I'm not mathematically trained. A module I'm taking this semester needs me to: Show that a square matrix with only diagonal values that are all positive is a positive definite matrix.
What is the most layman and simplistic way of answering this question? Thank you in advance!
On
Start writing down the defitions: A symmetric matrix is positive definite iff we have for all $x \in \def\R{\mathbf R}\R^n$, $x \ne 0$ that $$ x^tAx = \sum_{i,j=1}^n a_{ij}x_ix_j > 0 $$ If now $A$ is diagonal, that is $a_{ij} = 0$ for $i \ne j$, the sum simplifies (the off-diagonal terms vanish): $$ x^tAx = \sum_{i=1}^n a_{ii} x_ix_i = \sum_{i=1}^n a_{ii}x_i^2 $$ Now note, that for $x \ne 0$, there is some coordinate, $x_k$ say, with $x_k > 0$. As all $a_{ii}$ are positive by assumption, we have $$ x^tA x = \sum_{i=1}^n a_{ii}x_i^2 \ge a_{kk}x_k^2 > 0 $$ The $\ge$ sign holds as $a_{ii}x_i^2 \ge 0$ for all $i$, the $>$-sign as $a_{kk}>0$ and $x_k^2 > 0$. So $x^tAx > 0$ for $x \ne 0$ and $A$ is positive.
Starting from definition:
A symmetric matrix $A$ is positive defined if for all non zero vectors $x$ we have $x^T A x >0$.
Let: $$ A= \begin{bmatrix} a_1&0&0&\cdots &0\\ 0&a_2&0&\cdots &0\\ \cdots\\ 0&0&0&\cdots& a_n \end{bmatrix} $$ be a diagonal matrix with $a_i>0$ than, for a vector $x=[x_1,x_2,\cdots,x_n]^T$, we have:
$$ x^TAx= \begin{bmatrix} x_1&x_2&\cdots &x_n\\ \end{bmatrix} \begin{bmatrix} a_1&0&0&\cdots &0\\ 0&a_2&0&\cdots &0\\ \cdots\\ 0&0&0&\cdots& a_n \end{bmatrix} \begin{bmatrix} x_1\\x_2\\\cdots \\x_n\\ \end{bmatrix}= $$ $$=\begin{bmatrix} a_1x_1&a_2x_2&\cdots &a_nx_n\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2\\\cdots \\x_n\\ \end{bmatrix}= a_1x_1^2+a_2x_2^2+\cdots +a_nx_n^2 $$ that, since $a_i>0$, is a sum of squares, so it is always positive.