2.4 Is the Target 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 $\mathbf{v}_1 = (1, 0, 0.5)$ and $\mathbf{v}_2 = (0, 1, 0.5)$ , and let $\mathbf{x} = (x_1, x_2)$ . The matrix–vector product $A\mathbf{x}$ is defined to be the combination of the columns weighted by the entries of $\mathbf{x}$ :

$A\mathbf{x} = x_1\mathbf{v}_1 + x_2\mathbf{v}_2 = (x_1, \; x_2, \; 0.5x_1 + 0.5x_2)$

That is the whole definition, and it is the operation you have run since this chapter's first section. Letting $\mathbf{x}$ range over $\mathbb{R}^2$ produces every scaled sum of $\mathbf{v}_1$ and $\mathbf{v}_2$ — their span, nothing more. Writing $x = x_1$ and $y = x_2$ , the third coordinate is always $0.5x + 0.5y$ , so what $A$ can produce is the plane $z = 0.5x + 0.5y$ 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 $\mathbf{b}$ in $\mathbb{R}^3$ , is there an $\mathbf{x}$ with $A\mathbf{x} = \mathbf{b}$ ? The definition answers geometrically before any arithmetic starts: $A\mathbf{x} = \mathbf{b}$ has a solution exactly when $\mathbf{b}$ 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 segment from the origin to (1, 0, 0.5) and a short blue segment from the origin to (0, 1, 0.5), both lying flat in it. A third, amber segment runs from the origin to the target b, the point (1, 1, c), whose height is set by a slider c. A pink segment rises vertically from the point (1, 1, 1) on the plane to the amber segment's tip, measuring how far b misses the plane. The figure opens with c at 1, where the pink segment has zero length and the amber segment lies flat in the plane, exactly on the diagonal of the parallelogram the red and blue segments make; at every other c the amber tip floats off the plane and the pink miss reappears. — The plane $z = 0.5x + 0.5y$ — everything $A$ can reach — with the red column $\mathbf{v}_1 = (1, 0, 0.5)$ and the blue column $\mathbf{v}_2 = (0, 1, 0.5)$ lying flat inside it. The amber arrow is the target $\mathbf{b} = (1, 1, c)$ , and the pink segment runs from the plane's point $(1, 1, 1)$ to the amber tip: it is the miss, and its length is $|c - 1|$ .

Explore the figure

At the default $c = 1$ the pink 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: $\mathbf{b} = \mathbf{v}_1 + \mathbf{v}_2$ , so $\mathbf{x} = (1, 1)$ .
Drag $c$ up to $2$ . The tip lifts to $(1, 1, 2)$ while the plane above $(1, 1)$ stays at height $0.5(1) + 0.5(1) = 1$ , and a pink 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 $-2$ . 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 $z = 0.5x + 0.5y$ , so it has exactly one height above $(1, 1)$ 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 $(0, 1, 1.5)$ ?

Where the miss becomes a number

Elimination says the same thing arithmetically. The augmented matrix for $\mathbf{b} = (1, 1, c)$ has rows $(1, 0 \mid 1)$ , $(0, 1 \mid 1)$ and $(0.5, 0.5 \mid c)$ . 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 pink segment written as an equation — its length was $|c - 1|$ all along. At $c = 1$ that row is $0 = 0$ , a true statement constraining nothing, and the two pivot rows hand over the unique solution $\mathbf{x} = (1, 1)$ .

\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}

The consistent target ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The augmented matrix for $\mathbf{b} = (1, 1, 1)$ , 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 $0 =$ a nonzero number, and no choice of $x_1, x_2$ repairs it; reduced form scales that bad row to $0 = 1$ , 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 $0 = 1$ .

\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

A target one unit above the plane ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The same computation for $\mathbf{b} = (1, 1, 2)$ . The identity matrix is bad news here, not good: the pivot in the third column is the row $0 = 1$ — 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 $\mathbb{R}^3$ 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.

Example 1 Two targets, one test

Decide whether $\mathbf{b} = (2, -2, 0)$ and $\mathbf{b}' = (0, 4, 1)$ lie in the reach of $A$ , and give the solving $\mathbf{x}$ whenever one exists.

Show solution

The elimination above never used the specific entries $1, 1, c$ : for a general target $(b_1, b_2, b_3)$ the last row becomes $0 = b_3 - 0.5b_1 - 0.5b_2$ . So $\mathbf{b}$ is reachable exactly when $b_3 = 0.5b_1 + 0.5b_2$ — the equation of the plane, as it had to be — and then $\mathbf{x} = (b_1, b_2)$ .

For $\mathbf{b} = (2, -2, 0)$ : $0.5(2) + 0.5(-2) = 0 = b_3$ . Consistent, with $\mathbf{x} = (2, -2)$ , and indeed $2\mathbf{v}_1 - 2\mathbf{v}_2 = (2, 0, 1) - (0, 2, 1) = (2, -2, 0)$ .

For $\mathbf{b}' = (0, 4, 1)$ : $0.5(0) + 0.5(4) = 2$ , but $b_3 = 1$ . The last row reads $0 = -1$ , so the system is inconsistent and $\mathbf{b}'$ misses the plane by one unit — the $c = 2$ picture, moved sideways.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.