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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!

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3 Answers 3

up vote 3 down vote accepted

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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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).

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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?

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

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