2.4 Is b in the Span?

The previous section closed with a promise: everything a matrix produces is automatically a subspace, since it is assembled by adding and scaling and the subspace test asks for nothing else. Cashing the promise takes one piece of notation. Let $A$ be the $3 times 2$ matrix whose columns are $𝐯 sub 1 equals open parenthesis 1, 0, 0.5 close parenthesis$ and $𝐯 sub 2 equals open parenthesis 0, 1, 0.5 close parenthesis$ , and let $𝐱 equals open parenthesis x sub 1, x sub 2 close parenthesis$ . The matrix–vector product $A 𝐱$ is defined to be the combination of the columns weighted by the entries of $𝐱$ :

$A 𝐱 = x_{1} 𝐯_{1} + x_{2} 𝐯_{2} = (x_{1}, x_{2}, 0.5 x_{1} + 0.5 x_{2})$

That is the whole definition, and it is the operation you have run since this chapter's first section. Letting $𝐱$ range over $ℝ^{2}$ produces every scaled sum of $𝐯 sub 1$ and $𝐯 sub 2$ — their span, nothing more. Writing $x equals x sub 1$ and $y equals x sub 2$ , the third coordinate is always $0.5 x plus 0.5 y$ , so what $A$ can produce is the plane $z = 0.5 x + 0.5 y$ through the origin. Section 2.8 will call it the column space of $A$ ; for now, call it what $A$ can reach.

Now run the question backwards. Given a target $𝐛$ in $ℝ^{3}$ , is there an $𝐱$ with $A 𝐱 = 𝐛$ ? The definition answers geometrically before any arithmetic starts: $A 𝐱 = 𝐛$ has a solution exactly when $𝐛$ lies in the span of the columns of $A$ . Consistency is not an accident discovered at the bottom of an elimination; it is the question of whether one point sits on one plane. The figure below puts a target on a vertical track through that plane so you can watch consistency come and go.

A semi-transparent tilted plane through a marked origin: the set of all vectors A times x, spanned by a short red arrow to (1, 0, 0.5) and a short blue arrow to (0, 1, 0.5) that both lie flat in it. A third, amber arrow runs from the origin to the target b = (1, 1, c), whose height is set by a slider c. A thin dashed red segment rises vertically from the point (1, 1, 1) on the plane to the amber arrow's tip, measuring how far b misses the plane. Dragging c to 1 shrinks the dashed segment to nothing and lays the amber arrow flat in the plane, exactly on the diagonal between the red and blue arrows; at every other c the amber tip floats off the plane and the dashed miss reappears. — The plane $z = 0.5 x + 0.5 y$ — everything $A$ can reach — with the red column $𝐯 sub 1 equals open parenthesis 1, 0, 0.5 close parenthesis$ and the blue column $𝐯 sub 2 equals open parenthesis 0, 1, 0.5 close parenthesis$ lying flat inside it. The amber arrow is the target $𝐛 equals open parenthesis 1, 1, c close parenthesis$ , and the dashed red segment runs from the plane's point $open parenthesis 1, 1, 1 close parenthesis$ to the amber tip: it is the miss, and its length is $vertical bar c minus 1 vertical bar$ .

Explore the figure

At the default $c equals 1$ the dashed segment has vanished and the amber arrow lies flat in the plane. Orbit until you see its tip close the parallelogram built on the red and blue arrows: $𝐛 = 𝐯_{1} + 𝐯_{2}$ , so $𝐱 equals open parenthesis 1, 1 close parenthesis$ .
Drag $c$ up to $2$ . The tip lifts to $open parenthesis 1, 1, 2 close parenthesis$ while the plane above $open parenthesis 1, 1 close parenthesis$ stays at height $0.5 (1) + 0.5 (1) = 1$ , and a dashed segment one unit long opens up. No combination of the red and blue arrows can climb it — both are stuck in the sheet.
Drag $c$ down to $minus 2$ . The miss is three units and hangs below the plane, but nothing has changed in kind: unreachable is unreachable, above or below.
Sweep $c$ and count the values that make the system solvable. This plane is a graph $z = 0.5 x + 0.5 y$ , so it has exactly one height above $open parenthesis 1, 1 close parenthesis$ and the vertical line meets it exactly once: the answer is one value, not an interval. Where would the crossing sit if the second column were $open parenthesis 0, 1, 1.5 close parenthesis$ ?

Where the miss becomes a number

Elimination says the same thing arithmetically. The augmented matrix for $𝐛 equals open parenthesis 1, 1, c close parenthesis$ has rows $open parenthesis 1, 0 ∣ 1 close parenthesis$ , $open parenthesis 0, 1 ∣ 1 close parenthesis$ and $open parenthesis 0.5, 0.5 ∣ c close parenthesis$ . The first two are already pivots, so clear the third with them: subtract $0.5$ times row one and $0.5$ times row two from row three. Its coefficients die, its augmented entry becomes $c - 0.5 (1) - 0.5 (1)$ , and the last row is left reading

$0 = c - 1$

which is the dashed segment written as an equation — its length was $vertical bar c minus 1 vertical bar$ all along. At $c equals 1$ that row is $0 equals 0$ , a true statement constraining nothing, and the two pivot rows hand over the unique solution $𝐱 equals open parenthesis 1, 1 close parenthesis$ .

open bracket 1, 0, 1; 0, 1, 1; 0, 0, 0 close bracket

The consistent case, c = 1 ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The augmented matrix for $𝐛 equals open parenthesis 1, 1, 1 close parenthesis$ , reduced by a computer algebra system. The bottom row is all zeros: the third equation followed from the first two, no pivot lands in the augmented column, and the coefficient pivots read off $x_{1} = x_{2} = 1$ .

At any other $c$ the last row reads $0 equals$ a nonzero number, and no choice of $x sub 1, x sub 2$ repairs it; reduced form scales that bad row to $0 equals 1$ , a pivot in the augmented column. That is the general test: a system is inconsistent exactly when its reduced augmented matrix has a pivot in the last column, because such a pivot is a row asserting $0 equals 1$ .

open bracket 1, 0, 0; 0, 1, 0; 0, 0, 1 close bracket

The inconsistent case, c = 2 ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The same computation for $𝐛 equals open parenthesis 1, 1, 2 close parenthesis$ . The identity matrix is bad news here, not good: the pivot in the third column is the row $0 equals 1$ — elimination finding, from the numbers alone, the vertical miss of one unit visible in the figure.

Two consequences. $A$ has only two columns, so its reach is at most a plane and most vectors of $ℝ^{3}$ are unreachable. And a reachable target is reachable in exactly one way, since both coefficient columns hold pivots and no free variable is left to vary.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.