Rank–nullity theorem for linear maps

Linear maps: Linear maps

Rank–nullity theorem for linear maps

An explicit description of $\ker{L}$ can be found by solving a homogeneous system of linear equations. This is not as simple as for $\im{L}$ . Still, the dimension of $\ker{L}$ can be found quickly thanks to the following key result:

If $L :V\rightarrow W$ is a linear map with $\dim{V}\lt \infty$ , then
$\dim{V}=\dim{\ker{L}}+\dim{\im{L}}$

Write $n=\dim{V}$ and $p=\dim{\ker{L}}$ . Because $\ker{L}\subseteq V$ , we have $p\leq n$ .

Choose a basis $\basis{\vec{a}_1,\ldots ,\vec{a}_p}$ for $\ker{L}$ and extend it with $\vec{b}_{p+1},\ldots ,\vec{b}_n$ to a basis for $V$ :
$V=\linspan {\vec{a}_1,\ldots ,\vec{a}_p,\ \vec{b}_{p+1},\ldots ,\vec{b}_n}$ Because $L \vec{a}_1=\cdots= L \vec{a}_p=\vec{0}$ , it follows from Image as a spanned subspace that
$\begin{array}{rcl} \im{L}&=&\linspan{ L \vec{a}_1,\ldots , L \vec{a}_p, L \vec{b}_{p+1},\ldots , L \vec{b}_n} \\ &=&\linspan{ L \vec{b}_{p+1},\ldots , L \vec{b}_n} \end{array}$ If we can prove that the vectors $L \vec{b}_{p+1},\ldots , L \vec{b}_n$ are linearly independent, then we are done, because then $\dim{\im{L}}=n-p$ , so
$n=\dim{V}=p+(n-p)=\dim{\ker{L}}+\dim{\im{L}}$ In order to demonstrate the independence, we seek solutions for the unknowns $\alpha_{p+1},\ldots,\alpha_n$ in the equation
$\alpha_{p+1}\cdot L \vec{b}_{p+1}+\cdots +\alpha_n\cdot L \vec{b}_n=\vec{0}$ We reduce the equation to an equation without $L$ :
$\begin{array}{rcl} L (\alpha_{p+1}\vec{b}_{p+1}+\cdots +\alpha_n\vec{b}_n)&=&\vec{0} \\ \phantom{x}\color{blue}{\text{linearity of }L}&&\\ \alpha_{p+1}\vec{b}_{p+1}+\cdots +\alpha_n\vec{b}_n&\in&{\ker{L}} \\ \phantom{x}\color{blue}{\text{definition of }\ker{L}}&&\\ \alpha_{p+1}\vec{b}_{p+1}+\cdots +\alpha_n\vec{b}_n&=&\alpha_1\vec{a}_1+\cdots+\alpha_p\vec{a}_p\\ &&\text{for certain numbers}\\ && \alpha_1,\ldots ,\alpha_p\\ \phantom{x}\color{blue}{\basis{\vec{a}_1,\ldots,\vec{a}_p}\text{ is a basis of }\ker{L}}&&\\ -\alpha_1\vec{a}_1-\cdots -\alpha_p\vec{a}_p+\alpha_{p+1}\vec{b}_{p+1}+\cdots +\alpha_n\vec{b}_n&=&\vec{0}\\\phantom{x}\color{blue}{\text{all terms carried to the left }}&&\\\end{array}$ Because $\basis{\vec{a}_1,\ldots ,\vec{a}_p,\vec{b}_{p+1},\ldots ,\vec{b}_n}$ is a basis, it follows that the only solution is $\alpha_1=\cdots =\alpha_p=\alpha_{p+1}=\cdots =\alpha_n=0$ Therefore, according to the Dependence criterion, the system $L \vec{b}_{p+1},\ldots , L \vec{b}_n$ is linearly independent.

The image space of the perpendicular projection $P_U$ onto a subspace $U$ of the inner product space $\mathbb{R}^n$ is, of course, equal to $U$ , while the null space consists of all vectors which are perpendicular to $U$ , so it is equal to $U^{\perp}$ . The above Rank-nullity theorem implies that $n=\dim{U} + \dim{U^{\perp}}$ . This formula can be found in the theorem below.

