Some linear maps can be undone and some can't. Rotating the plane by $90^\circ$? Rotate back. Projecting the plane onto the $x$-axis? Good luck: once $(3, 5)$ and $(3, -2)$ both land on $(3, 0)$, nothing can tell you which one you started with. This page sorts out exactly which maps can be undone, how to undo them with matrices, and then uses that to answer a surprisingly deep question: when are two vector spaces really the same?
Some moves can be undone and some can't. A shear? Shear back. A turn? Turn back. A projection onto the $x$-axis? Once $(3, 5)$ and $(3, -2)$ both land on $(3, 0)$, nothing can tell you which one you started with. This page sorts out which maps can be undone, and how to undo them with matrices.
Undoing a map means a second map that brings every output back to where it came from, and that the first map undoes in turn.
Let $\V$ and $\W$ be vector spaces, and let $T\colon \V \to \W$ be linear. A function $U\colon \W \to \V$ is an inverse of $T$ if $TU = I_{\W}$ and $UT = I_{\V}$. If $T$ has an inverse, then $T$ is invertible.
A function $U$ is an inverse of the linear map $T$ if doing $T$ and then $U$ changes nothing, and doing $U$ and then $T$ changes nothing: $UT = I$ and $TU = I$. If $T$ has an inverse, then $T$ is invertible.
An invertible map has only one inverse (two inverses $U$ and $U'$ give $U = UI = U(TU') = (UT)U' = U'$), so it gets a name: $T^{-1}$. Three facts about inverses come straight from the definition.
Let $T$ and $U$ be invertible functions for which $TU$ makes sense. Then
(a) $TU$ is invertible, and $(TU)^{-1} = U^{-1}T^{-1}$;
(b) $T^{-1}$ is invertible, and $(T^{-1})^{-1} = T$;
(c) a function is invertible exactly when it is both one-to-one and onto.
(a) $(U^{-1}T^{-1})(TU) = U^{-1}(T^{-1}T)U = U^{-1}U = I$, and $(TU)(U^{-1}T^{-1}) = T(UU^{-1})T^{-1} = TT^{-1} = I$. (b) $T^{-1}T = I$ and $TT^{-1} = I$ say exactly that $T$ is an inverse of $T^{-1}$.
(c) If $T$ has an inverse, $T(x) = T(y)$ gives $x = T^{-1}T(x) = T^{-1}T(y) = y$, so $T$ is one-to-one, and every $y$ is $T(T^{-1}(y))$, so $T$ is onto. Conversely, if $T$ is one-to-one and onto, each output $y$ came from exactly one input; sending $y$ to that input defines $T^{-1}$. $\blacksquare$
This is a visual argument, not a formal proof. To undo "$U$, then $T$," the last move has to come off first: undo $T$, then undo $U$. For (b), $T$ undoes $T^{-1}$ just as $T^{-1}$ undoes $T$. For (c), going back works only if every output came from exactly one input (one-to-one) and every output was actually reached (onto).
The full proof is in the n version.
Fact (a) is the one you'll use most. Here it is on two simple moves.
Let $U$ be the quarter turn $U(a_1, a_2) = (-a_2, \ a_1)$ and $T$ the stretch $T(a_1, a_2) = (3a_1, \ a_2)$. Find a formula for $(TU)^{-1}$.
Undo each move: $T^{-1}(a_1, a_2) = (\tfrac{1}{3}a_1, \ a_2)$ squashes back, and $U^{-1}(a_1, a_2) = (a_2, \ -a_1)$ turns back. The last move done was $T$, so it comes off first:
$$(TU)^{-1}(a_1, a_2) = U^{-1}\big(T^{-1}(a_1, a_2)\big) = U^{-1}(\tfrac{1}{3}a_1, \ a_2) = (a_2, \ -\tfrac{1}{3}a_1).$$Check on $e_1$: $TU(1, 0) = T(0, 1) = (0, 1)$, and $(TU)^{-1}(0, 1) = (1, 0)$.
$(TU)^{-1} = U^{-1}T^{-1}$. To undo "socks, then shoes," you take off the shoes first. I did not need linear algebra for this, and yet here we are.
Fact (c) is where the last two pages pay off. One-to-one is about the null space, onto is about the range, and when the dimensions match, those two come together.
