Chapter 6 · Linear Transformations

6.2 Kernel and Range

What a linear map destroys and what it can reach: the kernel $\ker T$ (everything sent to $\mathbf{0}$) and the range $\operatorname{range}T$ (every possible output), tied together by $\dim\ker T+\dim\operatorname{range}T=\dim V$.

A camera turns the three-dimensional world into a flat picture. That is (approximately) a linear map from $\mathbb{R}^3$ to $\mathbb{R}^2$, and it is easy to see what it loses: every point along the same line of sight lands on the same pixel, so a single photograph cannot tell a small nearby object from a large distant one. It is just as natural to ask what the camera can produce: can every possible image be obtained?

For any linear transformation these are the two central questions. What is lost? — the vectors that are sent to $\mathbf{0}$ form the kernel, and the map is one-to-one exactly when nothing non-zero is lost. What can be reached? — the set of all outputs forms the range, and the map is onto when everything can be reached. For matrix maps these are old friends from Section 5.4: the null space and the column space. And they are linked by one of the most useful counting facts in the subject, the rank–nullity theorem.

Kernel and Range

📘 Definition 6.3 — Kernel and range
Let $T:V\to W$ be a linear transformation.
  • The kernel of $T$ is the set of vectors in $V$ that $T$ sends to zero: $$\ker T=\{\mathbf{v}\in V:\ T(\mathbf{v})=\mathbf{0}\}.$$
  • The range (or image) of $T$ is the set of all outputs: $$\operatorname{range}T=\{T(\mathbf{v}):\ \mathbf{v}\in V\}=\{\mathbf{w}\in W:\ \mathbf{w}=T(\mathbf{v})\text{ for some }\mathbf{v}\in V\}.$$
📐 Theorem 6.4 — Kernel and range are subspaces
If $T:V\to W$ is linear, then $\ker T$ is a subspace of $V$ (the domain) and $\operatorname{range}T$ is a subspace of $W$ (the codomain). The dimensions $\dim\ker T$ and $\dim\operatorname{range}T$ are called the nullity and the rank of $T$.
Kernel $T(\mathbf{0})=\mathbf{0}$, so $\mathbf{0}\in\ker T$. If $\mathbf{u},\mathbf{v}\in\ker T$ and $c$ is a scalar, then $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})=\mathbf{0}+\mathbf{0}=\mathbf{0}$ and $T(c\mathbf{u})=cT(\mathbf{u})=\mathbf{0}$. So $\ker T$ is closed under addition and scalar multiplication.
Range $\mathbf{0}=T(\mathbf{0})\in\operatorname{range}T$. If $\mathbf{w}_1=T(\mathbf{v}_1)$ and $\mathbf{w}_2=T(\mathbf{v}_2)$, then $\mathbf{w}_1+\mathbf{w}_2=T(\mathbf{v}_1+\mathbf{v}_2)$ and $c\mathbf{w}_1=T(c\mathbf{v}_1)$ are outputs too.

For a matrix transformation $T(\mathbf{x})=A\mathbf{x}$, the equation $T(\mathbf{x})=\mathbf{0}$ is the homogeneous system $A\mathbf{x}=\mathbf{0}$, and an output $A\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n$ is a combination of the columns $\mathbf{a}_j$ of $A$. So nothing new needs to be learned:

