MAT223: 11_Range & Nullspace

Range and Nullspace of a Linear Transformation §

Range and Nullspace of a Linear Transformation §

• Range - the range (or image) of a linear transformation $T:V\to W$ is the set of vectors $T$ can output, which is

  $$\text{range}(T)=\left\{\vec{y}\in W:\vec{y}=T\vec{x}\text{ for some }\vec{x}\in V\right\}$$

• The range of a linear transformation is the exact same as the definition of the range of a function

• The range of a linear transformation is always a subspace

• Rank a linear transformation $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$, the rank of $T$, denoted $\text{rank}(T)$, is the dimension of the range of $T$

  • Rank 0 transformation turns all vectors into $\vec{0}$
  • Rank 1 transformation turns all vectors into lines etc…

Let $\mathcal{P}$ be the plane given by $x+y+z=0$, and let $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be projection onto $\mathcal{P}$. Find $\text{range}(T),\text{rank}(T)$ §

First, find $\text{range}(T)$. Since $T$ is a projection onto $\mathcal{P}$, then $\text{range}(T)\subseteq \mathcal{P}$. And $T(\vec{p})=\vec{p}$ for all $\vec{p}\in \mathcal{P}$. Then get $\mathcal{P}\subseteq \text{range}(T)$, which implies that $\text{range}(T)=\mathcal{P}$ Since $\mathcal{P}$ is a plane, then $\text{dim}(\mathcal{P})=2=\text{dim}(\text{range}(T))=\text{rank}(T)$

• Null space (kernel) - of a linear transformation $T:V\to W$ is the set of vectors that get mapped to the zero vector under $T$. That is:
  $$\text{null}(T)=\left\{\vec{x}\in V:T\vec{x}=\vec{0}\right\}$$
• The null space of a linear transformation is always a subspace
• Akin to the rank-range connection, there is a special number called the nullity which specifies the dimension of the null space.
  • Nullity - a linear transformation $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$, the nullity of $T$ denoted $\text{nullity}(T)$ is the dimension of the null space of $T$

  Let $\mathcal{P}$ be the plane given by $x+y+z=0$, and let $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be projection onto $\mathcal{P}$. Find null$(T)$ and nullity$(T)$ §

  Since $T$ is a projection onto $P$, (and because $\mathcal{P}$ passes through $\vec{0}$), then every normal vector for $\mathcal{P}$ will get sent to $\vec{0}$ when $T$ is applied. And besides $\vec{0}$ itself, these are the only vectors that get sent to $\vec{0}$. Then:

  $$\text{null}(T)=\{\text{normal vectors}\cup \{\vec{0}\}\}=\text{span}\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]\right\}$$

  Since null$(T)$ is a line, then nullity$(T)=1$

Fundamental Subspaces of a Matrix §

• Every linear transformation has a range and a null space. Then, every matrix is associated with three fundamental subspaces
• Fundamental Subspaces - associated with any matrix $M$ are 3 fundamental subspaces: the row space, denoted row$(M)$, is the span of the rows of $M$; the column space, denoted col$(M)$, is the span of the columns fo $M$; and the null space, denoted null$(M)$, is the set of solutions to $M\vec{x}=\vec{0}$
  • The columns of $A$ corresponding to pivot columns of rref$(A)$ form a basis of col$(A)$
  • The non-zero rows of rref$(A)$ form a basis for row$(A)$
• To find a basis for the column/row space of matrix $M$
  1. Assume to pick a basis for the span $\{v_1,v_2,\dots \}$ where $v_1,v_2$ are columns/row of the matrix $M$
  2. Row reduce $M$ to get tis row-reducing-echelon-form
  3. Pick the pivot columns which are the original vectors form a maximal linearly independent subset (the independent vectors in this span) as the vectors in the span of column space $M$

Find the null space of $M=\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right]$ §

Need to solve the homogeneous matrix equation of $M\vec{x}=\vec{0}$, first row reduce

$$\text{rref}(M)=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\end{array}\right]$$

See that the $z$ column is a free variable column (let $t$ reps $z$)

$$\begin{gathered} \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=0 \\ \{\begin{array}{c}x+z=0\\ y+2z=0\end{array}\Rightarrow \{\begin{array}{c}x=-z=-t\\ y=-2z=-2t\end{array} \end{gathered}$$

Therefore the complete solution in vector form:

$$\text{null}(M)=\text{span}\left\{\left[\begin{array}{c}-1\\ -2\\ 1\end{array}\right]\right\}$$

