# strictly diagonally dominant matrices in numerical

I am trying to find all values for $$\alpha$$ and $$\beta$$ for which

$$A(\alpha, \beta)= \left[ \begin{matrix} 3&0&-2\\\alpha&3&2\\-2&2&\beta \end{matrix} \right]$$

is strictly diagonally dominant.

Here are my workings, I wanted to see if I was using the correct definition for strictly diagonally dominant as this is different then just diagonally dominant.

A $$nxn$$square matrix is strictly diagonally dominant matrix if

$$|a_{ii}| > \sum_{j=1,i\ne j}^{n}|a_{ij}|$$

for $$i = 1,2,...,n$$

So for each row the absolute value of the diagonal element is strictly greater than the sum of the absolute values of the rest of the elements of that row.

My workings:

row $$2$$: is $$|a_{22}|>|a_{21}|+|a_{23}|$$.

$$|a_{22}| = |3| = 3$$
$$|a_{21}|+|a_{23}| = |\alpha| + |2| < 3$$.
Then $$0\leq \alpha < 1$$

row $$3$$: is $$|a_{33}|>|a_{31}|+ |a_{32}|$$
$$|a_{33}| = |\beta|$$
$$|a_{31}|+ |a_{32}| = |-2| + |2| = 4$$
Then $$\beta >4$$ and $$\beta <-4$$

Am I using the definition correctly? Am I evaluating the absolute values correctly?

• Why do you assume $\alpha,\beta\in\mathbb{R}$? Can't they be in $\mathbb{C}$? – user10354138 Sep 30 '18 at 17:33
• Correct, but note that $\alpha \in (-1,1)$ – Villa Sep 30 '18 at 17:33
• @Villa yes you are right, I should have considered the negative value like I did for $\beta$ – jh123 Sep 30 '18 at 17:42

Assuming these are real valued matrices, for row 1, we need $$|\alpha | + |2| < |3|$$ $$\implies |\alpha| + 2< 3$$ $$\implies |\alpha| < 1$$ $$\implies -1 < \alpha < 1$$.
For row 2, you are are correct: $$|-2| + |2| < |\beta | \implies 4 < | \beta | \implies \beta < -4 \text{ or } 4 < \beta$$.
To be clear the matrix $$A(\alpha, \beta )$$ is strictly diagonally dominant if and only if both of these conditions are met: $$\alpha \in (-1, 1)$$ and $$\beta \in (-\infty, -4) \cup (4, \infty)$$.