📐 Theorem 6.5 — Kernel and range of a matrix transformation
If $A$ is an $m\times n$ matrix and $T(\mathbf{x})=A\mathbf{x}$ from $\mathbb{R}^n$ to $\mathbb{R}^m$, then $$\ker T=\operatorname{Nul}A\subseteq\mathbb{R}^n,\qquad\operatorname{range}T=\operatorname{Col}A\subseteq\mathbb{R}^m,$$ so the rank of $T$ is $\operatorname{rank}A$ and the nullity of $T$ is $\operatorname{nullity}A$. Moreover, $\mathbf{b}$ is in $\operatorname{range}T$ exactly when the system $A\mathbf{x}=\mathbf{b}$ is consistent.
🧭 Problem-solving strategy — kernel and range of $T(\mathbf{x})=A\mathbf{x}$
  1. Row reduce $A$ to reduced echelon form and mark the pivot columns.
  2. Kernel: solve $A\mathbf{x}=\mathbf{0}$; write the general solution in terms of the free variables. The vectors multiplying the free variables form a basis of $\ker T$ (there is one per free variable).
  3. Range: the pivot columns of the original matrix $A$ (not of its echelon form) form a basis of $\operatorname{range}T$.
  4. Check: multiply $A$ by each kernel basis vector — you must get $\mathbf{0}$ — and confirm (number of free variables) + (number of pivots) $=n$.
✏️ Example 1 — Kernel and range of a map on $\mathbb{R}^3$
Let $T(\mathbf{x})=A\mathbf{x}$ with $A=\begin{bmatrix}1&2&3\\2&5&7\\1&3&4\end{bmatrix}$. Find bases of $\ker T$ and $\operatorname{range}T$, describe both geometrically, and decide whether $\mathbf{b}=\begin{bmatrix}1\\3\\2\end{bmatrix}$ and $\mathbf{c}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ are in the range.
Row reduce $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$, then $R_3\to R_3-R_2$, then $R_1\to R_1-2R_2$: $$\begin{bmatrix}1&2&3\\2&5&7\\1&3&4\end{bmatrix}\to\begin{bmatrix}1&2&3\\0&1&1\\0&1&1\end{bmatrix}\to\begin{bmatrix}1&2&3\\0&1&1\\0&0&0\end{bmatrix}\to\begin{bmatrix}1&0&1\\0&1&1\\0&0&0\end{bmatrix}.$$ Pivots in columns 1 and 2; $x_3$ is free.
Kernel $x_1=-x_3$, $x_2=-x_3$, so $\mathbf{x}=x_3\begin{bmatrix}-1\\-1\\1\end{bmatrix}$. $\ker T=\operatorname{span}\left\{\begin{bmatrix}-1\\-1\\1\end{bmatrix}\right\}$ — a line through the origin in $\mathbb{R}^3$. Nullity $1$. (Check: $A\begin{bmatrix}-1\\-1\\1\end{bmatrix}=\begin{bmatrix}-1-2+3\\-2-5+7\\-1-3+4\end{bmatrix}=\mathbf{0}$ ✓.)
Range The pivot columns of $A$: $\operatorname{range}T=\operatorname{span}\left\{\begin{bmatrix}1\\2\\1\end{bmatrix},\begin{bmatrix}2\\5\\3\end{bmatrix}\right\}$ — a plane through the origin; rank $2$. A normal vector is $(1,2,1)\times(2,5,3)=(1,-1,1)$, so the range is the plane $x-y+z=0$.
Membership $\mathbf{b}$: $1-3+2=0$, so $\mathbf{b}$ is in the range — indeed $A\begin{bmatrix}-1\\1\\0\end{bmatrix}=\begin{bmatrix}1\\3\\2\end{bmatrix}$. $\mathbf{c}$: $1-0+0=1\ne0$, so $A\mathbf{x}=\mathbf{c}$ has no solution. Note $1+2=3=\dim\mathbb{R}^3$.
✏️ Example 2 — A map that is not a matrix map
Let $T:P_2\to\mathbb{R}$ be evaluation at $1$: $T(p)=p(1)$. Find $\ker T$ and $\operatorname{range}T$ and their dimensions.
Kernel Write $p(t)=a+bt+ct^2$. Then $p\in\ker T$ when $p(1)=a+b+c=0$, i.e. $a=-b-c$: $$p(t)=(-b-c)+bt+ct^2=b(t-1)+c(t^2-1).$$ So $\ker T=\operatorname{span}\{t-1,\ t^2-1\}$. These two polynomials are not multiples of each other, hence independent: nullity $2$. (These are exactly the polynomials with a root at $t=1$.)
Range Every real number $r$ is an output: the constant polynomial $p(t)=r$ has $p(1)=r$. So $\operatorname{range}T=\mathbb{R}$, rank $1$.
Count $2+1=3=\dim P_2$ — the same pattern as in Example 1. This is no accident (Theorem 6.8 below).

One-to-One and Onto

📘 Definition 6.4 — One-to-one and onto
A function $T:V\to W$ is
  • one-to-one (injective) if different inputs always give different outputs: $T(\mathbf{u})=T(\mathbf{v})$ implies $\mathbf{u}=\mathbf{v}$;
  • onto (surjective) if every vector of $W$ is an output: $\operatorname{range}T=W$.

For a general function, checking one-to-one means comparing every pair of inputs. For a linear map it is enough to look at what is sent to $\mathbf{0}$:

📐 Theorem 6.6 — One-to-one means trivial kernel
A linear transformation $T:V\to W$ is one-to-one if and only if $\ker T=\{\mathbf{0}\}$.
⇒ If $T$ is one-to-one and $T(\mathbf{v})=\mathbf{0}=T(\mathbf{0})$, then $\mathbf{v}=\mathbf{0}$. So only $\mathbf{0}$ is in the kernel.
⇐ Suppose $\ker T=\{\mathbf{0}\}$ and $T(\mathbf{u})=T(\mathbf{v})$. Then $T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}$, so $\mathbf{u}-\mathbf{v}\in\ker T$, i.e. $\mathbf{u}-\mathbf{v}=\mathbf{0}$ and $\mathbf{u}=\mathbf{v}$.
Picture The proof shows more: $T(\mathbf{u})=T(\mathbf{v})$ exactly when $\mathbf{u}-\mathbf{v}\in\ker T$. The inputs with the same output as $\mathbf{u}$ form the shifted copy $\mathbf{u}+\ker T$ of the kernel — a whole line or plane collapses to one point (see Interactive 1).

For a matrix map, Theorem 6.6 and Theorem 6.5 turn both properties into statements about pivots, which we read off from an echelon form.

📐 Theorem 6.7 — The pivot test for $T(\mathbf{x})=A\mathbf{x}$

Let $A$ be an $m\times n$ matrix and $T(\mathbf{x})=A\mathbf{x}$, $T:\mathbb{R}^n\to\mathbb{R}^m$.