For the same $M$ above, find a basis for the row space and the columns of $M$ §

For column space: row reduce columns of $M$ and find its independent subsets

$$\text{rrerf}\left(\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right]\right)=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\end{array}\right]$$

$\therefore$ The first 2 columns are linearly independent, thus

$$\text{col}(M)=\text{span}\left\{\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}2\\ -2\end{array}\right]\right\}={\mathbb{R}}^2\text{and a basis is }\left\{\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}2\\ -2\end{array}\right]\right\}$$

For the row space: row reduce the rows of $M$ and find its independent subsets

$$\text{rrerf}\left(\left[\begin{array}{cc}1& 2\\ 2& -2\\ 5& -2\end{array}\right]\right)=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$$

$\therefore$ This span is already linearly independent, thus

$$\text{row}(M)=\text{span}\left\{\left[\begin{array}{c}1\\ 2\\ 5\end{array}\right],\left[\begin{array}{c}2\\ -2\\ -2\end{array}\right]\right\}\text{and a basis is }\left\{\left[\begin{array}{c}1\\ 2\\ 5\end{array}\right],\left[\begin{array}{c}2\\ -2\\ -2\end{array}\right]\right\}$$

Transpose §

• Transpose, notated $M^T$ is a matrix where the rows and columns of $M$ are swapped

  $$M=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1m}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2m}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nm}\end{array}\right]M^T=\left[\begin{array}{cccc}{a}_{11}& a_{21}& \dots & a_{n1}\\ a_{12}& a_{22}& \dots & a_{n2}\\ a_{13}& a_{23}& \dots & a_{n3}\\ ⋮& ⋮& \ddots & ⋮\\ a_{1m}& a_{2m}& \dots & a_{nm}\end{array}\right]$$

• For a matrix $A$, the dimension of the row space equals the dimension of the column space

  $$\text{col}(M)=\text{row}(M^T)\text{and}\text{row}(M)=\text{col}(M^T)$$

  Every pivot of rref$(A)$ lies in exactly one row and one column. Thus the number of basis vectors in row$(A)$ is the same as the number of basis vectors in col$(A)$

Equations, Null Space, and Geometry §

• Let $A$ be a matrix, $\vec{b}$ be a vector, and let $\vec{p}$ be a particular solution to $A\vec{x}=\vec{b}$. Then the set of all solutions to $A\vec{x}=\vec{b}$ is

  $$\text{null}(A)+\{\vec{p}\}$$

  To right a complete solutions to $A\vec{x}=\vec{b}$, need ① nullspace of $A$ ② particular solution to $A\vec{x}=\vec{b}$

• Let $M$ be a matrix and let ${\vec{r}}_1,\dots ,{\vec{r}}_n$ be the rows of $M$. Then the solutions to $M\vec{x}=\vec{0}$ is precisely the vectors which are orthogonal to every row ($\vec{r}$) of $M$.

  • null$(M)$ consists of all vectors orthogonal to the rows of $M$

  • row$(M)$ consists of all vectors orthogonal to everything in null$(M)$

  Let $\vec{a}=\left[\begin{array}{c}1\\ 2\\ 5\end{array}\right]$, and $\vec{b}=\left[\begin{array}{c}2\\ -2\\ -2\end{array}\right]$. Find all vectors orthogonal to both of them §

  Let $M=\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right]$ be the matrix whose rows are $\vec{a},\vec{b}$. Then, calculate null$(M)$ to get all vectors orthogonal to row$(M)$. Thus the solution is:

  $$\text{null}(M)=\text{span}\left\{\left[\begin{array}{c}-1\\ -2\\ 1\end{array}\right]\right\}$$

Transformations and Matrices §

• Induced Transformations - Let $M$ be an $n\times m$ matrix. $M$ induces a linearly transformation ${\mathcal{T}}_M:{\mathbb{R}}^m\to {\mathbb{R}}^n$ defined by

  $$[{\mathcal{T}}_M\vec{v}{]}_{{\mathcal{E}}^{\prime }}=M[\vec{v}{]}_{\mathcal{E}}$$

  Where $\mathcal{E}$ is the standard bassi for $R^m$ and ${\mathcal{E}}^{\prime }$ is the standard basis for ${\mathbb{R}}^n$

  Let $\mathcal{T}$ be the transformation induced by the matrix $M=\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right]$, and let $\vec{v}=3{\vec{e}}_1-3{\vec{e}}_3$. Computer $\mathcal{T}(\vec{v})$ §

  Using the definition fo induced transformations:

  $$[{\mathcal{T}}_M\vec{v}{]}_{{\mathcal{E}}^{\prime }}=M[\vec{v}{]}_{\mathcal{E}}=\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right][\vec{v}{]}_{\mathcal{E}}$$

