1
$\begingroup$

To find: rank $A$ and nullity $A$ for $$A=\begin{pmatrix} 0 &0 &0 \\ 0 & 0.5&-0.5 \\ 0&-0.5 & 0.5 \end{pmatrix}$$ I know the nullity refers to the number of free variables in the matrix and the rank refers to the $dim(columnspace)$; where to from here?

$\endgroup$
1
  • 1
    $\begingroup$ Hint: Consider $A\mathbf x= \mathbf0$. Then $\mathbf x$ is in the nullspace. How many vectors does it take to span this nullspace? The number of vectors is the nullity, i.e. it's the dimension of the nullspace. $\endgroup$
    – jdods
    Jul 26, 2016 at 3:03

2 Answers 2

8
$\begingroup$

More Generally. First you are going to want to set this matrix up as an Augmented Matrix where $Ax=0$.

$1)$ To find the rank, simply put the Matrix in REF or RREF

$\left[\begin{array}{ccc|c} 0 & 0 & 0 &0 \\ 0 & 0.5 & -0.5 & 0 \\ 0 & -0.5 & 0.5 & 0 \end{array}\right] \longrightarrow RREF \longrightarrow \left[\begin{array}{ccc|c} 0 & 0 & 0 &0\\ 0 & 0.5 & -0.5 & 0\\ 0 & 0 & 0 & 0 \end{array}\right] $

Seeing that we only have one leading variable we can now say that the rank is 1.

$2)$ To find nullity of the matrix simply subtract the rank of our Matrix from the total number of columns.

So:

Null (A)=3 - 1=2

Hope this is helpful.

$\endgroup$
3
$\begingroup$

The matrix has one linearly independent row (take the negative of the second to get the third) implying that the rank is 1 and the nullity is 2.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .