A matrix comes with four subspaces, and once you can see them, most of what a matrix does becomes a picture. Two of them live on the input side: the span of the rows, and the inputs that get sent to $0$. Two live on the output side: the span of the columns, and the outputs' leftover directions. Each pair meets at a right angle, their dimensions add up to the size of the space, and the matrix carries the first space of one pair onto the first space of the other. This page builds all four, with the dimension tools from the last page doing the counting.
Take a $3 \times 3$ matrix $A$ and think of it as a machine: a point $x$ goes in, the point $Ax$ comes out. Four pictures come with it, all planes and lines through the origin. On the input side: the plane the rows span, and the line of inputs the machine squashes to $0$. On the output side: the plane the columns span (every output lands in it), and the line of directions no output ever reaches. The two planes are the machine's real business, and this page shows how it pairs them up.
Everything here uses one running example, the matrix
$$A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix},$$whose third row is the sum of the first two. Its rank is $2$, which is the number that decides every dimension on this page.
Elimination works by taking combinations of rows (Systems). Every row you ever write down during elimination is a combination of the rows you started with. So the right place to watch elimination is the space those combinations fill.
The row space of an $m \times n$ matrix $A$, written $\row(A)$, is the span of its rows: a subspace of $\R^n$.
The row space of a matrix $A$ with $3$ columns, written $\row(A)$, is the span of its rows, read as arrows in $\R^3$: a point, a line, a plane, or all of space. (A matrix with $2$ columns has its row space in $\R^2$.)
Its dimension is called the row rank, and the first job is to find a basis for it. Here's the smallest possible example.
Find the row space of $A = \begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}$.
The row space is $\{c_1(2, 3) + c_2(4, 6) : c_1, c_2 \in \R\}$. Since $(4, 6) = 2(2, 3)$, every combination is $(c_1 + 2c_2)(2, 3)$, a multiple of $(2, 3)$. So $\row(A)$ is the line through $(2, 3)$, with basis $\{(2, 3)\}$ and dimension $1$.
Both rows point the same way, so their combinations never leave one line.
For bigger matrices you don't want to eyeball multiples; you want elimination. And elimination is safe to use here, because of this.
If $A$ row reduces to $B$, then $\row(A) = \row(B)$. Row operations never change the row space.
One operation at a time. A swap leaves the set of rows alone. Scaling $R_i$ by $c \neq 0$ or replacing $R_i$ by $R_i + kR_j$ makes every row of $B$ a combination of rows of $A$, so every row of $B$ is in $\row(A)$, and then every combination of rows of $B$ is too: $\row(B) \subseteq \row(A)$. Each operation can be undone by another row operation (scale by $1/c$, add $-kR_j$), so the same argument run backward gives $\row(A) \subseteq \row(B)$. Equal after one operation, so equal after any number of them. $\blacksquare$
This is a visual argument, not a formal proof. The rows span a plane. Adding a multiple of one row to another slides a tip around inside that plane; scaling a row slides it along its own line; swapping changes nothing. The plane the rows span never moves, and since every move can be undone, it never shrinks either.
The full proof is in the n version.
So reduce all the way and read the row space off the echelon form, where the dead weight has already been cleared out.
The nonzero rows of a row echelon form of $A$ are a basis for $\row(A)$. In particular $\dim \row(A) = \rank(A)$, the number of pivots.
Let $E$ be an echelon form of $A$. By the theorem above, $\row(A) = \row(E)$, and the zero rows of $E$ contribute nothing to a span, so the nonzero rows span $\row(A)$.
They are independent: suppose $c_1E_1 + \cdots + c_kE_k = 0$ for the nonzero rows $E_1, \ldots, E_k$. Look at the pivot column of $E_1$. In echelon form every later row has a $0$ there, so that coordinate of the combination is $c_1 \cdot (\text{pivot}) = 0$, forcing $c_1 = 0$. Now look at $E_2$'s pivot column, where $E_3, \ldots, E_k$ are $0$: it forces $c_2 = 0$. Continue down the rows. Only the trivial relation, so the nonzero rows are a basis, and there is one per pivot. $\blacksquare$
This is a visual argument, not a formal proof. Elimination slides rows around inside their span and collapses any row that was a combination of the others onto $0$. What's left is a list of arrows that still spans the same plane (or line, or space) and that has no dead weight: each nonzero echelon row has a pivot where the rows below it are $0$, so it points in a direction the later rows can't reach. That's a basis, and the next example draws it.
