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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!

3 Answers 3

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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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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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    yes that is also true...2012-06-11