Let $T\colon \V \to \W$ be linear, where $\V$ and $\W$ are finite-dimensional vector spaces of equal dimension. Then $T$ is invertible if and only if $\N(T) = \{0\}$, if and only if $\rank(T) = \dim(\V)$.
By (c) of the last theorem, $T$ is invertible exactly when it is one-to-one and onto. With $\dim(\V) = \dim(\W)$, one-to-one, onto, and $\rank(T) = \dim(\V)$ are all the same thing (Null Space & Range), and one-to-one means $\N(T) = \{0\}$ (same page). $\blacksquare$
A linear map of $\R^2$ to itself, or of $\R^3$ to itself, is invertible exactly when it flattens nothing: the only arrow it sends to $0$ is $0$ itself. Equivalently, its rank is $2$ (or $3$).
This is a visual argument, not a formal proof. If nothing is flattened, the unit cube goes to a solid slanted box, and space is just stretched and slanted; every point has one place it came from, so the move can be played backward. If something is flattened, a whole segment of points lands on one point, and nothing can split them up again.
The full proof is in the n version.
So deciding invertibility is a null-space computation.
Which of $T(a_1, a_2) = (a_1 + 2a_2, \ 2a_1 + 4a_2)$ and $U(a_1, a_2) = (a_1 + 2a_2, \ 2a_1 + 3a_2)$ is invertible?
$T(2, -1) = (0, 0)$: a nonzero arrow is flattened, so $T$ is not invertible (its second output is always twice its first). For $U$, $U(a_1, a_2) = (0, 0)$ means $a_1 + 2a_2 = 0$ and $2a_1 + 3a_2 = 0$; subtracting twice the first from the second gives $-a_2 = 0$, so $a_2 = 0$ and $a_1 = 0$. Only $0$ goes to $0$, so $U$ is invertible.
The hypothesis about dimensions is easy to drop.
Don't drop "of equal dimension." A map $\R^2 \to \R^3$ can have rank $2$ and be one-to-one, but its range is only a plane in $\R^3$, so it isn't onto and can't be inverted. Rank equal to the input dimension is enough only when the input and output spaces are the same size.
Watch the difference. Each map sends the F somewhere; then we try to send it back.
An inverse is just a function, a priori. The good news is that for linear maps you never have to check linearity separately.
If $T$ is linear and invertible, then $T^{-1}$ is linear.
Let $y_1, y_2$ be outputs and $c \in \R$. Since $T$ is one-to-one and onto, there are unique $x_1, x_2$ with $T(x_1) = y_1$ and $T(x_2) = y_2$, so $x_1 = T^{-1}(y_1)$ and $x_2 = T^{-1}(y_2)$. Then
$$T^{-1}(cy_1 + y_2) = T^{-1}\big(T(cx_1 + x_2)\big) = cx_1 + x_2 = cT^{-1}(y_1) + T^{-1}(y_2). \quad\blacksquare$$This is a short visual argument, not a formal proof. $T$ sends the parallelogram on $x_1, x_2$ to the parallelogram on $T(x_1), T(x_2)$, far corner to far corner. Playing $T$ backward sends that image parallelogram straight back, far corner to far corner, so $T^{-1}$ sends sums to sums too (and stretched arrows to stretched arrows).
The full proof is in the n version.
A square matrix $A$ is invertible when $AB = BA = I$ for some $B$, and a $2 \times 2$ matrix has the formula $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ when $ad - bc \neq 0$.
Only square matrices get inverses, and that matches maps: an invertible map can't change the dimension.
Let $T\colon \V \to \W$ be linear and invertible. Then $\V$ is finite-dimensional if and only if $\W$ is, and in that case $\dim(\V) = \dim(\W)$.
If $\V$ has a finite basis $\beta$, then $T(\beta)$ spans $\Rg(T) = \W$ (Null Space & Range), so $\W$ is finite-dimensional; the same argument with $T^{-1}$ goes the other way. Then $\nullity(T) = 0$ and $\rank(T) = \dim(\W)$, so the Dimension Theorem gives $\dim(\V) = 0 + \dim(\W)$. $\blacksquare$
An invertible linear map goes from a space to one of the same dimension: $\R^2$ to $\R^2$, $\R^3$ to $\R^3$ (or a plane to a plane, a line to a line).
