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

What is the dimension of the vector space of all symmetric matrices of order $n\times n$ $(n\geq 2)$ with real entries and trace equal to zero?

share|cite|improve this question
up vote 3 down vote accepted

Think of how you normally calculate the dimension of the space of symmetric matrices. The only difference comes in choosing elements on the diagonal: if I choose the first $n -1$ diagonal entries, then is there a choice for the last diagonal entry which will make the trace zero? How many choices for this last entry are there?

share|cite|improve this answer

You can choose arbitrary values for the upper triangular values, that gives $\frac{n(n-1)}{2}$ degrees of freedom, and you can choose arbitrary values for $n-1$ of the diagonal elements, so the dimension is $\frac{n(n-1)}{2}+(n-1)$.

share|cite|improve this answer
dear sir what if matrix have complex entries? – srijan May 8 '12 at 16:37
@srijan: Same answer. The choices have nothing to do with the field in question. – copper.hat May 8 '12 at 16:44

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.