  The equation of $\vec{v}$ is given, calculate and substitute the values of $[\vec{v}{]}_{\mathcal{E}}$

  $$[{\mathcal{T}}_M\vec{v}{]}_{{\mathcal{E}}^{\prime }}\left[\begin{array}{ccc}1& 2& 5\\ 2& -2& -2\end{array}\right]\left[\begin{array}{c}3\\ 0\\ -3\end{array}\right]=\left[\begin{array}{c}-12\\ 12\end{array}\right]$$

  In other words $

• Rank of a matrix - let $M$ be a matrix, the rank of $M$, denoted rank$(M)$, is the rank of the linear transformation induced by $M$

• Nullity of a matrix - let $M$ be a matrix, the nullity of $M$, denoted nullity$(M)$, is the nullity of the linear transformation induced by $M$

Range vs Column Space & Null Space vs. Null Space §

• Let $M$ be a matrix and $\mathcal{T}$ a transformation. The column space of $M$ is the set of all linear combinations of the columns of $M$. However, the columns of $M$ are numbers, not vectors, so need to turn them into vectors first

  $$\text{col}(M)=\text{span}\{[C_1{]}_{\mathcal{E}},[C_2{]}_{\mathcal{E}},\dots ,[C_m{]}_{\mathcal{E}},\}$$

  This gets $M[{\vec{e}}_i]=C_i$ by the definition of induced transformation, know

  $$[\mathcal{T}({\vec{e}}_i){]}_{\mathcal{E}}=M[{\vec{e}}_i{]}_{\mathcal{E}}=C_i\Rightarrow \mathcal{T}({\vec{e}}_i)=[C_i{]}_{\mathcal{E}}$$

  Every input of $\mathcal{T}$ can be written as a linear combination of ${\vec{e}}_i$ (because $\mathcal{E}$ is a basis) and therefore, since $\mathcal{T}$ is linear, every output can be written as linear combinations of $[C_i{]}_{\mathcal{E}}$ Thus

  $$\text{range}(T)=\text{col}(M)$$

  Let $\mathcal{T}:{\mathbb{R}}^3\to {\mathbb{R}}^2$ be defined by $\mathcal{T}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2x-z\\ 4x-2z\end{array}\right]$. Find the range and rank of $\mathcal{T}$ §

  Let $M$ be a matrix for $\mathcal{T}$. And $M=\left[\begin{array}{ccc}1& 0& -1\\ 4& 0& -2\end{array}\right]$. From definitions, know:

  • $\text{range}(T)=\text{col}(M)$
  • $\text{rank}(T)=\text{dim}(\text{range}(T))=\text{dim}(\text{col}(M))$

  See that $\left\{\left[\begin{array}{c}2\\ 4\end{array}\right]\right\}$ is a basis for col$(M)$ and its one diminutional then:

  $$\text{range}(\mathcal{T})=\text{span}\left\{\left[\begin{array}{c}2\\ 4\end{array}\right]\right\}\text{and}\text{rank}(\mathcal{T})=1$$

• Let $M$ be a matrix, the rank of $M$ is equal to the number of pivots in rref$(M)$

• Let $\mathcal{T}$ be a linear transformation and let $M$ be a matrix for $\mathcal{T}$. Then nullity$(T)$ is equal to the number of free variable columns in rref$(M)$

• For a matrix $A$, $\text{rank}(A)=\text{rank}(A^T)$

Rank-Nullity Theorem §

• Define a matrix $A$

  \text{rank}(A) + \text{nullity}(A) = \text{# of columns in}A
  • Rank is the number of pivots of $A$ and Nullity is the number of non-pivots of $A$

• Define a transformation $\mathcal{T}$

  $$\text{rank}(\mathcal{T})+\text{nullity}(\mathcal{T})=\text{dim}(\text{domain of }\mathcal{T})$$

Let $\mathcal{P}\subseteq {\mathbb{R}}^4$ be a plane in ${\mathbb{R}}^4$ given in vector form by $\vec{x}=t\left[\begin{array}{c}1\\ 2\\ 2\\ 2\end{array}\right]+s\left[\begin{array}{c}-1\\ 1\\ -1\\ 1\end{array}\right]$ How many normal vectors does $\mathcal{P}$ have §

The matrix $A=\left[\begin{array}{cccc}1& 2& 2& 2\\ -1& 1& -1& 1\end{array}\right]$ is rank 2, and so has nullity 2

Therefore there exist linear independent normal directions for $\mathcal{P}$