The full proof is in the n version.
On the running example, that's one reduction.
Find a basis for the row space of $A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix}$.
Reduce:
$$\begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix} \xrightarrow{R_3 \to R_3 - R_1} \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 0 & -2 & 2 \end{pmatrix} \xrightarrow{R_3 \to R_3 - R_2} \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 0 & 0 & 0 \end{pmatrix} \xrightarrow[R_1 \to R_1 - \frac{1}{2}R_2]{R_1 \to \frac{1}{2}R_1, \ R_2 \to -\frac{1}{2}R_2} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}.$$Two pivots, two nonzero rows. A basis for $\row(A)$ is $\{(1, 0, 1), (0, 1, -1)\}$, and $\dim \row(A) = 2$: the row space is a plane through the origin. (The two original rows $(2, 1, 1)$ and $(0, -2, 2)$ are a basis for the same plane; the third row, $(2, -1, 3)$, is their sum, which is why it collapsed to $0$.)
Watch the third row collapse, and the echelon rows settle into the same plane.
One thing to watch when you write the basis down.
The nonzero echelon rows are a basis for the row space, but they are usually not rows of $A$: elimination mixed them. Any independent set of $A$'s own rows that spans the same space is a basis too. Columns will behave differently: the pivot columns of $A$ really are columns of $A$ (Bases & Dimension), while the columns of the echelon form are not.
Now the other direction. The columns of $A$ are what the column picture combines when you compute $Ax$, so their span is the set of everything $A$ can output.
The column space of an $m \times n$ matrix $A$, written $\col(A)$, is the span of its columns: a subspace of $\R^m$.
The column space of a matrix $A$ with $3$ rows, written $\col(A)$, is the span of its columns, read as arrows in $\R^3$: a point, a line, a plane, or all of space. (A matrix with $2$ rows has its column space in $\R^2$.)
Its dimension is the column rank. The reason to care about it is a system of equations.
$\col(A)$ is exactly the set of outputs $\{Ax : x \in \R^n\}$. So $Ax = b$ has a solution exactly when $b \in \col(A)$.
By the column picture, $Ax = x_1c_1 + \cdots + x_nc_n$ where $c_1, \ldots, c_n$ are the columns. So every output is a combination of the columns, and every combination of the columns is the output for the $x$ made of its coefficients. That's $\col(A)$, and $Ax = b$ is solvable exactly when $b$ is one of those outputs. $\blacksquare$
This is a visual argument, not a formal proof. $Ax$ is a chain of the columns, $x_1$ of the first, $x_2$ of the second, $x_3$ of the third, so the outputs are exactly the points the columns' chains reach: their span. A right side $b$ is reachable when it sits in that plane and unreachable when it doesn't, which the next example draws.
The full proof is in the n version.
So "is this system solvable?" is "is this right side in the column space?". Here it is on the running example, with one right side that works and one that doesn't.
For $A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix}$, find a basis for $\col(A)$, and decide whether $Ax = b$ is solvable for $b = (3, -2, 1)$ and for $b' = (1, 0, 0)$.
The columns are $c_1 = (2, 0, 2)$, $c_2 = (1, -2, -1)$, $c_3 = (1, 2, 3)$. From the reduction above, the pivots are in columns $1$ and $2$, so $\{c_1, c_2\}$ is a basis for $\col(A)$ and $\dim \col(A) = 2$: a plane. The third column is a combination of the first two, $c_3 = c_1 - c_2$ (check: $(2 - 1, 0 + 2, 2 + 1) = (1, 2, 3)$).
$b = (3, -2, 1)$ is $c_1 + c_2$, so it's in the column space and $x = (1, 1, 0)$ solves $Ax = b$. For $b' = (1, 0, 0)$, try $x_1c_1 + x_2c_2 = (1, 0, 0)$: the second coordinate gives $x_2 = 0$, then the first gives $x_1 = \tfrac{1}{2}$, and the third gives $2x_1 - x_2 = 1 \neq 0$. No solution: $b'$ is off the plane.
The column plane, one right side on it and one off it.
Here's a trap. Row operations preserved the row space, so you might hope they preserve the column space too. They don't.
