If $A \in \mathcal{M}_n$ is a strictly diagonally dominant matrix, then every principal minor is non-zero. 
The question states the following: Prove that if $A \in \mathcal{M}_n$ is a strictly diagonally dominant matrix, then every principal minor is non-zero.

My attempt
In order to prove this result, I have thought about using reduction to absurdity. Let's denote \begin{equation}\delta_k =\begin{vmatrix}
a_{11} & \ldots & a_{1k} \\\ \vdots & \ddots & \vdots \\\ a_{k1} & \ldots & a_{kk}
\end{vmatrix}\end{equation}
the principal minor of order $k$. Let's suppose that indeed there exists a $k$ such that $\delta_k = 0$.
I first notice that $k \neq n$ because the fact of $A$ being strictly diagonally dominant matrix implies being not singular. Then $k \in \lbrace 0,\ldots,n-1\rbrace$.
Using Laplace expansion of the determinant of a matrix:
$$0 = \sum_{j = 1}^k (-1)^{i+j} \cdot a_{ij} \cdot \beta_{ij}$$
Where $i$ is any row, $j$ represents the columns and $\beta_{ij}$ is the determinant of the submatrix when removing row $i$ and column $j$. Therefore:
$$0 = \left| \sum_{j = 1}^k (-1)^{i+j} \cdot a_{ij} \cdot \beta_{ij} \right| \leq \sum_{j = 1}^k   |a_{ij}| \cdot |\beta_{ij}| = |a_{ii}||\beta_{ii}| + \sum_{j \neq i}^k   |a_{ij}| |\beta_{ij}|$$
Denoting $\beta_{i\lambda} = \displaystyle \max_{1 \leq j \leq k} |\beta_{ij}|$ then
$$0 \leq |a_{ii}||\beta_{ii}| + \sum_{j \neq i}^k   |a_{ij}||\beta_{ij}| \leq |a_{ii}||\beta_{i\lambda}| + \sum_{j \neq i}^k   |a_{ij}||\beta_{i\lambda}|$$
Dividing by $\beta_{i\lambda}$ we get
$$0  \leq |a_{ii}| + \sum_{j \neq i}^k   |a_{ij}| \Rightarrow  |a_{ii}| \leq \sum_{j \neq i}^k   |a_{ij}|$$
So we have a contradiction.
However, I'm not sure about the proof, since I am dividing by $\beta_{i\lambda}$, which could be zero. I don't know where is the mistake, or I may have done an incorrect approach. What happens when  $\beta_{i\lambda}$ =0? Any help?
 A: The principal submatrices of a matrix that is strictly diagonally dominant are automatically strictly diagonally dominant. Hence every principal minor is nonzero.
EDIT: Suppose that $A \in \mathbb{R}^{n \times n}$ is strictly diagonally dominant by rows. Then $|a_{ii}| \geq \sum_{j \not = i} |a_{ij}|$ for all $i$. Let $A_k$ denote the principal submatrix of $A$ obtained by deleting the $k$ row and column of $A$. Then $A_k$ is strictlly diagonally dominant by rows because $$|a_{ii}| \geq \sum_{j \not = i} |a_{ij}| \geq \sum_{j \not = \{i,k\} } |a_{ij}|$$ for all $i \not = k$.
Remark: One could try to map the entries of $A_k$ to the entries of $A$, but in my opinion this adds complications without any tangible benefit. Depending on the intended reader I would lead with an example that is small enough to be understood quickly and large enough to exhibit the general pattern. Example: If $A$ is given by $$ A = \begin{bmatrix} -5 & 3 & 1 \\ 2 & 6 & 3 \\ -1 & 2 & 4\end{bmatrix}, $$ then $A_2$ is given by $$ A_2 = \begin{bmatrix} -5 & 1 \\ -1 & 4 \end{bmatrix}.$$ Here the proof reduces to the observations that $5 > 4$ implies $5 > 1$ and $4 > 3$ implies $4 > 1$.
EDIT: As a general rule, I prefer to complete the original proof rather than replace it with another approach. In this case, there is no choice. The matrix $A$ given by
$$ A = \begin{bmatrix} 4 & 2 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 5 \end{bmatrix}$$
is strictly diagonally dominant by rows, but if we delete column 1 and row 3, we are left with the submatrix $$\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$$
which is singular.
