# Prove: symmetric positive definite matrix

I'm studying for my exam of linear algebra.. I want to prove the following corollary:

If $A$ is a symmetric positive definite matrix then each entry $a_{ii}> 0$, ie all the elements of the diagonal of the matrix are positive.

My teacher gave a suggestion to consider the unit vector "$e_i$", but I see that is using it.

$a_{ii} >0$ for each $i = 1, 2, \ldots, n$. For any $i$, define $x = (x_j)$ by $x_i =1$ and by $x_j =0$, if $j\neq i$, since $x \neq 0$, then:

$0< x^TAx = a_{ii}$

But my teacher says my proof is ambiguous. How I can use the unit vector $e_1$ for the demonstration?

• Please use LaTeX formatting (I have a pending edit that formatted it for you). I am not sure what you mean by xj 00. Aug 8, 2012 at 4:14
• Your proof seems OK, generally speaking; I suspect that your teacher objected to the presentation. It doesn't make sense to start out with $a_{ii}\gt0$; this is what you want to prove, not something you're starting from. Aug 8, 2012 at 4:22
• @EdGorcenski Sorry is xj =0. Aug 8, 2012 at 4:25

Let $$e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$$, and so on, where $$e_i$$ is a vector of all zeros, except for a $$1$$ in the $$i^{\mathrm{th}}$$ place. Since $$A$$ is positive definite, then $$x^T A x > 0$$ for any non-zero vector $$x \in \Bbb R^n$$. Then, $$e_1^T A e_1 > 0$$, and likewise for $$e_2, e_3$$ and so on.

If the $$i^{\mathrm{th}}$$ diagonal entry of $$A$$ was not positive, $$a_{ii} < 0$$, then $$e_i^T A e_i = 0\cdot a_{11}\cdot 0 + 1\cdot a_{12}\cdot 0 + \cdots + 1\cdot a_{ii}\cdot 1 + \cdots + 0\cdot a_{nn} \cdot 0$$, since $$e_i$$ has zeros everywhere but in the $$i^{\rm th}$$ spot.

Thus, what would happen if $$a_{ii}$$ was negative?

Since $$A$$ is a positive definite $$n \times n$$ matrix, $$x^TAx > 0$$ $$\forall x \in \mathbb{R}^n$$.
Now, every unit vector is of the form $$e_i = \begin{cases} 1 \text{ if } i=j \\ 0 \text{ if } i \neq j \end{cases}$$
Since the above condition for positive definiteness applies for all $$x$$, $$e_i^T A e_i >0$$ $$\implies a_{ii} > 0$$ after matrix multiplication.