This is a visual argument, not a formal proof. Going down in dimension ($\R^3$ to $\R^2$) always flattens something, so it can't be undone. Going up ($\R^2$ to $\R^3$) can only reach a plane, so most of space is never reached and there's nothing to send back. Only same-size moves survive both tests.
The full proof is in the n version.
So undoing a map and inverting its matrix are the same job.
Let $\V$ and $\W$ be finite-dimensional vector spaces with ordered bases $\beta$ and $\gamma$, and let $T\colon \V \to \W$ be linear. Then $T$ is invertible if and only if $[T]_\beta^\gamma$ is invertible, and then $[T^{-1}]_\gamma^\beta = ([T]_\beta^\gamma)^{-1}$. In particular, for $T\colon \R^n \to \R^n$, $[T^{-1}] = [T]^{-1}$.
Suppose $T$ is invertible. By the last theorem $\dim(\V) = \dim(\W) = n$, so $[T]_\beta^\gamma$ is $n \times n$. By the matrix of a composition,
$$I_n = [I_\V]_\beta = [T^{-1}T]_\beta = [T^{-1}]_\gamma^\beta[T]_\beta^\gamma,$$and similarly $[T]_\beta^\gamma[T^{-1}]_\gamma^\beta = [I_\W]_\gamma = I_n$. So $[T]_\beta^\gamma$ is invertible with inverse $[T^{-1}]_\gamma^\beta$.
Conversely, suppose $A = [T]_\beta^\gamma$ has an inverse $B$. There is a linear $U\colon \W \to \V$ with $[U]_\gamma^\beta = B$ (send each vector of $\gamma$ where the columns of $B$ say; a map is decided by a basis). Then $[UT]_\beta = BA = I_n = [I_\V]_\beta$, so $UT = I_\V$, and likewise $TU = I_\W$. $\blacksquare$
A linear map $T$ of $\R^2$ (or $\R^3$) is invertible exactly when its matrix $[T]$ is, and then $[T^{-1}] = [T]^{-1}$.
This is a visual argument, not a formal proof. "Do $T$, then $T^{-1}$" is the move that changes nothing, and its matrix is $[T^{-1}][T]$ (the matrix of a chain is the product, first move on the right). The matrix of "change nothing" is $I$. So $[T^{-1}][T] = I$, and the same in the other order: $[T^{-1}]$ is the inverse matrix.
The full proof is in the n version.
For a map given by a formula, that's a recipe: read off the matrix, invert it, read the formula back.
Let $T\colon \R^2 \to \R^2$ be $T(a_1, a_2) = (3a_1 + a_2, \ 5a_1 + 2a_2)$. Show $T$ is invertible and find a formula for $T^{-1}$.
The standard matrix is $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$, and $ad - bc = 6 - 5 = 1 \neq 0$, so $A$ is invertible, and so is $T$. The formula gives
$$A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}, \quad\text{so}\quad T^{-1}(a_1, a_2) = (2a_1 - a_2, \ -5a_1 + 3a_2).$$Check one: $T(1, 0) = (3, 5)$, and $T^{-1}(3, 5) = (6 - 5, \ -15 + 15) = (1, 0)$. Back where we started.
Sometimes the move is simple enough that you can undo it without any formula.
Let $S\colon \R^3 \to \R^3$ be the shear $S(a_1, a_2, a_3) = (a_1 + 2a_3, \ a_2, \ a_3)$. Find $S^{-1}$ and check it with matrices.
$S$ slides each point sideways by twice its height and leaves the height alone. To undo it, slide back: $S^{-1}(a_1, a_2, a_3) = (a_1 - 2a_3, \ a_2, \ a_3)$. As matrices,
$$\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$Undo the move and the matrix takes care of itself.
Whichever way you get it, the check is cheap.
Multiply your answer by the original and look for $I$. For the $2 \times 2$ formula, the three moves each get forgotten: swap the diagonal, negate the off-diagonal, divide by $ad - bc$. And never use the transpose as an inverse unless you know the map is a rotation or a reflection.
Your turn: undo a map at one point. In free response, type the vector, like 2, -1.
