Given a matrix $A$, show that it is positive. 
Show that
  $$A := \begin{bmatrix}7 & 2 & -4\\2 & 4 & -2\\-4 & -2 & 7 \end{bmatrix}$$
  is positive definite.

Could this be proven by showing that each of the vectors of the standard basis gives a positive result, e.g.:
$$\begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}7 & 2 & -4\\2 & 4 & -2\\-4 & -2  & 7 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix} > 0.$$
The second part of the question asks me to diagonalize the matrix using an orthogonal matrix, which as I understand, is to use elementary matrices on the rows and columns of the matrix to get it to a diagonal form. Would it make a difference if Ifirstly only dealt with the rows and only afterward used the exact matrices only on the columns?
Thanks for your time.
 A: This article gives several methods to check for positivity.
Checking for positivity on basis vectors wont work. For ex. consider $A = \begin{bmatrix}1 & 5 \\ 4 & 2 \end{bmatrix}$. $x^t A x > 0 $ for $x \in \{e_1,e_2\}$. But, its eigenvalues are $6, -3$ ! 
A: No, checking the standard basis does not guarantee positive definiteness of your scalar product.  It would work if the standard basis was orthogonal with respect to your bilinear form (this is Sylvester's theorem): but, in our case, this is equivalent to have the matrix being diagonal.
By definition, a $n\times n$ matrix is positive definite if its signature is $(n,0,0)$. The first entry in the signature is defined as the number of vectors $v$ in an orthogonal basis (with respect to the form represented by the matrix) such that $\langle v,v\rangle>0$; the second entry is the number of $v$ such that $\langle v,v\rangle<0$; the last is the number of $v$ such that $\langle v,v\rangle=0$. Sylvester theorem guarantees that this definition indeed makes sense: in fact it says that the signature of a matrix is the same, no matter what orthogonal basis we choose.
To diagonalize a form, is the same as to find an orthogonal basis. In fact, when you have an orthogonal basis, the matrix associated to your form with respect to that basis is diagonal.
So, to solve both part of the exercise, you can orthogonalize the standard basis, getting  a new basis, say $\{\,v_1, \dots, v_n\,\}$. Then your diagonal matrix is the matrix that has on the diagonal the values $\langle v_i,v_i\rangle$; and you have positive definiteness if all elements on the diagonal are positive.
A: It's easyiest to just find the eigenvalues of the matrix and show they are positive.
A: I'm assuming we're thinking of $\Bbb R^3$.
If you have $v=\sum \alpha_ib_i$ written in terms of a basis of column vectors $b_i$, then 
$$
v^\top Av=(\sum_i \alpha_ib_i^\top)A(\sum_j \alpha_jb_j)=\sum_{i,j}\alpha_i\alpha_jb_i^\top Ab_j
$$
Clearly the diagonal terms ($\alpha_i^2 b_i^\top A b_i\geq 0$) but you see there is some potential for negatives among the terms with $i\neq j$.
The two main tools are Sylvester's criterion or checking eigenvalues. Sylvester's criterion clears this one up rather quickly.
A: There is a short and nice alternative possibility to show that your matrix is positiv definit (without much calculation):


*

*The determinant of your matrix is $108\neq 0$, hence $0$ is not an eigenvalue.

*This matrix is symmetric, thus all eigenvalues are real

*Now Gershgorin circle theorem gives you that the spectrum of your matrix holds $\sigma(A)\subset \left[0,13\right]$


All together we have: All eigenvalues are positiv!
A: There is another approach that cannot be always used, but is applicable here. Your matrix $A$ is diagonally dominant and has nonnegative diagonal elements, which makes her positive semidefinite (see here). Check the determinant to see that it's nonsingular, i.e., it is positive definite.
Had we been a wee bit more lucky with the second row, i.e., if it was strictly diagonally dominant, we wouldn't even need to check the determinant.
This doesn't apply to all positive (semi)definite matrices, but it's a very easy check to perform before other attempts.
