Chapter 5 · Vector Spaces

5.2 Spanning Sets and Linear Independence

Which vectors can be built from a given set, and is every vector in the set really needed? Both questions become a linear system — and the pivots answer them.

A printer makes every colour on the page from a few inks; a dietitian builds a meal plan from a few foods; a sound engineer recreates a voice from a few pure tones. In each case two questions come up. Can a given target be made from the ingredients we have? And are all the ingredients needed, or could one of them be mixed from the others and thrown away?

In the language of Section 5.1 the first question asks whether a vector lies in a span, and the second asks whether a set of vectors is linearly independent. Both questions turn into linear systems, so the row reduction of Chapter 1 answers them. Together they prepare the central idea of Section 5.3: a basis, a set that spans with nothing wasted.

Linear Combinations and Spanning Sets

Recall (Definition 5.3) that a linear combination of $\mathbf{v}_1,\dots,\mathbf{v}_k$ is a vector $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k$, and that $\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ is the set of all of them. In $\mathbb{R}^n$ the vector equation $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{b}$ is, row by row, a system of $n$ linear equations in the unknowns $c_1,\dots,c_k$.

📐 Theorem 5.5 — Span membership is a linear system

Let $\mathbf{v}_1,\dots,\mathbf{v}_k$ and $\mathbf{b}$ be vectors in $\mathbb{R}^n$, and let $A=\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\cdots&\mathbf{v}_k\end{bmatrix}$ be the $n\times k$ matrix with these vectors as its columns. Then $$\mathbf{b}\in\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\}\iff\text{the system with augmented matrix }\left[\begin{array}{c|c}A&\mathbf{b}\end{array}\right]\text{ is consistent}.$$ Any solution $(c_1,\dots,c_k)$ gives the weights in $\mathbf{b}=c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k$.

✏️ Example 1 — Is $\mathbf{b}$ in the span?

Let $\mathbf{v}_1=\begin{bmatrix}1\\2\\-1\end{bmatrix}$ and $\mathbf{v}_2=\begin{bmatrix}3\\1\\2\end{bmatrix}$. Decide whether (a) $\mathbf{b}=\begin{bmatrix}-1\\3\\-4\end{bmatrix}$ and (b) $\mathbf{b}'=\begin{bmatrix}1\\1\\1\end{bmatrix}$ lie in $\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}$.

💡 Show solution▶

(a) Set up We need $c_1,c_2$ with $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{b}$, i.e. the augmented matrix $\left[\begin{array}{cc|c}1&3&-1\\2&1&3\\-1&2&-4\end{array}\right]$.

(a) Reduce $$\xrightarrow[R_3\to R_3+R_1]{R_2\to R_2-2R_1}\left[\begin{array}{cc|c}1&3&-1\\0&-5&5\\0&5&-5\end{array}\right]\xrightarrow{R_3\to R_3+R_2}\left[\begin{array}{cc|c}1&3&-1\\0&-5&5\\0&0&0\end{array}\right]\xrightarrow[R_1\to R_1-3R_2]{R_2\to-\frac15R_2}\left[\begin{array}{cc|c}1&0&2\\0&1&-1\\0&0&0\end{array}\right].$$ Consistent, with $c_1=2$, $c_2=-1$. So $\mathbf{b}\in\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}$ and $\mathbf{b}=2\mathbf{v}_1-\mathbf{v}_2$.

Check $2(1,2,-1)-(3,1,2)=(2-3,\ 4-1,\ -2-2)=(-1,3,-4)$. ✓

(b) The same operations on $\left[\begin{array}{cc|c}1&3&1\\2&1&1\\-1&2&1\end{array}\right]$ give $R_2:\ [\,0\ \ -5\mid-1\,]$, $R_3:\ [\,0\ \ 5\mid2\,]$ and then $R_3\to R_3+R_2$: $[\,0\ \ 0\mid1\,]$, the equation $0=1$. Inconsistent, so $\mathbf{b}'$ is not in the span. Geometrically, $\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2\}$ is a plane through the origin, and $\mathbf{b}'$ is off it.