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And undo two maps in a row, which is fact (a) with matrices: $(AB)^{-1} = B^{-1}A^{-1}$. Type the matrix row by row, halves like 1/2.
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An invertible linear map matches up two spaces perfectly: every vector of one gets exactly one partner in the other, and sums and stretches go along for the ride. As far as linear algebra can tell, the two spaces are the same.
Vector spaces $\V$ and $\W$ are isomorphic, written $\V \cong \W$, if there is an invertible linear transformation $T\colon \V \to \W$. Such a $T$ is an isomorphism from $\V$ onto $\W$.
Two of our spaces (lines or planes through the origin, $\R^2$, $\R^3$) are isomorphic if some linear map from one onto the other is invertible. Such a map is an isomorphism.
So when are two spaces isomorphic? The answer is shockingly simple: count.
Let $\V$ and $\W$ be finite-dimensional vector spaces. Then $\V \cong \W$ if and only if $\dim(\V) = \dim(\W)$. In particular, $\V \cong \R^n$ exactly when $\dim(\V) = n$.
If $T\colon \V \to \W$ is an isomorphism, $\dim(\V) = \dim(\W)$ by the theorem on dimensions above.
Conversely, let $\beta = \{v_1, \ldots, v_n\}$ and $\gamma = \{w_1, \ldots, w_n\}$ be bases. There is a linear $T$ with $T(v_i) = w_i$ (a map is decided by a basis), and $\Rg(T) = \span(T(\beta)) = \span(\gamma) = \W$, so $T$ is onto. With equal dimensions it is also one-to-one, hence invertible. $\blacksquare$
Two of our spaces are isomorphic exactly when they have the same dimension. Every line through the origin is isomorphic to $\R$, every plane through the origin to $\R^2$.
This is a visual argument, not a formal proof. A plane through the origin has a basis $v_1, v_2$, and $(a, b) \mapsto av_1 + bv_2$ lays the square grid of $\R^2$ onto the plane's slanted grid, one point to one point, nothing missed: an isomorphism. Different dimensions can't be matched, since an invertible map can't change the dimension.
The full proof is in the n version.
Every plane through the origin in $\R^3$, every $2$-dimensional solution set, every space with a basis of two vectors: they're all "$\R^2$ wearing a costume." Dimension is the only thing that tells these spaces apart.
See it with a plane in $\R^3$. Two vectors span it, and $(a, b) \mapsto av_1 + bv_2$ is an isomorphism from $\R^2$ onto the plane. Drag the point on the left and its twin follows on the right.
Finding an isomorphism is finding a basis.
Is $\W = \{(x_1, x_2, x_3, x_4) \in \R^4 : x_1 + x_2 + x_3 + x_4 = 0\}$ isomorphic to $\R^3$? If so, write down an isomorphism.
Solve the one equation for $x_1 = -x_2 - x_3 - x_4$; then $x_2$, $x_3$, $x_4$ are free, so $\dim(\W) = 3$, and $\W \cong \R^3$. An isomorphism comes straight from the free variables: $T(a, b, c) = (-a - b - c, \ a, \ b, \ c)$.
Is the plane $x + y + z = 0$ isomorphic to $\R^2$? If so, write down an isomorphism.
Solve for $z = -x - y$: the plane's points are $(x, y, -x - y)$, two free numbers, so it has dimension $2$ and is isomorphic to $\R^2$. An isomorphism: $T(a, b) = (a, \ b, \ -a - b)$. It's linear, it reaches every point of the plane, and only $(0, 0)$ goes to $0$.
Two last results tie the chapter together. Picking bases turns every linear map into a matrix, and that assignment is itself an isomorphism.
Let $\V$ and $\W$ be finite-dimensional vector spaces of dimensions $n$ and $m$, with ordered bases $\beta$ and $\gamma$. Then $\Phi\colon \Lin(\V, \W) \to M_{m \times n}(\R)$, $\Phi(T) = [T]_\beta^\gamma$, is an isomorphism. In particular, $\dim \Lin(\V, \W) = mn$.
$\Phi$ is linear by the matrix of a sum. For each $m \times n$ matrix $A$ there is exactly one linear $T$ with $T(v_j) = \sum_i A_{ij}w_i$ for every $j$ (a map is decided by a basis), and that says $[T]_\beta^\gamma = A$. So $\Phi$ is one-to-one and onto. $M_{m \times n}(\R)$ has dimension $mn$, one for each entry. $\blacksquare$
So counting maps is counting matrix entries.
