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Prove that a diagonal matrix is positive definite if and only if all its diagonal entries are positive. Also write down and identify its associated inner product. (I can't use eigenvalues to prove this because I haven't learned them yet).

My attempt:

In order to be positive definite, matrix K must be symmetric and satisfy positivity. Since we have a diagonal matrix and all its diagonal entries are positive its determinant will be positive as well as its leading coefficient, but how can I show all this information formally using a proof?

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1 Answer 1

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If $K$ is diagonal, it is of the form $K=\mathrm{diag}(k_1,k_2,\ldots,k_n)$ for some real $k_i$. Let $x=(x_i)$ be a real $n\times 1$ vector; then

$$x^TKx = \sum_{1\leq i\leq n} k_i x_i^2$$

is the inner product associated with $K$.

Since $K$ is positive definite, we have that $x^TKx>0$ for all nonzero $x$. Suppose that some diagonal entry $k_{d}\leq 0$. Then if we take an $x$ such that $x_d=1$ and other components of $x$ are zero, then $x^TKx = k_{d}x_d^2 <0$. This contradicts the fact that $K$ is positive definite.

The converse should be fairly easy and proceed along similar lines.

Note:$K$ need not be symmetric to be positive definite.

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Hmm, so I should try and prove this question by contradiction? – diimension Oct 12 '12 at 1:36
Well, that's one possible route – Ganesh Oct 12 '12 at 1:36
How did you know proving by contradiction was the easy route? – diimension Oct 12 '12 at 1:37
I have studied this proof! – Ganesh Oct 12 '12 at 1:37
haha :). This is my first semester learning proofs and I am very raw it at. Just wanted to know .. Thank you very much though! – diimension Oct 12 '12 at 1:38

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