How to show that a symmetric matrix that is strictly diagonal dominant (sdd) is is positive and definite?

Question

Let $A\in M_n(\mathbb{R})$ be a real symmetric matrix that is strictly diagonal dominant (sdd) with positive diagonal values, that is, $$\forall i, a_{ii} = |a_{ii}|>\sum_{j\neq i}|a_{ij}|.$$ I know that a sdd matrix is invertible and that if $\lambda_i$ is an eigenvalue of $A$ then we have that $$\forall i,|\lambda_i-a_{ii}|\leq \sum_{j\neq i}|a_{ij}|.$$

I guess, that we could simply diagonalize $A = PDP^{t}$ with $D= \text{diag}(\lambda_1,\lambda_2,\cdots,\lambda_n).$ Then if we can show that $A$ is strictly diagonal dominant (with positive diagonal values) iff $D$ is strictly diagonal dominant (with positive diagonal values) then we are done. I am not sure how to proceed with this part of the proof.

Answer 1

You can do it explicitly: let $ 0 \neq x= (x_1,\ldots, x_n)^t \in \mathbb R^n$, and consider \begin{align*} x^t A x &= \sum^n_{i=1}\sum^n_{j=1} a_{ij}x_ix_j \\ &= \sum^n_{i=1} x_i^2a_{ii} + 2 \sum_{i=1}^n \sum^{n}_{j=i+1} x_i x_j a_{ij}\\ &\ge \sum^n_{i=1} x_{i}^2a_{ii} - \sum_{i=1}^n \sum^{n}_{j=i+1} (x_i^2 + x_j^2) \lvert a_{ij} \rvert \\ &= \sum_{i=1}^n \left[\left(a_{ii} - \sum_{i\neq j} \lvert a_{ij}\rvert \right)x_i^2\right] > 0, \end{align*} since $\left(a_{ii} - \sum_{i\neq j} \lvert a_{ij}\rvert \right) > 0$ for all $i$, and $x_i^2>0$ for at least one $i$.

Answer 2

From your observations, $$ \forall i, |\lambda_i - a_{ii}| < a_{ii} $$ with $a_{ii}>0$.

We need to show that $\lambda_i > 0$.

if $\lambda_i \leq 0$, i.e, $\lambda_i = -|\lambda_i|$, $$ |\lambda_i - a_{ii}| = |-1 \times (|\lambda_i| + a_{ii})| = |\lambda_i| + a_{ii} $$

Now,

$$ |\lambda_i| + a_{ii} \geq a_{ii} $$

Therefore, $\lambda_i>0$ by contradiction.