[Math] When dimension of a subspace is asked, what do we find nullity or rank

linear algebramatricesmatrix-rankvector-spaces

  1. 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 $\dim W$.

  2. 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.

Best Answer

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 $V=\ker(\phi)$.

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