In your mentioned proposition, since $A$ is real symmetric and diagonally dominant and it has a positive diagonal, it is positive semidefinite (Gershgorin discs in play here). The role of irreducibility (coupled with diagonal dominance) is to ensure that $A$ is nonsingular. Every nonsingular positive semidefinite matrix is, of course, positive definite.
To prove that $A$ is nonsingular, suppose $Ax=0$. Since $A$ is irreducible, it has no zero rows. Therefore the diagonal entries of $A$ are nonzero, because $A$ is diagonally dominant. (This fact, of course, also follows from the given assumption that $A$ has a positive diagonal, but here we wish to infer the invertibility of $A$ using only irreducibility and diagonal dominance. So, we deduce the nonzeroness of the diagonal of $A$ from these two assumptions too.)
Now, suppose $x_i$ has the largest modulus among all entries of $x$. By assumption, strict diagonal dominance of $A$ occurs on some row $j$ and there is a path $i_0=i\to i_1\to \ldots\to i_k=j$ such that $a_{i_mi_{m+1}}$ is nonzero for each $m$. Since $|x_{i_0}|$ has a maximum modulus, by (weak) diagonal dominance of row $i_0=i$, for every $\ell$ such that $a_{i_0\ell}\ne0$, we must have $|x_\ell|=|x_{i0}|$. In particular, $|x_{i1}|=|x_{i0}|$. Apply the same argument recursively, we get $|x_j|=|x_{i_k}|=|x_{i_{k-1}}|=\cdots=|x_{i_0}|=|x_i|$. Therefore $x_j$ also has a maximum modulus. But then the strict diagonal dominance of $A$ on row $j$ implies that the $j$-th entry of $Ax$ is nonzero, which is a contradiction. Therefore $A$ must be nonsingular.
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
\left|a_{2,2}-a_{1,2}{a_{2,1}\over a_{1,1}}\right|\ge\sum_{i=3}^n\left|a_{2,i}-a_{1,i}{a_{2,1}\over a_{1,1}}\right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|\ge\sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
\begin{eqnarray*}
\sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&\le&
|a_{1,1}|\sum_{i=3}^n|a_{2,i}|+|a_{2,1}|\sum_{i=3}^n|a_{1,i}| \\
&\le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \\
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\\
&\le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
\end{eqnarray*}
Best Answer
It seems to me that the following computation proves the result.
Let ${{a'}}_{j , k}$ be the coefficients after the first step i.e. for $j > 1$
$${{a'}}_{j , k} = {a}_{j , k}-{a}_{j , 1} \frac{{a}_{1 , k}}{{a}_{1 , 1}}$$
Let us compute
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl}\displaystyle \sum\limits _{\substack{j = 2\\j \neq k}}^{n} \left|{{a'}}_{j , k}\right|&=&\displaystyle \sum\limits _{\substack{j = 2\\j \neq k}}^{n} \left|{a}_{j , k}-{a}_{j , 1} \frac{{a}_{1 , k}}{{a}_{1 , 1}}\right|\\ & \leqslant &\displaystyle \sum\limits _{\substack{j = 2\\j \neq k}}^{n} \left|{a}_{j , k}\right|+\sum\limits _{\substack{j = 2\\j \neq k}}^{n} \left|{a}_{j , 1}\right| \frac{\left|{a}_{1 , k}\right|}{\left|{a}_{1 , 1}\right|}\\ & < &\displaystyle \left|{a}_{k , k}\right|-\left|{a}_{1 , k}\right|+\left|{a}_{1 , k}\right| \sum\limits _{\substack{j = 2\\j \neq k}}^{n} \frac{\left|{a}_{j , 1}\right|}{\left|{a}_{1 , 1}\right|}\\ & \leqslant &\displaystyle \left|{a}_{k , k}\right|-\left|{a}_{1 , k}\right|+\left|{a}_{1 , k}\right| \left(1-\frac{\left|{a}_{k , 1}\right|}{\left|{a}_{1 , 1}\right|}\right)\\ & \leqslant &\displaystyle \left|{a}_{k , k}\right|-\left|{a}_{1 , k}\right| \frac{\left|{a}_{k , 1}\right|}{\left|{a}_{1 , 1}\right|}\\ & \leqslant &\displaystyle \left|{a}_{k , k}-{a}_{1 , k} \frac{{a}_{k , 1}}{{a}_{1 , 1}}\right|\\ & \leqslant &\displaystyle \left|{{a'}}_{k , k}\right| \end{array}$$