Modules › Linear transformations
Linear transformations
Linear maps, standard matrix, kernel and range, rank–nullity, one-to-one and onto, composition, inverse, matrices relative to bases.
Lecture 6 · Exercise Sheet 6 · Linear transformations
Textbook sections: Elementary Linear Algebra ch. 6 §6.1–6.4 · Strang ch. 7 §7.1–7.3
Textbook sections: Elementary Linear Algebra ch. 6 §6.1–6.4 · Strang ch. 7 §7.1–7.3
Linear transformations
Linear Algebra · Module 6
After this module you can
- Test linearity, find the standard matrix, image and preimage.
- Find bases of the kernel and range, rank and nullity; decide one-to-one / onto / invertible.
- Compute the matrix of $T$ relative to bases ($P^{-1}AP$ or $P_{B'}^{-1}AP_B$) and of compositions.
Linear maps and the standard matrix
- $T(u+v)=T(u)+T(v)$, $T(cu)=cT(u)$; then $T(0)=0$. $(x_1+3,x_2)$, $(|x|,|y|)$, $(xy,x)$ are not linear.
- Standard matrix: columns are $T(e_1),\dots,T(e_n)$; $T(\mathbf x)=A\mathbf x$; for $T:\mathbb{R}^n\to\mathbb{R}^m$, $A$ is $m\times n$.
- Image of $v$: compute $Av$. Preimage of $w$: solve $A\mathbf x=w$ (a line/plane if $T$ is not one-to-one).
- Composition: $[T_2\circ T_1]=[T_2][T_1]$ (the first map is on the right).
Kernel, range, rank–nullity
| Notion | How to find it |
|---|---|
| $\ker T$ | null space of $A$: solve $A\mathbf x=0$ |
| range $T$ | column space of $A$: pivot columns |
| rank / nullity | $\dim\operatorname{range}$, $\dim\ker$; rank + nullity $=\dim(\text{domain})=n$ |
| one-to-one | $\ker T=\{0\}$ (nullity $0$) |
| onto | rank $=\dim(\text{codomain})=m$ |
| invertible | square and $\det A\ne0$; $[T^{-1}]=A^{-1}$ |
Examples: $T:\mathbb{R}^5\to\mathbb{R}^3$ with nullity $2$ has rank $3$; $T:P_5\to P_3$ with nullity $4$ has rank $6-4=2$.
Matrix relative to bases
- $P_B$ = matrix with the basis vectors of $B$ as columns. $[T]_{B,B'}=P_{B'}^{-1}\,A\,P_B$ and $[T(x)]_{B'}=[T]_{B,B'}[x]_B$.
- Same basis in domain and codomain: $[T]_B=P^{-1}AP$ — $A$ and $[T]_B$ are similar (same eigenvalues, trace, determinant).
- Final 2024–25 (Shift 1): $A=\begin{bmatrix}1&-2&1\\2&-3&5\\3&-2&1\end{bmatrix}$ and $B=\{(1,2,0),(1,0,-1),(-2,1,2)\}$ give $[T]_B=\begin{bmatrix}-4&2&-8\\9&-16&44\\4&-7&19\end{bmatrix}$.
- Proving that $B$ is a basis: $\det P_B\ne0$.
Beyond the syllabus: the Jacobian
- A differentiable map $F:\mathbb{R}^n\to\mathbb{R}^m$ is approximated near $p$ by a linear map: $F(p+h)\approx F(p)+J(p)h$.
- $J=\left[\frac{\partial F_i}{\partial x_j}\right]$ is the Jacobian matrix; for $m=n$, $|\det J|$ is the local scaling of area/volume.
- Polar coordinates $(r,\theta)\mapsto(r\cos\theta,r\sin\theta)$: $\det J=r$, so $dx\,dy=r\,dr\,d\theta$.
- This is Calculus II material; the lab has a Jacobian tool so you can see that the derivative IS a linear transformation.
Typical mistakes
Standard matrixColumns, not rows, are the images of the basis vectors.
Composition order$[T_1\circ T_2]=[T_1][T_2]$.
Relative matrix$P^{-1}AP$, not $PAP^{-1}$ (when $P$ has the new basis as columns).
Practice next
- Chapter quizzes Ch. 6 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Matrix of a linear map relative to bases
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Image and preimage (Exercise Sheet 6, Ex. 1(b))
For $T(v_1,v_2,v_3)=(0,\ v_1+v_2,\ v_2+v_3)$, which vector belongs to the preimage of $w=(0,2,5)$?
Show the answer and the solution
Answer: $(0, 2, 3)$
Need $v_1+v_2=2$ and $v_2+v_3=5$: $v=(t,\ 2-t,\ 3+t)$ — a whole line of preimages ($T$ is not one-to-one). $t=0$ gives $(0,2,3)$.
Worked example 2 · Composition (Exercise Sheet 6, Ex. 25)
$T_1(x,y)=(x,\ x+y,\ y)$ and $T_2(x,y,z)=(0,\ y)$. The standard matrix of $T_1\circ T_2$ is
Show the answer and the solution
Answer: $\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0\end{bmatrix}$
$[T_1\circ T_2]=[T_1][T_2]=\begin{bmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0\end{bmatrix}$ (a $3\times3$ matrix: $\mathbb{R}^3\to\mathbb{R}^2\to\mathbb{R}^3$).
Worked example 3 · Matrix relative to bases (Exercise Sheet 6, Ex. 26)
$T(x,y)=(-x,\ y,\ x+y)$ with $B=\{(1,1),(1,-1)\}$ and $B'=\{(0,1,0),(0,0,1),(1,0,0)\}$. The matrix $[T]_{B,B'}$ is
Show the answer and the solution
Answer: $\begin{bmatrix}1 & -1 \\ 2 & 0 \\ -1 & -1\end{bmatrix}$
$[T]_{B,B'}=P_{B'}^{-1}AP_B$ with $A=\begin{bmatrix}-1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix}$, $P_B=\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$, $P_{B'}=\begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$: $[T]_{B,B'}=\begin{bmatrix}1 & -1 \\ 2 & 0 \\ -1 & -1\end{bmatrix}$. Check on $v=(0,1)$: $[v]_B=(\frac{1}{2}, - \frac{1}{2})$, $[Tv]_{B'}=(1, 1, 0)$, i.e. $Tv=(0, 1, 1)$.
Worked example 4 · One-to-one and onto (Final 2024–25 Shift 2, Q3(c))
Exam: $T:\mathbb{R}^3\to\mathbb{R}^4$, $T(x_1,x_2,x_3)=(x_1-x_2+x_3,\ 6x_1-2x_2,\ 3x_1-4x_2+x_3,\ 5x_2-4x_3)$. Is $T$ one-to-one? onto?
Show the answer and the solution
Answer: one-to-one, not onto
$A=\begin{bmatrix}1 & -1 & 1 \\ 6 & -2 & 0 \\ 3 & -4 & 1 \\ 0 & 5 & -4\end{bmatrix}$ has rank $3$: nullity $3-3=0$, so $\ker T$ is $\{0\}$ (one-to-one); the range has dimension $3<4$, so $T$ is not onto.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Matrix of a linear map relative to bases