Row operations can change the column space of a matrix, but they never change its dimension: if $A$ row reduces to $B$, then $\dim \col(A) = \dim \col(B)$.
That the column space can change: $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \to \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$ by $R_2 \to R_2 - 2R_1$. The first column space is the line $y = 2x$; the second is the $x$-axis.
That the dimension doesn't: $Ax = 0$ and $Bx = 0$ have the same solutions (row operations preserve solutions), and by the column picture a solution $x$ is a relation $x_1c_1 + \cdots + x_nc_n = 0$ among the columns. So $A$'s columns and $B$'s columns satisfy exactly the same relations, in the same positions. The pivot columns of $B$ are independent and every other column of $B$ is a combination of them; the same relations say the same about $A$'s columns in those positions. Both column spaces have a basis of the same size, the number of pivots. $\blacksquare$
This is a visual argument, not a formal proof. A row operation moves the columns (it rewrites one coordinate of every column at once), so the line or plane they span can swing somewhere else. But a chain of columns that came home before still comes home after, because row operations keep the solutions of $Ax = 0$. Same relations, same number of independent columns, same dimension.
The full proof is in the n version.
So to find a basis for the column space, reduce, find the pivot columns, and go back to the original columns in those positions. Never take the echelon form's columns. Try both spaces now. In free response, type any vector in the space asked for, like 3, 3, 6; the checker tests it.
Loading practice engine...
The row space of our $A$ had dimension $2$. So did its column space. Rows and columns of a non-square matrix don't even live in the same space, and yet this always happens.
For every matrix $A$, $\dim \row(A) = \dim \col(A) = \rank(A)$. In particular $\rank(A^t) = \rank(A)$.
Reduce $A$ to its reduced row echelon form $R$. The row rank is the number of nonzero rows of $R$ (the theorem above), which is the number of pivots. The column rank is the number of pivot columns (the previous theorem, or the Bases & Dimension page), which is also the number of pivots. Both equal $\rank(A)$, the pivot count.
Transposing turns the rows of $A$ into the columns of $A^t$, so $\col(A^t) = \row(A)$ as sets of vectors, and $\rank(A^t) = \dim \col(A^t) = \dim \row(A) = \rank(A)$. $\blacksquare$
This is a visual argument, not a formal proof. In the reduced echelon form, the nonzero rows are counted by the pivots, and so are the pivot columns. One count, two bases.
The full proof is in the n version.
The Rank & Inverses page defined the rank as a pivot count and promised it was a dimension. Now it's two: the dimension of the row space and of the column space. And since $\rank(A^t) = \rank(A)$, a square matrix is invertible exactly when its rows are independent, exactly when its columns are.
Here's the theorem doing work on a matrix that isn't square, where the two spaces really do live in different places.
Find the row rank and the column rank of $B = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 7 \end{pmatrix}$.
Reduce: $R_2 \to R_2 - 2R_1$ gives $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 1 \end{pmatrix}$, two pivots. The row space is a plane in $\R^3$ with basis $\{(1, 2, 3), (0, 0, 1)\}$: row rank $2$. The pivot columns are the first and third, so the column space has basis $\{(1, 2), (3, 7)\}$, two independent vectors in $\R^2$: column rank $2$, which means the column space is all of $\R^2$. Rows in $\R^3$, columns in $\R^2$, and both spaces have dimension $2$. (The second column, $(2, 4)$, is twice the first, which is why only two of the three columns are needed.)
Third subspace: the inputs that $A$ throws away.
The null space of an $m \times n$ matrix $A$, written $\N(A)$, is the set of all $x$ in $\R^n$ with $Ax = 0$: the solution set of the homogeneous system.
The null space of a matrix $A$ with $3$ columns, written $\N(A)$, is the set of all points $x$ in $\R^3$ that $A$ sends to $0$: the solutions of $Ax = 0$.
You've met this set twice: as the solutions of a homogeneous system on Solution Sets, where every solution set of $Ax = b$ was a shifted copy of it, and as the null space of a linear map. It is a subspace, because $Ax = 0$ and $Ay = 0$ give $A(x + y) = 0$ and $A(cx) = 0$.
Its dimension is the number you've been counting as "free unknowns" since the Systems page. Now it's a theorem.