A set $\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ is a spanning set for a vector space $V$ (we also say the vectors span $V$) if $\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\}=V$ — every vector of $V$ is a linear combination of them. For $V=\mathbb{R}^n$ there is a clean test.

📐 Theorem 5.6 — When do the columns span $\mathbb{R}^n$?

Let $A=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_k\end{bmatrix}$ be an $n\times k$ matrix. The vectors $\mathbf{v}_1,\dots,\mathbf{v}_k$ span $\mathbb{R}^n$ if and only if an echelon form of $A$ has a pivot in every row. In particular, fewer than $n$ vectors can never span $\mathbb{R}^n$.

Why. If every row has a pivot, no row of $\left[\begin{array}{c|c}A&\mathbf{b}\end{array}\right]$ can reduce to $[\,0\ \cdots\ 0\mid\text{nonzero}\,]$, so $A\mathbf{c}=\mathbf{b}$ is consistent for every $\mathbf{b}$. If some row has no pivot, choosing $\mathbf{b}$ to put a nonzero number in that row of the reduced augmented matrix and undoing the row operations gives a $\mathbf{b}$ that cannot be reached. Since there are at most $k$ pivots, $k\lt n$ vectors leave some row without one.

✏️ Example 2 — Spanning $\mathbb{R}^3$, or only a plane

(a) Do $(1,1,0)$, $(0,1,1)$ and $(1,0,1)$ span $\mathbb{R}^3$? (b) Do $(1,1,0)$, $(0,1,1)$ and $(1,2,1)$?

💡 Show solution▶

(a) $$\begin{bmatrix}1&0&1\\1&1&0\\0&1&1\end{bmatrix}\xrightarrow{R_2\to R_2-R_1}\begin{bmatrix}1&0&1\\0&1&-1\\0&1&1\end{bmatrix}\xrightarrow{R_3\to R_3-R_2}\begin{bmatrix}1&0&1\\0&1&-1\\0&0&2\end{bmatrix}.$$ A pivot in every row, so the three vectors span $\mathbb{R}^3$.

(b) $$\begin{bmatrix}1&0&1\\1&1&2\\0&1&1\end{bmatrix}\xrightarrow{R_2\to R_2-R_1}\begin{bmatrix}1&0&1\\0&1&1\\0&1&1\end{bmatrix}\xrightarrow{R_3\to R_3-R_2}\begin{bmatrix}1&0&1\\0&1&1\\0&0&0\end{bmatrix}.$$ Row 3 has no pivot, so these vectors do not span $\mathbb{R}^3$.

What they span In (b) the third vector is the sum of the first two: $(1,1,0)+(0,1,1)=(1,2,1)$. It adds nothing new, and the span is only the plane through $(1,1,0)$ and $(0,1,1)$. Its normal is $(1,1,0)\times(0,1,1)=(1,-1,1)$, so the span is the plane $x-y+z=0$; for example $(0,0,1)$ is not in it.

The same method works in any vector space once we compare coefficients.

✏️ Example 3 — A span in $P_2$

Write $p(x)=2+4x+4x^2$ as a linear combination of $p_1=1+x$, $p_2=x+x^2$ and $p_3=1+x^2$.

💡 Show solution▶

Compare coefficients $a(1+x)+b(x+x^2)+c(1+x^2)=(a+c)+(a+b)x+(b+c)x^2$. Matching with $2+4x+4x^2$: $$a+c=2,\qquad a+b=4,\qquad b+c=4 .$$

Solve Adding the three equations, $2(a+b+c)=10$, so $a+b+c=5$. Then $b=5-2=3$, $c=5-4=1$, $a=5-4=1$.

Check $(1+x)+3(x+x^2)+(1+x^2)=2+4x+4x^2$. ✓ (The same steps solve the system for any target $p$, so $p_1,p_2,p_3$ span all of $P_2$.)

The grid of slanted lines shows the combinations $a\mathbf{u}+c\mathbf{v}$ with whole-number $a$ or $c$; between them every point of the plane is reached. Drag the tip of $\mathbf{v}$ (it snaps to half-units), or press Play to swing $\mathbf{v}$ round. When $\mathbf{v}$ lands on the line of $\mathbf{u}$ the grid collapses onto that line: the span drops from the whole plane to a line, and the target $\mathbf{b}$ can no longer be reached.

