A transformation $T$ from $U$ to $V$ assigns to each vector $x \in U$ a vector $T(x) \in V$, called the image of $x$ under $T$.
For $T: U \to V$, comparing solutions of $T(x) = b$ across every $b \in V$ tells us which of these $T$ is.
Consider $T:\mathbb{R}^3\to\mathbb{R}^2$, $(x,y,z)^\top \mapsto (2x-y+z,\ 2x-z)^\top$.
Recall $T:\mathbb{R}^3\to\mathbb{R}^2$, $(x,y,z)^\top \mapsto (2x-y+z,\ 2x-z)^\top$, shown to be linear in Section 02.
Column $i$ of $A$ is the image of the $i$-th standard basis vector $e_i$. Finding $x$ such that $T(x) = b$ is exactly solving the linear system $Ax = b$.
Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be linear with standard matrix $A$.
| Subspace of T | Equals |
|---|---|
| ker T | the null space of $A$, Nul $A$ |
| Im T | the column space of $A$, Col $A$ |
| coker T | the null space of $A^\top$, Nul $A^\top$ |
So every question about $T$ (whether it is injective, surjective, what it maps onto) reduces to a question about the four fundamental subspaces of its standard matrix $A$.
Let $T:\mathbb{R}^n \to \mathbb{R}^m$ be linear with standard matrix $A$, of size $m \times n$.
| T is injective when | T is surjective when | T is bijective when |
|---|---|---|
| ker T = {0} | – | ker T = {0} |
| – | Im T = $\mathbb{R}^m$ | Im T = $\mathbb{R}^m$ |
| – | coker T = {0} | coker T = {0} |
| columns of A are linearly independent | rows of A are linearly independent | columns and rows of A are linearly independent |
| rank(A) = n | rank(A) = m | rank(A) = n = m |
| Ax = b has at most one solution, for every b | Ax = b has at least one solution, for every b | Ax = b has exactly one solution, for every b |
A is $3\times2$ (more rows than columns), so $T:\mathbb{R}^2 \to \mathbb{R}^3$. Rank decides whether T is injective; it can never be surjective here.
A is $2\times3$ (more columns than rows), so $T:\mathbb{R}^3 \to \mathbb{R}^2$. Rank decides whether T is surjective; it can never be injective here.
Let $U$ have basis $B=\{b_1,\dots,b_n\}$ and $V$ have basis $C=\{c_1,\dots,c_m\}$, and let $T: U \to V$ be linear.
$A_{BC}$ turns the $B$-coordinates of any $v \in U$ into the $C$-coordinates of $T(v)$:
When $U=\mathbb{R}^n$, $V=\mathbb{R}^m$ and both use the standard basis, $A_{BC}$ is exactly the standard matrix $A$.
Every $P_{X\leftarrow Y}$ here denotes a change of basis matrix, converting $Y$-coordinates into $X$-coordinates. For more on how these matrices are built and used, see the Vector Spaces visual guide.
Follow the diagram: go up from B to E₃ (multiply by $P_{E_3\leftarrow B}$), across with $A$, then down from E₂ to C (multiply by $P_{C\leftarrow E_2}$). In the formula below, that same order reads right to left.
The transformation. Recall $T:\mathbb{R}^3\to\mathbb{R}^2$, $(x,y,z)^\top \mapsto (2x-y+z,\ 2x-z)^\top$, with standard matrix $A = \begin{pmatrix}2&-1&1\\2&0&-1\end{pmatrix}$.
Basis of $\mathbb{R}^3$. $B=\{b_1,b_2,b_3\}$, with $b_1=(1,1,-1)^\top$, $b_2=(0,1,1)^\top$, $b_3=(1,0,2)^\top$.
Basis of $\mathbb{R}^2$. $C=\{c_1,c_2\}$, with $c_1=(2,0)^\top$, $c_2=(1,-1)^\top$.
$P_{E_3\leftarrow B}$. Built from the columns of $B$ directly: $P_{E_3\leftarrow B} = \begin{pmatrix}1&0&1\\1&1&0\\-1&1&2\end{pmatrix}$.
$P_{C\leftarrow E_2}$. The inverse of $P_{E_2\leftarrow C} = \begin{pmatrix}2&1\\0&-1\end{pmatrix}$, giving $P_{C\leftarrow E_2} = \frac{1}{2}\begin{pmatrix}1&1\\0&-2\end{pmatrix}$.
Result. $A_{BC} = P_{C\leftarrow E_2}\, A\, P_{E_3\leftarrow B} = \frac{1}{2}\begin{pmatrix}3&-1&4\\-6&2&0\end{pmatrix}$.
If $T: U \to V$ and $S: V \to W$ are both linear, chaining them gives a new map $U \to W$.
$S\circ T$ is linear too. If $A$ is the standard matrix of $T$ and $B$ is the standard matrix of $S$, the standard matrix of $S\circ T$ is just the product $BA$.
Order matters: in $S\circ T$, $T$ acts first (it is closest to $u$), then $S$ acts on the result. That is why $A$ appears on the right in $BA$.
The straight arrows go step by step, $T$ then $S$. The curved arrow underneath is the shortcut $S\circ T$, doing both steps at once, first $T$ then $S$, read right to left in the name.
$T$. $T:\mathbb{R}^3\to\mathbb{R}^2$, $(x,y,z)^\top \mapsto (x+y,\ y+z)^\top$.
$S$. $S:\mathbb{R}^2\to P_2$, $(a,b)^\top \mapsto a + bt + (a-b)t^2$.
Apply $T$ first. $T(x,y,z) = (x+y,\ y+z)$.
Then apply $S$. $(S\circ T)(x,y,z) = (x+y) + (y+z)t + \bigl((x+y)-(y+z)\bigr)t^2$.
Simplify. $(x+y)-(y+z) = x - z$, so $(S\circ T)(x,y,z) = (x+y) + (y+z)t + (x-z)t^2$.
Matrix of $T$. Columns $T(e_1),T(e_2),T(e_3)$: $A = \begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}$.
Matrix of $S$. With respect to the basis $\{1,t,t^2\}$ of $P_2$, columns $S(1,0)$ and $S(0,1)$: $B = \begin{pmatrix}1&0\\0&1\\1&-1\end{pmatrix}$.
Product. The matrix of $S\circ T$ is $BA$, with $A$ on the right since $T$ acts first: $BA = \begin{pmatrix}1&0\\0&1\\1&-1\end{pmatrix}\begin{pmatrix}1&1&0\\0&1&1\end{pmatrix} = \begin{pmatrix}1&1&0\\0&1&1\\1&0&-1\end{pmatrix}$.
Reading off the rows of $BA$ against $(x,y,z)^\top$ gives back the same coefficients as the mapping approach: $x+y$, $y+z$, $x-z$.