  • $T$ is one-to-one $\iff$ $A\mathbf{x}=\mathbf{0}$ has only the trivial solution $\iff$ the columns of $A$ are linearly independent $\iff$ there is a pivot in every column $\iff\operatorname{rank}A=n$.
  • $T$ is onto $\iff$ $A\mathbf{x}=\mathbf{b}$ is consistent for every $\mathbf{b}$ in $\mathbb{R}^m$ $\iff$ the columns of $A$ span $\mathbb{R}^m$ $\iff$ there is a pivot in every row $\iff\operatorname{rank}A=m$.
✏️ Example 3 — Classifying matrix maps
Decide whether $T(\mathbf{x})=A\mathbf{x}$ is one-to-one and whether it is onto: (a) $A=\begin{bmatrix}1&0\\2&1\\0&3\end{bmatrix}$; (b) $A=\begin{bmatrix}1&2&3\\2&4&7\end{bmatrix}$; (c) the matrix of Example 1; (d) $A=\begin{bmatrix}2&1\\1&1\end{bmatrix}$.
(a) $R_2\to R_2-2R_1$, then $R_3\to R_3-3R_2$: $\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}$. Pivots in both columns but only $2$ of the $3$ rows. $T:\mathbb{R}^2\to\mathbb{R}^3$ is one-to-one, not onto (its range is a plane in $\mathbb{R}^3$).
(b) $R_2\to R_2-2R_1$: $\begin{bmatrix}1&2&3\\0&0&1\end{bmatrix}$. Pivots in columns 1 and 3, and in both rows. $T:\mathbb{R}^3\to\mathbb{R}^2$ is onto, not one-to-one ($x_2$ is free; the kernel is the line spanned by $(-2,1,0)$).
(c) Example 1 found rank $2$ for a $3\times3$ matrix: a pivot is missing in column 3 and in row 3. Neither.
(d) $\det A=2-1=1\ne0$, so $A$ is invertible and has a pivot in every row and column: both one-to-one and onto.
⚠️ Size alone can rule things out

A matrix has at most one pivot per row and per column, so $\operatorname{rank}A\le\min(m,n)$.

Wide matrices ($n\gt m$, more inputs than outputs) are never one-to-one: there are too few pivots to fill every column, so a free variable — and a non-zero kernel — must exist. Tall matrices ($m\gt n$) are never onto: $n$ columns cannot span the bigger space $\mathbb{R}^m$. Only a square matrix can be both.

And keep the two spaces apart: the kernel lives in the domain $\mathbb{R}^n$, the range in the codomain $\mathbb{R}^m$. In Example 3(a) a kernel vector would have $2$ entries, a range vector $3$.

Interactive 1 — Squashing the Plane: the Kernel Goes to Zero

Each dot $\mathbf{x}$ slides to its image $A\mathbf{x}$ (press Play, or drag the slider). For a matrix of rank $1$ the whole plane is squashed onto the blue line $\operatorname{range}T$, while the red line $\ker T$ shrinks to the origin. Dots of the same colour lie on a line parallel to the kernel — and all of them land on the same point, which is why the map is not one-to-one.

$\ker T$—
$\operatorname{range}T$—
nullity + rank—
one-to-one? onto?—

The Rank–Nullity Theorem

In Examples 1 and 2 the dimension of what is lost plus the dimension of what is reached equals the dimension of the domain. Every input dimension is either "crushed" into the kernel or "survives" into the range.

