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

Let $D$ be a $n\times n$ real matrix, $n\ge 2$. Which of the following is valid?

  1. $\det(D)=0\Rightarrow \mathrm{rank}(D)=0$

  2. $\det(D)=1\Rightarrow \mathrm{rank}(D)\neq 1$

  3. $\det(D)=1\Rightarrow \mathrm{rank}(D)\neq0$

  4. $\det(D)=n\Rightarrow \mathrm{rank}(D)\neq 1$

Well, (1) is wrong because there is a $3\times 3$ matrix with rank $2$ and determinant $0$, namely $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

I am confused about the other three: please help!

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


  1. The following are equivalent
    • A square matrix is invertible
    • A square matrix has full rank
    • A square matrix has non-zero determinant
  2. The rank of the matrix is between $0$ and $n$
  3. More than one may be correct
share|cite|improve this answer

Your answer to #1 is fine!

For #2, #3 and #4, we should make sure you are aware there is a simple fact: a matrix $A$ over a field is invertible iff it has nonzero determinant iff it has full rank (rank $n>1$ in this case).

share|cite|improve this answer

Hint: What do you know about the relation between invertibility (non-singularity) of a given matrix and its determinant? What can you say about the rank of an invertible (or non-singular) matrix?

share|cite|improve this answer
If it has non zero determinant and full rank. – Un Chien Andalou Jun 11 '12 at 11:51
so $3$ is the correct answer. – Un Chien Andalou Jun 11 '12 at 11:52
@Mex: Yes, if it has non-zero determinant, then it has full rank (a fact that has been mentioned twice in the other answers). And yes, #3 is correct, but it is not the only one... – Martin Wanvik Jun 11 '12 at 11:54
also $4$ may be correct. – Un Chien Andalou Jun 11 '12 at 11:55
@Mex: How about #2? (And #4 may be correct?) – Martin Wanvik Jun 11 '12 at 11:56

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.