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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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    $\begingroup$ Please use LaTeX formatting (I have a pending edit that formatted it for you). I am not sure what you mean by xj 00. $\endgroup$
    – Emily
    Aug 8, 2012 at 4:14
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    $\begingroup$ 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. $\endgroup$
    – joriki
    Aug 8, 2012 at 4:22
  • $\begingroup$ @EdGorcenski Sorry is xj =0. $\endgroup$ Aug 8, 2012 at 4:25

2 Answers 2

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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?

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I guess your teacher wanted a more structured answer.

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.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Feb 15 at 12:50

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