📐 Theorem 6.8 — Rank–nullity theorem
Let $V$ be a finite-dimensional vector space, $W$ any vector space, and $T:V\to W$ a linear transformation. Then $$\dim\ker T+\dim\operatorname{range}T=\dim V,\qquad\text{i.e.}\qquad\operatorname{nullity}T+\operatorname{rank}T=\dim V.$$ For $T(\mathbf{x})=A\mathbf{x}$ with $A$ of size $m\times n$ this is $\operatorname{nullity}A+\operatorname{rank}A=n$: (free variables) + (pivot variables) = (number of columns).
Set up Let $\{\mathbf{k}_1,\dots,\mathbf{k}_p\}$ be a basis of $\ker T$ and extend it to a basis $\{\mathbf{k}_1,\dots,\mathbf{k}_p,\mathbf{v}_1,\dots,\mathbf{v}_r\}$ of $V$, so $\dim V=p+r$.
Spanning Any output is $T(a_1\mathbf{k}_1+\cdots+b_1\mathbf{v}_1+\cdots)=b_1T(\mathbf{v}_1)+\cdots+b_rT(\mathbf{v}_r)$, because the $T(\mathbf{k}_i)$ are $\mathbf{0}$. So $T(\mathbf{v}_1),\dots,T(\mathbf{v}_r)$ span $\operatorname{range}T$.
Independence If $b_1T(\mathbf{v}_1)+\cdots+b_rT(\mathbf{v}_r)=\mathbf{0}$, then $b_1\mathbf{v}_1+\cdots+b_r\mathbf{v}_r\in\ker T$, so it is a combination of the $\mathbf{k}_i$. Independence of the whole basis forces every $b_j=0$. Hence $\dim\operatorname{range}T=r=\dim V-p$.
✏️ Example 4 — Symmetric and skew-symmetric parts
Let $T:M_{2\times2}\to M_{2\times2}$, $T(A)=A-A^{T}$. Find $\ker T$, $\operatorname{range}T$ and check the rank–nullity theorem.
Compute For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$: $$T(A)=\begin{bmatrix}a&b\\c&d\end{bmatrix}-\begin{bmatrix}a&c\\b&d\end{bmatrix}=\begin{bmatrix}0&b-c\\c-b&0\end{bmatrix}=(b-c)\begin{bmatrix}0&1\\-1&0\end{bmatrix}.$$
Kernel $T(A)=O$ exactly when $b=c$, i.e. $A=A^{T}$: the kernel is the space of symmetric matrices, with basis $\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}$. Nullity $3$.
Range Every output is a multiple of $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$, and every multiple $k$ occurs (take $b=k$, $c=0$). The range is the $1$-dimensional space of skew-symmetric $2\times2$ matrices ($A^{T}=-A$). Rank $1$.
Check $3+1=4=\dim M_{2\times2}$ ✓. $T$ is neither one-to-one nor onto.
✏️ Example 5 — Counting without computing
(a) $T:\mathbb{R}^7\to\mathbb{R}^4$ is linear and onto. Find $\dim\ker T$. (b) $T:\mathbb{R}^4\to\mathbb{R}^6$ is linear with $\dim\ker T=1$. Find the rank. Is $T$ onto? (c) Can a linear map $\mathbb{R}^3\to\mathbb{R}^5$ be onto? Can a linear map $\mathbb{R}^5\to\mathbb{R}^3$ be one-to-one?
(a) Onto means $\operatorname{range}T=\mathbb{R}^4$, rank $4$. So $\dim\ker T=7-4=3$.
(b) $\operatorname{rank}T=4-1=3$. The range is a $3$-dimensional subspace of $\mathbb{R}^6$, so $T$ is not onto.
(c) $\mathbb{R}^3\to\mathbb{R}^5$: rank $\le\dim V=3\lt5$, so never onto. $\mathbb{R}^5\to\mathbb{R}^3$: rank $\le3$, so nullity $=5-\text{rank}\ge2\gt0$ — never one-to-one. (Some information must be lost when squeezing $5$ dimensions into $3$.)
Interactive 2 — One-to-One? Onto? Read It from the Pivots

Watch $A$ being row reduced to echelon form, one row operation at a time (exact fractions). Then read the pivot pattern: a pivot in every column means $T(\mathbf{x})=A\mathbf{x}$ is one-to-one; a pivot in every row means it is onto. Free columns (no pivot) give the kernel; zero rows show that some $\mathbf{b}$ cannot be reached. Try Random for new matrices.

size $m\times n$—
rank = pivots—
nullity = free columns—
one-to-one / onto—

Isomorphisms

A linear map that is both one-to-one and onto loses nothing and reaches everything; it can be undone. Such a map shows that two vector spaces are "the same space with different names for the vectors".

📘 Definition 6.5 — Isomorphism
A linear transformation $T:V\to W$ that is both one-to-one and onto is called an isomorphism. If an isomorphism from $V$ to $W$ exists, $V$ and $W$ are isomorphic, written $V\cong W$. An isomorphism has an inverse $T^{-1}:W\to V$ with $T^{-1}(T(\mathbf{v}))=\mathbf{v}$ and $T(T^{-1}(\mathbf{w}))=\mathbf{w}$; we also call $T$ an invertible linear map.
📐 Theorem 6.9 — Facts about isomorphisms
  1. If $T:V\to W$ is an isomorphism, then $T^{-1}:W\to V$ is also linear (and an isomorphism).
  2. If $\dim V=\dim W=n$ is finite, a linear map $T:V\to W$ is one-to-one $\iff$ it is onto $\iff$ it is an isomorphism. (By rank–nullity: nullity $0$ $\iff$ rank $n$.)
  3. For an $n\times n$ matrix $A$, $T(\mathbf{x})=A\mathbf{x}$ is an isomorphism of $\mathbb{R}^n$ $\iff$ $A$ is invertible $\iff$ $\det A\ne0$, and then $T^{-1}(\mathbf{y})=A^{-1}\mathbf{y}$.
📐 Theorem 6.10 — Every $n$-dimensional space looks like $\mathbb{R}^n$
Let $V$ be a real vector space of finite dimension $n$, with basis $\mathcal B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$. The coordinate map $\mathbf{v}\mapsto[\mathbf{v}]_{\mathcal B}$ (Section 5.5) is an isomorphism from $V$ onto $\mathbb{R}^n$. Consequently $V\cong\mathbb{R}^n$, and two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension. (We state this without proof.)