Let $A$ be $m \times n$ with $\rank(A) = r$. Then $\dim \N(A) = n - r$, and a basis is obtained from the reduced row echelon form by setting one free unknown to $1$ and the others to $0$, once for each free unknown.
Reduce $A$ to $R$; the solutions don't change. $R$ has $r$ pivots, so $r$ pivot unknowns and $n - r$ free unknowns $x_{f_1}, \ldots, x_{f_{n-r}}$. Each pivot row of $R$ says one pivot unknown equals a combination of the free unknowns (reading the reduced form), so the general solution is $x = x_{f_1}v_1 + \cdots + x_{f_{n-r}}v_{n-r}$, where $v_j$ is the solution with $x_{f_j} = 1$ and the other free unknowns $0$. So $v_1, \ldots, v_{n-r}$ span $\N(A)$.
They are independent: in the combination $x_{f_1}v_1 + \cdots + x_{f_{n-r}}v_{n-r}$, the coordinate in position $f_j$ is exactly $x_{f_j}$ (only $v_j$ has a nonzero entry there). So the combination is $0$ only when every coefficient is $0$. A basis of $n - r$ vectors. $\blacksquare$
This is a visual argument, not a formal proof. Each free unknown is a dial you can turn; set it to $1$ with the other dials at $0$ and you get one arrow of solutions. One dial, one arrow, a line of solutions; two dials, two arrows pointing different ways, a plane of solutions. The arrows can't be combinations of each other, because each one has a $1$ in a slot where the others have $0$.
The full proof is in the n version.
For the running example there's one free unknown, so one basis vector.
Find a basis for $\N(A)$, with $A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix}$ as before.
The reduced form is $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$: pivots in columns $1$ and $2$, so $x_3$ is free, $x_1 = -x_3$, and $x_2 = x_3$. Setting $x_3 = 1$ gives $(-1, 1, 1)$, so
$$\N(A) = \operatorname{span}\{(-1, 1, 1)\}, \qquad \dim \N(A) = 3 - 2 = 1.$$Check: $A(-1, 1, 1) = (-2 + 1 + 1, \ -2 + 2, \ -2 - 1 + 3) = (0, 0, 0)$. The null space is a line through the origin.
Slide a point along that line and every row's dot product with it stays $0$.
Your turn. In free response, type any nonzero vector with $Ax = 0$, like -2, 1, 0; the checker multiplies it out.
Loading practice engine...
In the example, the null space was the line through $(-1, 1, 1)$ and the row space was the plane of $(2, 1, 1)$ and $(0, -2, 2)$. Dot them: $(2, 1, 1) \cdot (-1, 1, 1) = 0$ and $(0, -2, 2) \cdot (-1, 1, 1) = 0$. That's not luck; it's what $Ax = 0$ says.
Every vector of $\row(A)$ is orthogonal to every vector of $\N(A)$, the only vector in both is $0$, and their dimensions add up to $n$: $\dim \row(A) + \dim \N(A) = r + (n - r) = n$.
$Ax = 0$ says, entry by entry, that $r_i \cdot x = 0$ for every row $r_i$. So a null-space vector $x$ is orthogonal to every row, and then to every combination of rows: $(a_1r_1 + \cdots + a_mr_m) \cdot x = a_1(r_1 \cdot x) + \cdots + a_m(r_m \cdot x) = 0$.
If $x$ is in both spaces, it is orthogonal to itself: $x \cdot x = 0$, which forces $x = 0$. The dimensions are the two theorems above: $\dim \row(A) = r$ and $\dim \N(A) = n - r$. $\blacksquare$
This is a visual argument, not a formal proof. "$Ax = 0$" is a list of dot products: each row dotted with $x$ is $0$. So a solution stands square to every row, and square to everything in the plane the rows span. Plane and line, at a right angle, meeting only at the origin: $2 + 1 = 3$.
The full proof is in the n version.
The same thing happens on the output side, with the fourth subspace. It's the null space of $A^t$, and the name comes from where the vector sits when you write the equation.
The left null space of an $m \times n$ matrix $A$ is $\N(A^t)$: the set of all $y$ in $\R^m$ with $y^tA = 0$, equivalently $A^ty = 0$. It is a subspace of $\R^m$.