The word rank in the name Rank-nullity theorem refers to the dimension of the image of $L$ . Later we will see that the rank as we know it for a matrix corresponding to $L$ is the same as this dimension.

The Rank-nullity theorem implies statements on dimensions of some spaces featuring no linear maps.

We recall that, if $U$ and $W$ are linear subspaces of a vector space $V$ , the following two subsets of $V$ are also linear subspaces of $V$ :

The span $\linspan{U\cup W}$ of $U$ and $W$ . Because each element of it can be written as $\vec{u}+\vec{w}$ for certain $\vec{u}\in U$ and $\vec{w}\in W$ , we also denote this linear subspace by $U+W$ . It contains both $U$ and $W$ (and is the smallest subspace with this property).
The intersection $U\cap W$ of $U$ and $W$ . This subspace is contained in both $U$ and $W$ (and is the greatest with this property).

Dimension theorem for subspaces and orthoplements

Let $U$ and $W$ be linear subspaces of $\mathbb{R}^n$ and denote by $U^\perp$ the subspace of $\mathbb{R}^n$ consisting of all vectors perpendicular to each vector of $U$ (with respect to the standard inner product on $\mathbb{R}^n$ ). The following two equalities between dimensions hold: $\begin{array}{rcl}n &=& \dim{U}+\dim{U^\perp}\\ \dim{U}+\dim{W} &=& \dim{U\cap W} + \dim{U+W}\end{array}$

Proof of $n = \dim{U}+\dim{U^\perp}$ : Let $P_U$ be the orthogonal projection from $\mathbb{R}^n$ onto $U$ .

The kernel of $P_U$ consists of the orthoplement $U^\perp$ of $U$ in $\mathbb{R}^n$ and the image is $U$ . Thus, it follows from the Rank-nullity theorem that $n = \dim{\mathbb{R}^n}=\dim{\ker{P_U}}+\dim{\im{P_U}}=\dim{U^\perp}+\dim{U}$

Proof of $\dim{U}+\dim{W} = \dim{U\cap W} + \dim{U+W}$ : We will use the statement just proved a number of times. We also apply the statement to a linear subspace $W$ of $\mathbb{R}^n$ in the following version:

$\dim{W} = \dim{W\cap U} + \dim{W\cap U^\perp}$

The proof of this runs just like the proof above with the restriction of $P_U$ to $W$ replacing $P_U$ itself; the image of this linear transformation is the same as $U\cap W$ and the kernel is $U^\perp\cap W$ .

We now derive the equality that needs to be proven:

$\begin{array}{rcl} \dim{U}+\dim{W} &=&\dim{U\cap W}+\dim{U\cap W^\perp}+\dim{W}\\&&\phantom{x}\color{blue}{\dim{U} = \dim{U\cap W} + \dim{U\cap W^\perp}}\\ &=&\dim{U\cap W}+\dim{W^\perp}-\dim{U^\perp\cap W^\perp}+\dim{W}\\&&\phantom{x}\color{blue}{\dim{W^\perp} = \dim{W^\perp\cap U} + \dim{ W^\perp\cap U^\perp}}\\ &=&\dim{U\cap W}+\dim{ W^\perp}-\dim{(U+W)^\perp}+\dim{W}\\&&\phantom{x}\color{blue}{ U^\perp\cap W^\perp =\left(U+ W\right)^\perp}\\ &=&\dim{U\cap W}+n-\dim{ W}-n+\dim{U+ W}+\dim{W}\\ &&\phantom{x}\color{blue}{\dim{X^\perp} + \dim{X}=n}\\ &=&\dim{U\cap W}+\dim{U+W}\\ \end{array}$

Suppose the linear map $L:\mathbb{R}^{4}\to\mathbb{R}^{3}$ is surjective.

What is the dimension of the kernel of $L$ ?

$\dim{\ker{L}}=$ $1$

According to the rank-nullity theorem we have
$\begin{array}{rcl} \dim{\ker{L}}&=&\dim{\mathbb{R}^{4}}-\dim{\im{L}}\\ &&\phantom{xx}\color{blue}{\text{rank-nullity theorem}}\\ &=& 4- 3\\ &&\phantom{xx}\color{blue}{L\text{ is surjective}}\\ &=& 1 \end{array}$

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