This is why so much of linear algebra can be done with columns of numbers: $P_n\cong\mathbb{R}^{n+1}$ via $a_0+a_1t+\cdots+a_nt^n\mapsto(a_0,a_1,\dots,a_n)$, and $M_{m\times n}\cong\mathbb{R}^{mn}$ by listing the entries. Questions about independence, spanning and bases can be translated into $\mathbb{R}^n$, answered by row reduction, and translated back — an isomorphism preserves all of them.

✏️ Example 6 — Using isomorphisms
(a) Show that $T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}2x+y\\x+y\end{bmatrix}$ is an isomorphism of $\mathbb{R}^2$ and find $T^{-1}$. (b) Use the isomorphism $P_2\cong\mathbb{R}^3$ to decide whether $\{1+t,\ t+t^2,\ 1+t^2\}$ is a basis of $P_2$.
(a) The standard matrix is $A=\begin{bmatrix}2&1\\1&1\end{bmatrix}$ with $\det A=1\ne0$, so $T$ is an isomorphism (Theorem 6.9). Using $\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$: $$A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix},\qquad T^{-1}\left(\begin{bmatrix}u\\v\end{bmatrix}\right)=\begin{bmatrix}u-v\\-u+2v\end{bmatrix}.$$
(a) Check $T(1,0)=(2,1)$ and $T^{-1}(2,1)=(2-1,\ -2+2)=(1,0)$ ✓.
(b) Coordinates relative to $\{1,t,t^2\}$: $1+t\mapsto(1,1,0)$, $t+t^2\mapsto(0,1,1)$, $1+t^2\mapsto(1,0,1)$. $$\det\begin{bmatrix}1&0&1\\1&1&0\\0&1&1\end{bmatrix}=1(1-0)-0+1(1-0)=2\ne0,$$ so the three coordinate vectors are independent and span $\mathbb{R}^3$.
(b) Conclude The coordinate map is an isomorphism, so the polynomials are independent and span $P_2$: they form a basis. Three independent vectors in a $3$-dimensional space is enough.
⚠️ Isomorphic does not mean equal

$P_3$ and $M_{2\times2}$ are isomorphic (both have dimension $4$), but a polynomial is not a matrix. "Isomorphic" means that the vector-space structure — addition and scaling — matches perfectly under some one-to-one correspondence. Also, an isomorphism is about the map: $\mathbb{R}^2\cong\mathbb{R}^2$, yet the projection $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ is not an isomorphism.

