# Prove that $\left \{ v_1,v_2,...,v_n\right \}$ is an orthogonal basis of $V$.

Let be $$V$$ a finite dimensional inner space over the $$\mathbb{R}$$. $$Dim(V)=n$$ with $$n>1$$. Let be $$T$$ a symmetric lineal operator in $$V$$, and $$\left \langle , \right \rangle$$ a inner product in $$V$$.

If $$v_1,v_2,...,v_n$$ are eigenvectors of $$T$$ associated to distinct eigenvalues, prove that $$\left \{ v_1,v_2,...,v_n\right \}$$ is an orthogonal basis of $$V$$.

If we propose a basis $$\mathcal{B}=\left \{ v_1,v_2,...,v_n\right \}$$, we know that:

\begin{align*} Tv_1=c_1 v_1 \ \ \ \ , \ \ \ Tv_2=c_2 v_2 \ \ \ \cdots \ \ Tv_n=c_n v_n \end{align*} And, \begin{align*} \left [ T \right ]_{\mathcal{B}}=\begin{pmatrix} c_1 & 0 & \cdots &0 \\ 0 & c_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & c_n \end{pmatrix} \end{align*}

But, I'm not sure how can I continue. Can you help me please? I would really appreciate your help!

• So far you haven't used the assumption that $T$ is symmetric. Try computing $\langle v_i, Tv_j \rangle$ in two different ways. Oct 6, 2020 at 18:19

Since we have a real inner product, the symmetry of $$T$$ wrt the inner product means $$\langle x,Ty\rangle = \langle Tx,y\rangle,\quad \forall x,y \in V$$ If $$T$$ has the set $$\{ v_1,\cdots,v_n \}$$ of eigenvestors to the distinct eigenvalues $$\{ c_1,\cdots,c_n \}$$, then we have $$c_j\langle v_i,v_j\rangle = \langle v_i,Tv_j\rangle = \langle Tv_i,v_j\rangle = c_i\langle v_i,v_j\rangle\\[10pt] (c_i-c_j)\langle v_i,v_j\rangle = 0 \quad \implies \quad \langle v_i,v_j\rangle = 0,\quad i\ne j$$ That means the eigenvectors to different eigenvalues are orthogonal. Even if one of the eigenvalues is zero the other in the equation cannot be (since $$\{ c_1,\cdots,c_n \}$$ are distinct). It remains to orthonormalise them.
Hint: Because $$T$$ is symmetric (self-adjoint), $$\langle Tv_1,v_2 \rangle = \langle v_1,Tv_2 \rangle$$.
We can assume $$c_1. Take $$i\ne j$$ and consider $$\langle v_i,v_j\rangle$$. Suppose, by contradiction that it is not zero.
One of the numbers $$c_i, c_j$$ is not zero. WLOG assume that $$c_j\ne 0$$. Then
$$0\ne c_j\langle v_i,v_j\rangle=\langle v_i,c_jv_j\rangle=\langle v_i,Av_j\rangle=\langle A^Tv_i, v_j\rangle=\langle Av_i,v_j\rangle=c_i\langle v_i,v_j\rangle.$$ Therefore $$c_i=c_j$$ (dividing both sides of $$c_j\langle v_i, v_j\rangle= c_i\langle v_i, v_j\rangle$$ by the same nonzero $$\langle v_i,v_j\rangle$$), a contradiction.
Hence $$\langle v_i,v_j\rangle=0$$.