The Row and Column Pictures

The same system of equations can be drawn two ways: as intersecting hyperplanes, or as a question about combining columns. Different questions are easy in each picture.

Both pictures describe the same solutions. Half of this subject is argued in one of them and half in the other, so fluency in only one leaves the rest looking harder than it is.

The two pictures answer different questions, and choosing between them is what the unit is for.

Existence is a column question. “Is there a solution?” asks whether $b$ is reachable by mixing the columns. In the row picture the same question means deciding whether $m$ flats in ${\mathbb{R}}^n$ share a point, which beyond three unknowns cannot be drawn and generally needs the elimination the picture was supposed to illuminate. The column reading often answers immediately: one column a multiple of another settles it without arithmetic.

Location is a row question. “Where are the solutions?” is what the row picture shows, and it is the space a linear program's feasible region lives in. Replace each equation by an inequality and you keep one side of each flat instead of the flat itself; the region is the intersection of those sides. Every graphical solution in this course is drawn in the row picture.

They never disagree. Both describe the same solution set, so a contradiction between them is an arithmetic error rather than a discovery. What differs is which question is cheap: reachability in one, position in the other.

In the row picture, each equation $a_i^{\mathsf{T}}x=b_i$ describes a hyperplane in ${\mathbb{R}}^n$, and

The solution set is where all $m$ flats meet.

In the column picture, the same system asks for weights $x_1,\dots ,x_n$ with

that is, whether $b\in \mathrm{span}\{A_1,\dots ,A_n\}\subseteq {\mathbb{R}}^m$.

The two are the same question asked from opposite sides: one asks where flats intersect, the other asks what the columns can reach.

Take

Row picture. Two lines in the plane: $x+2y=5$ and $3x-y=1$. They are not parallel, their normals $(1,2)$ and $(3,-1)$ point in different directions, so they cross at exactly one point. Solving gives $(1,2)$.

Column picture. The columns are $A_1=(1,3)$ and $A_2=(2,-1)$, and the question is which weights reach $b=(5,1)$:

With $x=1$ and $y=2$: $(1,3)+(4,-2)=(5,1)$. The same answer, arrived at by mixing rather than by intersecting.

What each made visible. The row picture showed where the solution sits. A point in the plane of the unknowns. The column picture showed why one exists: the two column arrows point in independent directions, so together they reach everything in ${\mathbb{R}}^2$, and $b$ was never in doubt.

A one-word change. Make the second column $(2,6)$ instead. Now both columns lie along the direction $(1,3)$, they can only reach that line, and whether a solution exists depends entirely on whether $b$ happens to lie on it. The column picture settles that instantly; the row picture needs elimination to discover the lines are parallel.

Problem. A system has

Does $Ax=b$ have a solution? Answer it with whichever picture does the least work, and say why the other is worse here.

Goal. Use the reading that matches the question, rather than the habitual one.

Relevant principle. The column picture answers existence directly, because existence is reachability.

Step 1: look at the columns. $A_1=(1,2,3)$ and $A_2=(2,4,6)=2A_1$.

Reason: the second column is a multiple of the first, so both point along the same direction.

Step 2: describe what is reachable. Every combination $x{A}_1+y{A}_2=(x+2y)A_1$ is a multiple of $(1,2,3)$. The columns reach exactly one line through the origin in ${\mathbb{R}}^3$.

Step 3: test $b$ against it. Is $(1,2,4)$ a multiple of $(1,2,3)$? The first two entries suggest the multiple is $1$, but then the third entry would be $3$, not $4$.

Reason: a multiple must scale every entry by the same factor, and no single factor works.

Step 4: conclude. $b$ is not reachable, so the system has no solution.

Result. Inconsistent, established in three lines without elimination.

Check. Row picture, for confirmation: the three equations are $x+2y=1$, $2x+4y=2$ and $3x+6y=4$. The first two are the same equation. The third demands $x+2y=4/3$, contradicting $x+2y=1$. Same verdict.