Exercises

I
For $T(\mathbf{x})=A\mathbf{x}$, find a basis of $\ker T$ and of $\operatorname{range}T$, and state the rank and nullity.
  • 1 $A=\begin{bmatrix}1&2\\3&6\end{bmatrix}$
  • 2 $A=\begin{bmatrix}1&0&-1\\0&1&2\end{bmatrix}$
  • 3 $A=\begin{bmatrix}1&2\\0&1\\2&1\end{bmatrix}$ (also give an equation of the range)
  • 4 $A=\begin{bmatrix}1&-1&2\\-2&2&-4\end{bmatrix}$
  • 5 $A=\begin{bmatrix}1&2&1\\2&4&3\\3&6&4\end{bmatrix}$ (also give an equation of the range)
  • 6 $A=\begin{bmatrix}1&1&0&2\\0&1&1&1\\1&2&1&3\end{bmatrix}$
  • 7 For the map in part 6, decide whether $(3,1,5)$ and $(2,3,5)$ are in $\operatorname{range}T$; for any that is, find an $\mathbf{x}$ with $A\mathbf{x}$ equal to it.
Check: multiply A by your kernel vectors (must give 0) and count dimensions
1 $R_2\to R_2-3R_1$ gives $\begin{bmatrix}1&2\\0&0\end{bmatrix}$; $x_2$ free, $x_1=-2x_2$. $\ker T=\operatorname{span}\{(-2,1)\}$, $\operatorname{range}T=\operatorname{span}\{(1,3)\}$. Rank $1$, nullity $1$.
2 Already reduced; $x_3$ free, $x_1=x_3$, $x_2=-2x_3$. $\ker T=\operatorname{span}\{(1,-2,1)\}$; pivots in both rows, so $\operatorname{range}T=\mathbb{R}^2$ (basis $(1,0),(0,1)$). Rank $2$, nullity $1$.
3 $R_3\to R_3-2R_1$: $\begin{bmatrix}1&2\\0&1\\0&-3\end{bmatrix}$, then $R_3\to R_3+3R_2$: $\begin{bmatrix}1&2\\0&1\\0&0\end{bmatrix}$. Both columns are pivot columns: $\ker T=\{\mathbf{0}\}$, nullity $0$. $\operatorname{range}T=\operatorname{span}\{(1,0,2),(2,1,1)\}$, rank $2$. Normal $(1,0,2)\times(2,1,1)=(-2,3,1)$: the range is the plane $-2x+3y+z=0$.
4 $R_2\to R_2+2R_1$ gives $\begin{bmatrix}1&-1&2\\0&0&0\end{bmatrix}$; $x_1=x_2-2x_3$. $\ker T=\operatorname{span}\{(1,1,0),(-2,0,1)\}$ (a plane), $\operatorname{range}T=\operatorname{span}\{(1,-2)\}$ (a line). Rank $1$, nullity $2$.
5 $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: $\begin{bmatrix}1&2&1\\0&0&1\\0&0&1\end{bmatrix}$, then $R_3\to R_3-R_2$, $R_1\to R_1-R_2$: $\begin{bmatrix}1&2&0\\0&0&1\\0&0&0\end{bmatrix}$. Pivots in columns 1 and 3; $x_2$ free, $x_1=-2x_2$, $x_3=0$. $\ker T=\operatorname{span}\{(-2,1,0)\}$; $\operatorname{range}T=\operatorname{span}\{(1,2,3),(1,3,4)\}$, the plane $z=x+y$ (normal $(-1,-1,1)$). Rank $2$, nullity $1$.
6 $R_3\to R_3-R_1$, $R_3\to R_3-R_2$, $R_1\to R_1-R_2$: $\begin{bmatrix}1&0&-1&1\\0&1&1&1\\0&0&0&0\end{bmatrix}$. $x_3,x_4$ free: $x_1=x_3-x_4$, $x_2=-x_3-x_4$. $\ker T=\operatorname{span}\{(1,-1,1,0),(-1,-1,0,1)\}$, nullity $2$. $\operatorname{range}T=\operatorname{span}\{(1,0,1),(1,1,2)\}$, rank $2$ — the plane $z=x+y$. ($2+2=4$ ✓.)
7 Range is the plane $z=x+y$. $(3,1,5)$: $3+1=4\ne5$ — not in the range. $(2,3,5)$: $2+3=5$ ✓. Solving $x_1+x_2+2x_4=2$, $x_2+x_3+x_4=3$ with $x_3=x_4=0$ gives $x_2=3$, $x_1=-1$: $A\begin{bmatrix}-1\\3\\0\\0\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}$ ✓. (Any vector $(-1,3,0,0)+\mathbf{k}$ with $\mathbf{k}\in\ker T$ works too.)
II
Is $T$ one-to-one? Is it onto? Justify with pivots, determinants or rank–nullity.
  • 1 $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
  • 2 $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$
  • 3 $A=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}$
  • 4 $A=\begin{bmatrix}1&2&3\\0&1&4\end{bmatrix}$
  • 5 $A=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}$
  • 6 $A=\begin{bmatrix}1&0&1\\0&1&1\\1&1&0\end{bmatrix}$
  • 7 Any linear map $T:\mathbb{R}^5\to\mathbb{R}^2$ (one-to-one is the question)
  • 8 $T:P_2\to\mathbb{R}^2$, $T(p)=\big(p(0),\ p(1)\big)$; also find $\ker T$.
1 $\det A=4-6=-2\ne0$: invertible, so both one-to-one and onto.
2 $\det A=0$, rank $1$ (row 2 = 2 × row 1). Neither; the kernel is $\operatorname{span}\{(-2,1)\}$.
3 Rank $2=n$: pivots in both columns, so one-to-one; $3\times2$ is tall, so not onto (e.g. $(0,0,1)$ is not reached, since every output has third entry = first + second).
4 Already in echelon form with pivots in both rows: onto. Column 3 is free, so not one-to-one (wide matrix).
5 Row 2 = 2 × row 1, rank $1$: not onto ($1\lt2$) and not one-to-one ($1\lt3$). Neither.
6 $\det A=1(0-1)-0+1(0-1)=-2\ne0$: both (an isomorphism of $\mathbb{R}^3$).
7 Never one-to-one: rank $\le2$, so nullity $\ge5-2=3$. (It can be onto, e.g. $(x_1,\dots,x_5)\mapsto(x_1,x_2)$.)
8 For $p=a+bt+ct^2$: $T(p)=(a,\ a+b+c)$. Any $(r,s)$ is reached by $p=r+(s-r)t$, so onto. Kernel: $a=0$, $b=-c$, so $\ker T=\operatorname{span}\{t^2-t\}$ — not one-to-one. ($1+2=3=\dim P_2$ ✓.)
III
Rank–nullity and maps on other spaces.
  • 1 $T:\mathbb{R}^6\to\mathbb{R}^4$ is linear with rank $4$. Find its nullity. Is $T$ onto?
  • 2 $T:\mathbb{R}^4\to\mathbb{R}^6$ is linear with $\ker T=\{\mathbf{0}\}$. Find its rank. Is $T$ onto?
  • 3 $T:M_{2\times2}\to\mathbb{R}$, $T(A)=\operatorname{tr}A$. Find a basis of $\ker T$ and the range; check rank–nullity.
  • 4 (uses calculus) $D:P_3\to P_3$, $D(p)=p^{\prime}$. Find $\ker D$ and $\operatorname{range}D$, and check rank–nullity.
  • 5 $T:M_{2\times2}\to M_{2\times2}$, $T(A)=A+A^{T}$. Find $\ker T$ and $\operatorname{range}T$ (compare Example 4).
  • 6 $T:\mathbb{R}^3\to\mathbb{R}^3$, $T(\mathbf{x})=\mathbf{x}\times\mathbf{e}_3$ (cross product with $(0,0,1)$). Find the standard matrix, $\ker T$ and $\operatorname{range}T$.
