I have a linear operator $T:R^n \rightarrow R^n$ satisfying $T\vec{x}\cdot \vec{y}=\vec{x} \cdot T\vec{y} , \forall\vec{x},\vec{y}\in{R^n}$. And i'm given that $\vec{v_1},\vec{v_2}$ are eigenvectors of $T$ with corresponding egienvalues ${\lambda}_1 , {\lambda}_2$ when ${\lambda}_1\neq{\lambda}_2$ ,trying to show that $\vec{v_1},\vec{v_2}$ are orthonormal.

I have shown that they are orthogonal this way : since $\vec{v_1},\vec{v_2}$ are eigenvectors then $Tv_1={\lambda}_1v_1 , Tv_1={\lambda}_2v_2$. Let $x=v_1 ,y=v_2$ then $Tv_1 \cdot v_2=v_1 \cdot Tv_2 \implies {\lambda}_1(v_1 \cdot v_2)={\lambda}_2 (v_1 \cdot v_2) \implies {\lambda}_1(v_1 \cdot v_2)-{\lambda}_2 (v_1 \cdot v_2)=0 \implies (v_1 \cdot v_2)({\lambda}_1-{\lambda}_2)=0.$ since i'm given ${\lambda}_1 \neq {\lambda}_2 \implies v_1 \cdot v_2 =0 \implies v_1\bot v_2$. Now i'm having trouble proving that $||{v_1}||=||{v_2}||=1$ to complete the proof.

  • $\begingroup$ You can’t prove anything about the norms of the two vectors (other than that they’re non-zero) from the given conditions since any non-zero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue. You can show that the matrix of $T$ is symmetric, though. $\endgroup$ – amd Jun 9 '17 at 18:54

You can not prove that $\Vert{v_1}\Vert=\Vert{v_2}\Vert=1$ !

Let $v=3v_1$ and $u=7v_2$. Then it still holds:

$u*v=0$ and $Tv=\lambda_1v$ and $Tu= \lambda_2u$.

  • $\begingroup$ Not sure i fully understand what you mean by that last part ,But i just noticed that $T$ is representing an orthogonal matrix if i'm not mistaken,i'm trying to reach a solution from that now. $\endgroup$ – user3133165 Jun 9 '17 at 10:38
  • $\begingroup$ Never mind i see what you mean ,thank you! $\endgroup$ – user3133165 Jun 10 '17 at 13:29

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