When dimension of a subspace is asked, what do we find nullity or rank? 
*

*Consider the following subspace:
$W = \{(x,y,z)\in \mathbb{R}^3\ \mid 2x+2y+z=0, 3x+3y-2z=0, x+y-3z=0\}$.
Find $\dim W$.

*Let $A = 
        \begin{bmatrix}
        1 & 1 & 1 \\
        3 & -1 & 1 \\
        1 & 5 & 3 \\
        \end{bmatrix}$ and $V = \{ X \in \mathbb{R}^3 \mid AX=0 \}$. Then $\dim V$ is?
I don't get whether to find rank or nullity in both of the cases?
I think the answer should be dimension of null space since $AX=0$ is asked in both the cases, i.e. all those vectors which when multiplied with given matrix give $0$, i.e. those vectors will then belong to kernel.
 A: Recall: 


*

*Let $A$ be an $m \times n$ matrix with real entries. The null space of $A$ is,
$$\operatorname{nullspace}(A) = \{X \in \mathbb{R}^n: AX=0\}\tag{1}$$
and the dimension of the null space is $\operatorname{null}(A)$.

*The rank of $A$ is given by, 
$$\dim (\operatorname{rowspace}(A))  = \dim (\operatorname{colspace}(A))= \operatorname{rank}(A)\tag{2}$$
where the row space is the subspace of $\mathbb{R}^n$ spanned by the rows, and the column space is the subspace of $\mathbb{R}^m$ spanned by the columns.

*The Rank-Nullity Theorem states: For any $m \times n$ matrix $A$,
$$\operatorname{rank}(A)+\operatorname{null}(A)=n\tag{3}$$



1) Find the dimension of the homogeneous system of linear equations,
\begin{align*}
2x+2y+z&=0\\
3x+3y-2z&=0\tag{4}\\
x+y-3z&=0
\end{align*}
Let the coefficient matrix be: 
$$C=\left[\begin{array}{ccc}
        2 & 2 & 1 \\
        3 & 3 & -2\\
        1 & 1 & -3 \\\end{array}\right]$$
Now form the Augmented Matrix:
$$C'=\left[\begin{array}{ccc|c}
        2 & 2 & 1 & 0\\
        3 & 3 & -2 & 0\\
        1 & 1 & -3 & 0\\\end{array}\right]$$
This reduces to row-echelon form:
$$C'_r=\left[\begin{array}{ccc|c}
        1 & 1 & 0 & 0\\
        0 & 0 & 1 & 0\\
        0 & 0 & 0 & 0\\\end{array}\right]$$
So the system of equations (4) reduces to $x+y=0$, and $z=0$, allowing us to parametrise the solution set as: 
$$W=\{(-x,x,0):x\in\mathbb{R}\}\tag{5}$$
Thus the dimension of the solution set is $1$, hence $\dim(W)=1$, since its solution set forms a line through the origin in $\mathbb{R}^3$. Here the nullity of $C$ is the dimension of the null space of $C$, which is the same as the dimension of the solution space of $CX = 0$. That is $\operatorname{nullspace}(C)=W$.
Note the rank of $C$ is given by 
$$\dim (\operatorname{rowspace}(C)) = \dim (\operatorname{colspace}(C))=\operatorname{rank}(C)=2$$
and so by (3) we can find, 
$$\operatorname{null}(C)=3-\operatorname{rank}(C)=3-2=1$$
and this is what we want, the dimension of the null space, as defined in (1), as $W$ is exactly the null space of $C$, and so $\dim(W)=1$.



*

*To answer the question in terms of the kernel of a linear transformation, $\theta$: 