  • 7 Explain why no linear $T:\mathbb{R}^3\to\mathbb{R}^3$ can have $\ker T=\operatorname{range}T$, and give an example of a linear $T:\mathbb{R}^4\to\mathbb{R}^4$ with $\ker T=\operatorname{range}T$.
  • 8 A $5\times7$ matrix $A$ has rank $4$. Find $\dim\operatorname{Nul}A$ and $\dim\operatorname{Col}A$, and say whether $\mathbf{x}\mapsto A\mathbf{x}$ is one-to-one or onto.
1 Nullity $=6-4=2$. The range is a $4$-dimensional subspace of $\mathbb{R}^4$, i.e. all of $\mathbb{R}^4$: onto.
2 Nullity $0$, so rank $=4-0=4\lt6$: not onto (but one-to-one).
3 $\operatorname{tr}\begin{bmatrix}a&b\\c&d\end{bmatrix}=a+d=0\iff d=-a$. Basis of $\ker T$: $\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}$ (nullity $3$). Range $=\mathbb{R}$ ($\operatorname{tr}\begin{bmatrix}r&0\\0&0\end{bmatrix}=r$), rank $1$. $3+1=4$ ✓.
4 $p^{\prime}=0$ exactly for constants: $\ker D=\operatorname{span}\{1\}$, nullity $1$. $D(a+bt+ct^2+dt^3)=b+2ct+3dt^2$ reaches every polynomial of degree $\le2$: $\operatorname{range}D=P_2$, rank $3$. $1+3=4=\dim P_3$ ✓. $D$ is neither one-to-one nor onto $P_3$ ($t^3$ is not reached).
5 $T(A)=\begin{bmatrix}2a&b+c\\b+c&2d\end{bmatrix}$. Kernel: $a=d=0$, $c=-b$ — the skew-symmetric matrices $\operatorname{span}\left\{\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\}$, nullity $1$. Range: all symmetric matrices (any symmetric $S$ equals $T(\tfrac12S)$), dimension $3$. $1+3=4$ ✓ — the roles are swapped compared with Example 4.
6 $(x,y,z)\times(0,0,1)=(y,\ -x,\ 0)$, so $A=\begin{bmatrix}0&1&0\\-1&0&0\\0&0&0\end{bmatrix}$. $\ker T=\operatorname{span}\{(0,0,1)\}$ (vectors parallel to $\mathbf{e}_3$), $\operatorname{range}T$ is the $xy$-plane $z=0$. Rank $2$, nullity $1$.
7 If $\ker T=\operatorname{range}T$ they have the same dimension $k$, and rank–nullity gives $2k=3$ — impossible for a whole number $k$. On $\mathbb{R}^4$: $T(x_1,x_2,x_3,x_4)=(x_3,x_4,0,0)$ has $\ker T=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}=\operatorname{range}T$ ($2+2=4$).
8 $\dim\operatorname{Col}A=4$, $\dim\operatorname{Nul}A=7-4=3$. Not one-to-one (nullity $3$), not onto ($4\lt5$).
IV
Isomorphisms and applications.
  • 1 Show that $T(x,y)=(3x+y,\ 5x+2y)$ is an isomorphism of $\mathbb{R}^2$ and find a formula for $T^{-1}$.
  • 2 Which pairs are isomorphic? (a) $\mathbb{R}^4$ and $M_{2\times2}$; (b) $P_3$ and $\mathbb{R}^4$; (c) $P_2$ and $M_{2\times2}$; (d) $\mathbb{R}^6$ and $M_{2\times3}$.
  • 3 Use the isomorphism $P_2\cong\mathbb{R}^3$ to decide whether $1+2t$, $3-t+t^2$, $5+3t+t^2$ are linearly independent in $P_2$.
  • 4 Coding. Letters are numbered $A=1,\dots,Z=26$ and a message is sent in pairs $\mathbf{x}$ as $\mathbf{y}=A\mathbf{x}$ with $A=\begin{bmatrix}2&3\\1&2\end{bmatrix}$. (a) Encode the word HI. (b) Decode the received pair $(63,37)$. (c) Why must the coding map be one-to-one?
  • 5 Imaging. A simple camera model sends a point $(x,y,z)$ to the image point $(x,y)$. Find the kernel and range of this map, check rank–nullity, and explain in words why one photograph cannot determine depth.
  • 6 Proof. Let $T:V\to W$ be a one-to-one linear map and $\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ linearly independent in $V$. Prove that $\{T(\mathbf{v}_1),\dots,T(\mathbf{v}_k)\}$ is linearly independent in $W$.
1 $A=\begin{bmatrix}3&1\\5&2\end{bmatrix}$, $\det A=6-5=1\ne0$, so $T$ is an isomorphism. $A^{-1}=\begin{bmatrix}2&-1\\-5&3\end{bmatrix}$, i.e. $T^{-1}(u,v)=(2u-v,\ -5u+3v)$. Check: $T(1,0)=(3,5)$ and $T^{-1}(3,5)=(6-5,\ -15+15)=(1,0)$ ✓.
2 Compare dimensions (Theorem 6.10): (a) $4=4$ yes; (b) $4=4$ yes; (c) $3\ne4$ no; (d) $6=6$ yes.
3 Coordinates $(1,2,0)$, $(3,-1,1)$, $(5,3,1)$. $\det\begin{bmatrix}1&3&5\\2&-1&3\\0&1&1\end{bmatrix}=0$, so they are dependent; indeed $5+3t+t^2=2(1+2t)+(3-t+t^2)$.
4 (a) H $=8$, I $=9$: $A\begin{bmatrix}8\\9\end{bmatrix}=\begin{bmatrix}16+27\\8+18\end{bmatrix}=\begin{bmatrix}43\\26\end{bmatrix}$. (b) $\det A=1$, $A^{-1}=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$, $A^{-1}\begin{bmatrix}63\\37\end{bmatrix}=\begin{bmatrix}126-111\\-63+74\end{bmatrix}=\begin{bmatrix}15\\11\end{bmatrix}$, i.e. O, K: the word "OK". (c) If two different messages gave the same code, the receiver could not tell which was sent; decoding needs $T^{-1}$, which exists only for a one-to-one (here: invertible) map.
5 Matrix $\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}$. $\ker T=\operatorname{span}\{(0,0,1)\}$ (the viewing direction), $\operatorname{range}T=\mathbb{R}^2$. $1+2=3$ ✓. All points $(x,y,z)$ that differ only in $z$ — i.e. differ by a kernel vector — produce the same image point, so depth is exactly the information in the kernel, which the map throws away. (Two cameras looking from different directions have different kernels — that is how stereo vision recovers depth.)
6 Suppose $c_1T(\mathbf{v}_1)+\cdots+c_kT(\mathbf{v}_k)=\mathbf{0}$. By linearity $T(c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k)=\mathbf{0}$, so $c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k\in\ker T=\{\mathbf{0}\}$ (Theorem 6.6). Independence of the $\mathbf{v}_i$ gives $c_1=\cdots=c_k=0$. Hence the images are independent.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. How many measurements does a scanner need before an image is uniquely determined — and which images are invisible to it?

