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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?

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Please use LaTeX formatting (I have a pending edit that formatted it for you). I am not sure what you mean by xj 00. – Emily Aug 8 '12 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. – joriki Aug 8 '12 at 4:22
@EdGorcenski Sorry is xj =0. – franvergara66 Aug 8 '12 at 4:25
up vote 8 down vote accepted

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 positive?

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