $v$ and $w \in \mathbb K^n$. This matrix $A^*$ is called the dual (or adjoint) matrix of $A$. For more details see "Linear Algebra" by Greub.

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