Find $\dim \Lin(\R^2, \R^3)$ and a basis for it.
With standard bases, $\Phi$ matches $\Lin(\R^2, \R^3)$ with the $3 \times 2$ matrices, so the dimension is $3 \cdot 2 = 6$. A basis: the six maps $E_{ij}$ whose matrix has a single $1$ in row $i$, column $j$, like $E_{21}(a_1, a_2) = (0, \ a_1, \ 0)$. Every linear map is one combination of them: $T = \sum_{i,j} [T]_{ij}E_{ij}$.
The other result is the isomorphism we've used without naming it since coordinate vectors showed up.
Let $\beta$ be an ordered basis for an $n$-dimensional vector space $\V$. The standard representation of $\V$ with respect to $\beta$ is $\phi_\beta\colon \V \to \R^n$, $\phi_\beta(x) = [x]_\beta$.
Taking coordinates is a linear map, and it loses nothing.
For any finite-dimensional vector space $\V$ with ordered basis $\beta$, $\phi_\beta$ is an isomorphism.
$\phi_\beta$ is linear because coordinates respect sums and multiples (Bases & Dimension). If $[v]_\beta = 0$, every coefficient of $v$ is $0$, so $v = 0$: $\phi_\beta$ is one-to-one, and with $\dim(\V) = n = \dim(\R^n)$ it is also onto. $\blacksquare$
Now picture it. A linear $T\colon \V \to \W$ with bases $\beta$, $\gamma$ and $A = [T]_\beta^\gamma$ gives two routes from $\V$ down to $\R^m$: apply $T$ and then take $\gamma$-coordinates, or take $\beta$-coordinates and then multiply by $A$. The matrix computes the map, so the routes agree: $A\phi_\beta = \phi_\gamma T$.
That square is why matrices are worth all the fuss. Any question about a linear map between finite-dimensional spaces can be carried down to $\R^n$ and $\R^m$, answered there with a matrix, and carried back up.
An invertible matrix is a lock with a key. Turn letters into numbers, scramble pairs of them with a matrix, and only someone with the inverse can unscramble them.
Number the letters $A = 1, B = 2, \ldots, Z = 26$, and code each pair of letters $x$ with $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$. Code the word MATH, and then decode $(25, 17)$.
MATH is the pairs $(13, 1)$ and $(20, 8)$. Coding:
$$A\begin{pmatrix} 13 \\ 1 \end{pmatrix} = \begin{pmatrix} 27 \\ 14 \end{pmatrix}, \qquad A\begin{pmatrix} 20 \\ 8 \end{pmatrix} = \begin{pmatrix} 48 \\ 28 \end{pmatrix}.$$To decode, undo $A$. Here $ad - bc = 1$, so $A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ has whole entries, and
$$A^{-1}\begin{pmatrix} 25 \\ 17 \end{pmatrix} = \begin{pmatrix} 8 \\ 9 \end{pmatrix}: \text{ the letters H and I.}$$Check: $A^{-1}(27, 14) = (13, 1)$, back to M and A.
The lock has to be one you can open.
A coding matrix must be invertible, or two messages get the same code. With $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ ($ad - bc = 0$), the pairs $(2, 1)$ and $(4, 0)$ both code to $(4, 8)$, and no key can tell them apart. Matrices with $ad - bc = \pm 1$ are the friendliest locks: their keys have whole-number entries.
The matrix of a map depends on the bases. What happens to it when you switch from one basis to another? The switch is itself an invertible matrix, and the next page works out exactly how it acts.
The toolkit for this page: $T$ is invertible when it's one-to-one and onto; with equal dimensions, that's $\N(T) = \{0\}$, or full rank. Undo in reverse order, $(TU)^{-1} = U^{-1}T^{-1}$. $T$ is invertible exactly when its matrix is, and $[T^{-1}] = [T]^{-1}$. Spaces of the same dimension are isomorphic.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. This set is multiple choice.
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Made it this far? Time for something with a little heat. Three moves in a row, and you have to undo all of them. In free response, type the matrix row by row, halves like 1/2.
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