Why it matters. CT scanners in hospitals do not photograph the inside of the body; they measure sums of tissue density along many lines (projections) and reconstruct the image by solving a linear system. The measurement map is linear, and its kernel consists of exactly the "ghost" images that produce all-zero measurements and so can never be seen. Engineers design scanning patterns so that this kernel is as small as possible.

What to do.

  1. Model a tiny $2\times2$ image $X=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ of densities. The scanner measures the two row sums and the two column sums. Write this map $T:M_{2\times2}\to\mathbb{R}^4$ as a $4\times4$ matrix acting on $(a,b,c,d)$.
  2. Row reduce to find $\operatorname{rank}T$ and a basis of $\ker T$. Describe the "ghost" image in words.
  3. Verify that $X_1=\begin{bmatrix}5&3\\2&6\end{bmatrix}$ and $X_2=\begin{bmatrix}7&1\\0&8\end{bmatrix}$ give identical measurements, and explain using $X_2-X_1$.
  4. Add one diagonal measurement $a+d$. Show that the new map $M_{2\times2}\to\mathbb{R}^5$ is one-to-one.
  5. For a $3\times3$ image with row and column sums only (6 measurements of 9 unknowns), use rank–nullity to find the smallest possible dimension of the kernel; then compute the actual rank.

Tools: Pencil and paper; a matrix calculator or Python (sympy) for step 5.

Hand in: A one-page note with your matrix, the kernel, the two look-alike images, and a sentence explaining why real scanners take projections from many angles.

Going further. Read about the Radon transform and the 1979 Nobel Prize in Physiology or Medicine (Cormack and Hounsfield). The same "kernel = invisible" idea appears in seismic imaging of the Earth and in electrical impedance tomography.

Instructor note: matrix rows $(1,1,0,0),(0,0,1,1),(1,0,1,0),(0,1,0,1)$; rank $3$ (row sums and column sums both total $a+b+c+d$), $\ker T=\operatorname{span}\{(1,-1,-1,1)\}$ — the checkerboard $\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$. $X_1$, $X_2$ both give row sums $8,8$ and column sums $7,9$; $X_2-X_1=2\times$ checkerboard. With $a+d$ added the rank is $4$. For $3\times3$: nullity $\ge9-6=3$, but the actual rank is $5$, so the nullity is $4$.

🧪 Self-check quiz

0/5 correct

Q1. For $T(\mathbf{x})=A\mathbf{x}$, the kernel $\ker T$ is

Q2. $T:\mathbb{R}^5\to\mathbb{R}^3$ is linear and onto. Then $\dim\ker T=$

Q3. A linear map $T:V\to W$ is one-to-one exactly when

Q4. A $4\times3$ matrix $A$ has $3$ pivots. The map $\mathbf{x}\mapsto A\mathbf{x}$ is

Q5. Which space is isomorphic to $P_3$?

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