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

Question

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!

Answer 1

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.

Answer 2

Hint: Because $T$ is symmetric (self-adjoint), $\langle Tv_1,v_2 \rangle = \langle v_1,Tv_2 \rangle$.

Answer 3

We can assume $c_1<c_2<...<c_n$. 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 $\langle v_i,v_j\rangle=0$.