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

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.

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 $$i$$.

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.