Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to prove the above statement but not able to find an elegant way.... any thoughts would be appreciated.

share|cite|improve this question
yes 2x2 can be shown as sum of squared expressions with positive coefficients – dasman Feb 14 '13 at 4:11
What if two variables are perfectly correlated? – hardmath Feb 14 '13 at 4:26

That statement is not true in general.

For example,

$$\begin{pmatrix} 1 & 0.9 & 0.2 \\ 0.9 & 1 & 0.7 \\ 0.2 & 0.7 & 1 \end{pmatrix} $$

share|cite|improve this answer
A correlation matrix must be positive semidefinite. Your example is not. – user1551 Feb 14 '13 at 9:09

You cannot. The statement is false. The most obvious example is $$ \begin{pmatrix}1&1\\1&1\end{pmatrix}, $$ where the random variables are perfectly correlated. For a less obvious example, take any $n\times k$ random matrix $X$ such that (a) $n\ge k$, (b) all row sums of $X$ are zero and (c) in each column, all entries have the same sign. Then the covariance matrix $XX^T$ has deficient rank but it almost always have all positive entries. And so does the correlation matrix. For a concrete example, consider $$ X=\begin{pmatrix} 1& 0&-1& 0\\ 1& 1& 0&-2\\ 0& 2&-1&-1\\ 1& 1&-2& 0\\ 1& 1&-1&-1 \end{pmatrix}, \ XX^T=\begin{pmatrix} 2&1&1&3&2\\ 1&6&4&2&4\\ 1&4&6&4&4\\ 3&2&4&6&4\\ 2&4&4&4&4 \end{pmatrix}. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.