The left null space of a matrix $A$ with $3$ rows is $\N(A^t)$: the set of all $y$ in $\R^3$ with $y^tA = 0$, that is, $y$ dotted with every column of $A$ is $0$.
Before the matching theorem, find it for the running example.
Find a basis for the left null space of $A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & 3 \end{pmatrix}$.
Solve $A^ty = 0$, that is, $y \cdot c_j = 0$ for each column: $2y_1 + 2y_3 = 0$, $y_1 - 2y_2 - y_3 = 0$, $y_1 + 2y_2 + 3y_3 = 0$. The first gives $y_1 = -y_3$; the second then gives $-y_3 - 2y_2 - y_3 = 0$, so $y_2 = -y_3$; and the third checks out: $-y_3 - 2y_3 + 3y_3 = 0$. With $y_3 = -1$: $y = (1, 1, -1)$.
$$\N(A^t) = \operatorname{span}\{(1, 1, -1)\}, \qquad \dim \N(A^t) = 3 - 2 = 1.$$There's a shortcut hiding here. $y^tA = 0$ says $y_1r_1 + y_2r_2 + y_3r_3 = 0$: a left null vector is a relation among the rows. Our third row is $r_1 + r_2$, so $r_1 + r_2 - r_3 = 0$ and $y = (1, 1, -1)$ reads straight off the relation.
Three dot products with the columns, three zeros: $y$ stands square to the column plane.
The pattern is exactly the one from the input side, with the matrix turned on its side.
Every vector of $\col(A)$ is orthogonal to every vector of $\N(A^t)$, the only vector in both is $0$, and $\dim \col(A) + \dim \N(A^t) = r + (m - r) = m$.
Apply the previous theorem to $A^t$, which is $n \times m$ with $\rank(A^t) = r$: $\row(A^t)$ is orthogonal to $\N(A^t)$, they meet only at $0$, and their dimensions add to $m$. And $\row(A^t) = \col(A)$, since the rows of $A^t$ are the columns of $A$. $\blacksquare$
This is a visual argument, not a formal proof. It's the same picture with the matrix flipped on its side: the columns of $A$ are the rows of $A^t$, so the left null space stands square to the column plane the way the null space stood square to the row plane. The next example draws it for our $A$. The full proof is in the n version.
"Left null space." Named for which side of $A$ the $y$ sits on. The null space should have been called the right null space. The committee was clearly in a hurry.
Count all four at once. In free response, type the four dimensions in order, separated by commas.
Loading practice engine...
Put the four pieces together and you can describe the whole action of $A$ in one sentence: it ignores the null space and carries the row space onto the column space, without doubling up.
Let $A$ be $m \times n$. Every $x$ in $\R^n$ can be written in exactly one way as $x = x_r + x_n$ with $x_r \in \row(A)$ and $x_n \in \N(A)$, and then $Ax = Ax_r$. Restricted to the row space, $A$ is one-to-one and its outputs are all of $\col(A)$: distinct vectors of $\row(A)$ have distinct outputs, and every vector of $\col(A)$ is the output of exactly one vector of $\row(A)$.
The split. Take a basis $u_1, \ldots, u_r$ of $\row(A)$ and a basis $v_1, \ldots, v_{n-r}$ of $\N(A)$. Together they are $n$ vectors, and they're independent: if $\sum a_iu_i + \sum b_jv_j = 0$, then $\sum a_iu_i = -\sum b_jv_j$ is a vector in both $\row(A)$ and $\N(A)$, hence $0$; so $\sum a_iu_i = 0$ and $\sum b_jv_j = 0$, and each basis forces its coefficients to be $0$. So they are a basis of $\R^n$ ($n$ independent vectors), every $x$ is $\sum a_iu_i + \sum b_jv_j = x_r + x_n$, and the coordinates are unique, so the split is unique. Then $Ax = Ax_r + Ax_n = Ax_r + 0$.
One-to-one on the row space. If $x, x' \in \row(A)$ and $Ax = Ax'$, then $A(x - x') = 0$, so $x - x'$ is in $\N(A)$ and in $\row(A)$, hence $0$.