Let $\theta\colon  \mathbb{R}^3\rightarrow\mathbb{R}^3$, be defined
   by:
$$\theta\colon 
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        =
        \begin{bmatrix}
        2x+2y+z \\
        3x+3y-2z \\
        x+y-3z \\
        \end{bmatrix}
$$
Note that:
$$\dim(\mathbb{R}^3)=\dim(\ker(\theta))+\dim(\operatorname{im}(\theta))$$
and that:
$$\operatorname{null}(C)=\dim(\ker(\theta))\qquad
\operatorname{rank}(C)=\dim(\operatorname{im}(\theta))$$
where $C$ is the $3\times3$ matrix of the linear transformation $\theta$. The kernel of $\theta$, is given by:
\begin{align*}
\ker(\theta)
&=
\left\{
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        :  2x+2y+z =3x+3y-2z =x+y-3z =0
\right\}\\
&=
\left\{
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        :
        \begin{bmatrix}
        2 & 2 & 1 \\
        3 & 3 & -2 \\
        1 & 1 & -3 \\
        \end{bmatrix}
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        =
        \begin{bmatrix}
        0 \\
        0 \\
        0 \\
        \end{bmatrix}
\right\}\\
&= 
\left\{
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        :
\begin{bmatrix}
        1 & 1 & 0\\
        0 & 0 & 1 \\
        0 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
                =
         \begin{bmatrix}
        0 \\
        0 \\
        0 \\
        \end{bmatrix}
\right\}\\
&=
\left\{
        \begin{bmatrix}
        -x \\
        x \\
        0 \\
        \end{bmatrix}:x\in\mathbb{R}
\right\}
\end{align*}
The dimension of the whole space is $\dim(\mathbb{R}^3)=3$, so by Rank-Nullity:
$$\dim(\mathbb{R}^3)=\operatorname{rank}(C)+\operatorname{null}(C)=2+1=3$$
Since the kernel of a matrix transformation is simply the null space of the matrix, and $W=\ker(\theta)$, then $\dim(W)=\dim(\ker(\theta))=1$.
Note that $W$ is your solution set, it is the thing you have to work out to give (5), it is not a linear transformation itself that you can use The Rank-Nullity Theorem on to find its dimension; rather it is the kernel of some linear transformation whose dimension is seen by the size of the subspace of $\mathbb{R}^3$ it spans, which in this case is a line through the origin isomorphic to $\mathbb{R}^1$.

2) Define $V = \{X \in \mathbb{R}^3 \mid AX=0 \}$. Hence $\dim V$ is the nullity of $A$, or the kernel of the linear transformation $\phi\colon  \mathbb{R}^3\rightarrow\mathbb{R}^3$ defined by,
$$\phi\colon 
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        =
        \begin{bmatrix}
        x+y+z \\
        3x-y+z \\
        x+5y+3z \\
        \end{bmatrix}
$$ 
The kernel of the linear transformation $\phi$, is given by:
\begin{align*}
\ker(\phi)
&=
\left\{
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        :
        \begin{bmatrix}
        1 & 1 & 1 \\
        3 & -1 & 1 \\
        1 & 5 & 3 \\
        \end{bmatrix}
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        =
        \begin{bmatrix}
        0 \\
        0 \\
        0 \\
        \end{bmatrix}
\right\}\\
&= 
\left\{
        \begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
        :
\begin{bmatrix}
        1 & 0 & \tfrac12 \\
        0 & 1 & \tfrac12 \\
        0 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
        x \\
        y \\
        z \\
        \end{bmatrix}
                =
         \begin{bmatrix}
        0 \\
        0 \\
        0 \\
        \end{bmatrix}
\right\}\\
&=
\left\{
        \begin{bmatrix}
        -x \\
        -x \\
        2x \\
        \end{bmatrix}:x\in\mathbb{R}
\right\}
\end{align*}
Where we have reduced $A$ to row-echelon form in the second step. Hence the solution set is $x+\tfrac12 z=0$, $y+\tfrac12 z=0$, so $x=y$, and $2x=-z$. Here the nullity of $A$ is the dimension of null space of $A$, or $\dim(\ker(\phi))=1$, which is the same as the dimension of the solution space of $AX = 0$, which is $\dim(V)=1$, since $V=\ker(\phi)$.
A: Both are similar problems. In first case you can form matrix A, and find rank for both cases. It will give dimension of subspace. Nullity gives number of dimension of the space which is doesn't intersect with your required subspace.
A: Again, to answer the question asked,
 "1) Find the dimension of the homogeneous system of linear equations, 
2x+2y+z= 0
3x+3y−2z= 0
x+y−3z=0"
 I would not think in terms of "matrices" or  "nullity" at all.  I would just solve the given system of equations!  If we multiply the first equation by 2 we get 4x+ 4y+ 2z= 0.  Adding that to the second equation, 3x+ 3y- 2z= 0, eliminates z:  7x+ 7y= 0 which is the same as y= -x.  Setting y= -x in the third equation gives -3z= 0 or z= 0.  But then y= -x, z= 0 satisfies all three equations for all x.  The solutions space of that system of equation consists of vectors of the form (x, y, z)= (x, -x, 0)= x(1, -1, 0) and so has dimension 1.
