I have the matrix:

$$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{pmatrix} $$

I have to find the kernel, nullity, range and rank for it.

Since the matrix is underdetermined (the determinant is 0), I am not really sure for the answer for the problem.

1) Is the kernel empty in this case?

2) Is the nullity infinite? Since nullity is just the dimension of the nullspace of the matrix and we have infinite solutions for the matrix above.

3) Is the range 3 since the column space of the matrix is 3?

4) Is the rank 0 since there are no linearly independent vectors? (they are all dependent)?

Thank you

  • $\begingroup$ It is ok to include your working even when you are not so sure about your answer. $\endgroup$ Mar 18 '18 at 3:05
  • $\begingroup$ @SiongThyeGoh I have explained my arguments for the last 3 points, but I totally have no idea for the first one. Empty doesn't make sense to me because there are infinite solutions. 0 makes sense to me because it is a valid value, but not the only one, and infinite(if this makes sense), but I don't know how to write that. $\endgroup$
    – zeeks
    Mar 18 '18 at 3:09


  • the zero vector is in the kernel, the kernel is not empty.

  • Note that a square matrix has full rank if and only if it is invertible.

  • $$\operatorname{rank}(A)+\operatorname{nullity}(A)=n=3$$

  • If the rank is zero, the matrix is the zero matrix, is it the case here?

  • If every row are multiples of each other, then the rank is $1$, does it happen here?

  • From all the information above, you should be able to deduce the rank and the nullity.

  • The range is the column space, it is not a number. It is the span of the columns, try to find a basis for the column space.

  • You might like to perform elimentary row operations to reduce the matrix to RREF.

  • Alternatively, by observation, we note that $2$ times the second column, subtract the first column gives you the third column.

  • $\begingroup$ so ker(A) = {0}, nullity(A) = 1, range(A) = {(100),(210),(320)} and rank(A)=2. (the vectors in the range are vertical). Are these correct? $\endgroup$
    – zeeks
    Mar 18 '18 at 3:46
  • 1
    $\begingroup$ kernel of $A$ should be a span of a non-zero vector in this case (since the nullity is $1$). nullity is correct. rank is correct. range should be a span of two non-zero vector (since the rank if $2$). The current solution is not right as it implies that the third coordinate must be $0$. $\endgroup$ Mar 18 '18 at 3:47

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