# Rank and determinant of $D$ , an $n\times n$ real matrix, $n\ge 2$

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 $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$

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HINT

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
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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).
 If it has non zero determinant and full rank. – Taxi Driver Jun 11 '12 at 11:51 so $3$ is the correct answer. – Taxi Driver 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. – Taxi Driver Jun 11 '12 at 11:55 @Mex: How about #2? (And #4 may be correct?) – Martin Wanvik Jun 11 '12 at 11:56