A Problem about Quadratic Forms and Eigenvalues In one of my linear algebra lectures, the lecturer gave a theorem:  
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a quadratic form, with matrix $A$. Then $f$ is positive definite if and only if all eigenvalues of $A$ are positive.  
Here is her proof of $(\Rightarrow)$: 
She tried to prove the contrapositive of $(\Rightarrow)$. Suppose one of the eigenvalues of $A$ is not positive, since $A$ is symmetric, then $A$ is orthogonally diagonalisable, i.e., $\exists$ orthogonal matrix $Q$ such that $Q^TAQ=D$, where $D$ is a diagonal matrix of which the diagonal entries are eigenvalues of $A$, $\lambda_1,\lambda_2,\dots,\lambda_n$. By the Principal Axes Theorem, 
$$
f({\bf{x}})={\bf{x}}^TA{\bf{x}}=(Q{\bf{y}})^TA(Q{\bf{y}})={\bf{y}}^TD{\bf{y}}=\lambda_1y_1^2+\lambda_2y_2^2+\cdots+\lambda_ny_n^2, 
$$
where 
$$
{\bf{y}}=(y_1y_2\dots y_n)^T=Q^T{\bf{x}}=Q^T(x_1x_2\dots x_n)^T.
$$
Let ${\bf{y}}={\bf{e}}_i=(00\dots010\dots0)^T$, a unit vector of which the $i^{\text{th}}$ coordinate is $1$. Then ${\bf{x}}=Q{\bf{y}}$ and we get 
$$
f({\bf{x}})=\lambda_i\leqslant0. 
$$
Then $f$ is not positive definite.  
Is this proof right? Could somebody tell me if her proof is appropriate? The proof of contrapositive should be without any examples I suppose?
 A: As written, this is not a proof by contradiction. It is a proof of the contrapositive.
The form of the argument is "If not $B$ implies not $A$." Since this is logically equivalent to "If $A$ then $B$", that has been proven also.
It could be rewritten as an argument by contradiction. One would say "Suppose all eigenvalues are positive but also $x^\top Qx\leq 0$ for some $x$." Then one can rewrite $x$ as a linear combination of orthogonal eigenvectors and show that one of the eigenvalues is nonpositive, causing a contradiction.
Edit: The poster has since rewritten the question and made this part useless.

OK, let me try to address what you're saying about "examples."  The argument is of the form "If for all eigenvalues of $Q$...something happens, then for all vectors, something else happens." The proof by contrapositive would then go this way: 


*

*OK, assume there is a vector such that something else doesn't happen.

*(insert valid steps of reasoning)

*Aha, I produced an eigenvalue for which the first something doesn't happen.


You can see the negations of the original universal quantifiers become existential quantifiers in the negation. 
That is, the contrapositive of "$\forall \lambda(\phi(\lambda))\implies \forall x(\psi (x))$" turns into
$\exists x(\neg\psi(x))\implies \exists \lambda(\neg\phi(\lambda))$
My guess is that you are saying that using these existential quantifiers to assert the existence of "an example" makes you uncomfortable. But it shouldn't... this is a perfectly natural use of logic :)
A: The proof is OK, but to be honest, I would prove the direct statement rather than the contrapositive. Actually, with no extra effort that proof shows that 
$$\tag{1}
\lambda_{\mathrm{min}}\sum_{j=1}^n x_j^2\le f(x)\le \lambda_{\mathrm{max}}\sum_{j=1}^n x_j^2,\quad \forall x=(x_1\ldots x_n)\in \mathbb{R}^n.$$
Here $\lambda_{\mathrm{min}}$ and $\lambda_{\mathrm{max}}$ are, respectively, the minimum and maximum eigenvalue of $A$. So if $\lambda_{\mathrm{min}}>0$, then $f$ is positive definite. Moreover, the inequalities are sharp, meaning that there exist vectors $x_{\mathrm{min}}$ and $x_{\mathrm{max}}$ such that 
$$\tag{2}
\lambda_{\mathrm{min}}\sum_{j=1}^n x_j^2=f(x_{\mathrm{min}})
$$
and
$$\tag{3}
f(x_{\mathrm{max}})=\lambda_{\mathrm{max}}\sum_{j=1}^n x_j^2,
$$
precisely, one must take eigenvectors corresponding to $\lambda_{\mathrm{min}}$ and $\lambda_{\mathrm{max}}$ respectively. Therefore, if $f$ is positive definite, then $\lambda_{\mathrm{min}}>0$, because the right hand side in (2) is positive by assumption.
