The nullity of a square matrix with linearly dependent rows is at least one. True or False?

Here is the answer my textbook gives:

True; if the rows are linearly dependent, then the rank is at least $1$ less than the number of rows, so since the matrix is square, its nullity is at least $1$.

I wonder: Why does the author ask this question specifically for square matrices? Is it different for other matrices.


If the matrix is not square, then the matrix is non -invertible, and so the nullity is at least one.

More details: If your matrix is $m$ x $n$, and say $m>n$. We know that the rank of a matrix is equal to the row rank, which is equal to the column rank. So, rank of the matrix is at most $n$- the number of columns, which is less than $m$, and hence the nullity is at least $1$. Similar arguments apply for $m<n$.

  • $\begingroup$ Hmm. But the answer already says that the nullity it at least one. So why not generalize to all matrices? $\endgroup$ – yolo123 Aug 30 '14 at 2:42
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    $\begingroup$ @yolo123: for a square matrix, the nullity need not be $1$. You are given an additional condition, namely that the rows are linearly dependent. For non square matrices you do not need that condition- they would automatically have nullity at least one. $\endgroup$ – voldemort Aug 30 '14 at 2:44
  • $\begingroup$ But is rank(A)+nullity(A)=n? If rank(A)=n, then nullity(A)=0? no? $\endgroup$ – yolo123 Aug 30 '14 at 2:52
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    $\begingroup$ @yolo123: yes: But for any matrix, rank $\leq$ max {number of rows, number of columns}, and thus it can never be full rank. The nullity has to be at least $1$. A non square matrix can never be invertible for similar reasons. $\endgroup$ – voldemort Aug 30 '14 at 2:56

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