Onto the column space. Every $b \in \col(A)$ is $Ax$ for some $x$ (the column space is the set of outputs), and $Ax = Ax_r$ with $x_r \in \row(A)$. Exactly one such $x_r$, by the previous paragraph. $\blacksquare$
This is a visual argument, not a formal proof. The row plane and the null line are perpendicular, so every input is a step in the plane plus a step along the line. $A$ kills the line step, so only the plane step matters, and sliding an input along the null direction doesn't move its output. Two different plane steps can't give the same output, because their difference would be a plane vector that $A$ kills, and the only such vector is $0$. So as the plane step sweeps the row plane, the output sweeps the column plane once.
The full proof is in the n version.
Here's the count on both sides, for our $A$ and for a wide matrix whose column space is everything.
Now drive it yourself. The sliders build an input from a row-space part and a null-space part; the right panel shows where $A$ sends it.
Mission 3 is the solution-set picture from a new angle: every input with the same output lies on one line parallel to the null space, a particular solution plus the homogeneous ones. And here are eight vectors of $\R^3$ to file into the four rooms of our running example.
A word on names before the application.
Other books write the column space as $C(A)$ or $R(A)$ (for "range") and the null space as $\ker A$ (for "kernel"); the row space is sometimes $C(A^t)$, which is the same set. Here it's $\col(A)$, $\row(A)$, $\N(A)$, $\N(A^t)$, and the trap to avoid is mixing the sides: $\row(A)$ and $\N(A)$ live with the inputs in $\R^n$, $\col(A)$ and $\N(A^t)$ with the outputs in $\R^m$. For a non-square matrix those are different spaces, and "is the row space perpendicular to the column space?" isn't even a question.
The next topic treats $A$ as a function, a linear map $x \mapsto Ax$. Its null space is $\N(A)$ and its range is $\col(A)$, so the count $\dim \N(A) + \dim \col(A) = n$ becomes the Dimension Theorem: nullity plus rank equals the dimension of the inputs.
Measuring more things than there are unknowns sounds safe: extra data. The column space says which readings are even possible, and the left null space gives a one-line check for a broken instrument.
A tank is filled with $x_1$ liters of water and $x_2$ liters of brine. Three gauges report a volume reading $x_1 + x_2$, a salt reading $3x_2$ (the brine carries $3$ units of salt per liter), and a conductivity meter calibrated to read $x_1 + 4x_2$. One batch reports $(5, 6, 11)$; the next reports $(5, 6, 10)$. What was in each tank?
The gauges are the matrix $A = \begin{pmatrix} 1 & 1 \\ 0 & 3 \\ 1 & 4 \end{pmatrix}$ acting on $(x_1, x_2)$, so a set of readings $b$ is possible exactly when $b \in \col(A)$, the plane spanned by $(1, 0, 1)$ and $(1, 3, 4)$. The third row is the first plus the second, so $y = (1, 1, -1)$ spans the left null space, and $b$ is in the column plane exactly when $y \cdot b = b_1 + b_2 - b_3 = 0$.
First batch: $5 + 6 - 11 = 0$. Consistent. Solve the first two equations: $3x_2 = 6$ gives $x_2 = 2$, then $x_1 = 3$, and the meter agrees: $3 + 8 = 11$. Three liters of water, two of brine.
Second batch: $5 + 6 - 10 = 1 \neq 0$. No amounts $(x_1, x_2)$ produce these readings; some gauge is off. The left null vector is the consistency check, and it's one dot product.
Real sensor systems, surveying networks, and accounting ledgers all carry checks like this: when there are more readings than unknowns, the left null space says which combinations of readings have to agree, and a violation points at the faulty instrument before anyone solves anything.
The toolkit for this page: four subspaces, $\row(A)$ and $\N(A)$ among the inputs, $\col(A)$ and $\N(A^t)$ among the outputs. Reduce once: the nonzero echelon rows are a basis for the row space, the pivot columns of $A$ for the column space, the free unknowns give the null space, and the relations among the rows give the left null space. Each pair is perpendicular with dimensions adding to the size of its space, and $A$ carries the row space onto the column space one-to-one.
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.
Loading practice engine...
Next, a mixed set: every question type on this page, shuffled. Type vectors and dimension lists on one line with commas, like -2, 1, 0.
Loading practice engine...
Made it this far? Time for something with a little heat. A $3 \times 3$ matrix of rank $2$ and a right side with one unknown entry $k$. Find the left null vector from the relation among the rows, and the one $k$ that puts $b$ in the column plane falls out of a single dot product. In free response, type the number.
Loading practice engine...