Interpretation. The row picture needed three planes in ${\mathbb{R}}^2$, already awkward, since the flats live in the plane of the unknowns while there are three of them, and then elimination to expose the contradiction. The column picture needed one observation about a multiple. When the question is "does a solution exist", the columns answer it; when the question is "where is the solution", the rows do.

Treating them as a matter of taste. They answer different questions with different effort. Choosing by preference rather than by question is what makes half the subject feel obscure.

Expecting them in the same space. The row picture lives in ${\mathbb{R}}^n$, the column picture in ${\mathbb{R}}^m$. For a $3\times 2$ system one diagram is in the plane and the other in three dimensions.

Reading parallel rows as dependent columns. In the square case the two coincide; in general the link runs through rank, not through a visual analogy between the drawings.

Using a picture as a proof. They make the case legible and memorable. The pivot structure of the reduced system is what decides it.

Assuming a disagreement is a discovery. Both describe the same solution set. If they seem to disagree, one of them has been drawn wrong.

Drawing the column picture for large $n$. With twenty columns there is nothing to draw. The reading still works as a way of thinking, which vectors can be reached, long after the diagram has stopped being possible.

The asymmetry to remember. Existence is a column question. Location is a row question. Asking the wrong picture for the wrong one is not an error, merely slow, and sometimes very slow, as when a five-second observation about a multiple replaces a full elimination.

Where each is met again. A graphical solution of a linear program draws the row picture and slides a contour across it. The simplex method works the column picture, swapping one column for another and asking what the new mixture reaches. Both appear in this subject within a few lessons of each other.

The same solution set, always. The two readings group the same equation differently, so the solution set is identical by construction rather than by coincidence.

1: fully structured. Consider $x+y=4$ and $x-y=0$.

(a) Describe the row picture. (b) Describe the column picture. (c) Give the solution and confirm it in both.

Check: (a) two lines in the plane, crossing at one point; (b) columns $(1,1)$ and $(1,-1)$, asking which weights reach $(4,0)$; (c) $(2,2)$. The lines meet there, and $2(1,1)+2(1,-1)=(4,0)$.

2: partly structured. A system has columns $A_1=(1,1)$ and $A_2=(3,3)$, with $b=(2,2)$.

(a) What can the columns reach? (b) Does a solution exist? (c) How many are there, and why does the row picture give the same answer?

Check: (a) only multiples of $(1,1)$, since $A_2=3A_1$; (b) yes, $(2,2)=2(1,1)$ is on that line; (c) infinitely many, since any $x+3y=2$ works. In rows, both equations reduce to the same line, so the lines coincide rather than crossing.

3: unstructured. An engineer says: "I always work in the row picture because I can see it. For this $4\times 7$ system I will sketch the seven planes and find where they meet."

Assess the plan and say what you would do instead.

Check: the plan cannot be carried out and misreads both pictures. A $4\times 7$ system has four equations in seven unknowns, so the row picture consists of four hyperplanes in ${\mathbb{R}}^7$, not seven planes, and not sketchable. The count of seven belongs to the columns, not the rows. More usefully: with four equations and seven unknowns there are at most four pivots, so at least three variables are free and the solution set, if non-empty, is a flat of dimension at least three. The right move is to reduce and read the pivots; the column picture then says whether $b$ is reachable from seven vectors in ${\mathbb{R}}^4$, which it usually is precisely because there are more columns than rows.

Row picture. Each equation is a hyperplane in ${\mathbb{R}}^n$; the solution set is their intersection.

Column picture. The columns are vectors in ${\mathbb{R}}^m$; the question is which weights reach $b$, that is whether $b$ lies in their span.

Same solutions, always. One equation grouped two ways. A disagreement between the pictures is an error, not a finding.

Different spaces. ${\mathbb{R}}^n$ for rows, ${\mathbb{R}}^m$ for columns. For a non-square system these differ in dimension.

Existence is a column question. If the columns cannot reach $b$, there is no solution, and that is often visible immediately.

Location is a row question. Where the solution sits, and what a feasible region looks like.

Where each returns. A graphical solution works the row picture; bases and simplex pivots work the column picture.

Neither proves anything alone. The pivot structure settles the case; the pictures make it legible.

The recurring error. Choosing a picture by preference instead of by question.