Target $\mathbf{b}$:

$\mathbf{u}$, $\mathbf{v}$—

$\det\begin{bmatrix}\mathbf{u}&\mathbf{v}\end{bmatrix}$—

$\operatorname{span}\{\mathbf{u},\mathbf{v}\}$—

$\mathbf{b}=a\mathbf{u}+c\mathbf{v}$—

Linear Independence

In Example 2(b) the vector $(1,2,1)$ was redundant: it was already a combination of the other two, so removing it did not shrink the span. Linear independence is the precise way to say that a set has no redundant vectors.

📘 Definition 5.4 — Linear independence and dependence

Vectors $\mathbf{v}_1,\dots,\mathbf{v}_k$ in a vector space $V$ are linearly independent if the vector equation $$c_1\mathbf{v}_1+c_2\mathbf{v}_2+\dots+c_k\mathbf{v}_k=\mathbf{0}$$ has only the trivial solution $c_1=c_2=\dots=c_k=0$. They are linearly dependent if there are weights, not all zero, with $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{0}$; such an equation is called a dependence relation.

📐 Theorem 5.7 — The homogeneous-system test in $\mathbb{R}^n$

Let $A=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_k\end{bmatrix}$ have the vectors $\mathbf{v}_1,\dots,\mathbf{v}_k\in\mathbb{R}^n$ as columns. The vectors are linearly independent if and only if $A\mathbf{c}=\mathbf{0}$ has only the trivial solution, that is, if and only if every column of $A$ is a pivot column (there are no free variables). Each free variable gives a dependence relation, and the reduced row echelon form displays it: row operations do not change the linear relations among the columns, so if a non-pivot column of the reduced matrix has entries $r_1,\dots,r_p$ in its pivot rows, then $\mathbf{v}_j=r_1\times(\text{first pivot column of }A)+\dots+r_p\times(\text{$p$-th pivot column of }A)$.

🧭 Problem-solving strategy — testing independence

Quick looks first. Is $\mathbf{0}$ in the set? Are two vectors multiples of each other? Are there more vectors than entries ($k\gt n$)? Any "yes" means dependent (Theorems 5.8 and 5.9 below).
Otherwise put the vectors in the columns of a matrix (for polynomials or matrices, use their coefficients or entries) and row reduce.
Every column a pivot column $\Rightarrow$ independent. A free column $\Rightarrow$ dependent: set that free variable to $1$ (others to $0$) and read off a dependence relation.
Check the relation by substituting the original vectors.

✏️ Example 4 — Finding a dependence relation

Are $\mathbf{v}_1=(1,2,3)$, $\mathbf{v}_2=(4,5,6)$, $\mathbf{v}_3=(7,8,9)$ linearly independent? If not, find a dependence relation and write one vector in terms of the others.

💡 Show solution▶

Reduce $$\begin{bmatrix}1&4&7\\2&5&8\\3&6&9\end{bmatrix}\xrightarrow[R_3\to R_3-3R_1]{R_2\to R_2-2R_1}\begin{bmatrix}1&4&7\\0&-3&-6\\0&-6&-12\end{bmatrix}\xrightarrow{R_3\to R_3-2R_2}\begin{bmatrix}1&4&7\\0&-3&-6\\0&0&0\end{bmatrix}\xrightarrow[R_1\to R_1-4R_2]{R_2\to-\frac13R_2}\begin{bmatrix}1&0&-1\\0&1&2\\0&0&0\end{bmatrix}.$$ Column 3 is not a pivot column, so the vectors are dependent.

Relation The reduced system is $c_1-c_3=0$, $c_2+2c_3=0$. Take $c_3=1$: $c_1=1$, $c_2=-2$. So $$\mathbf{v}_1-2\mathbf{v}_2+\mathbf{v}_3=\mathbf{0},\qquad\text{i.e.}\qquad\mathbf{v}_3=-\mathbf{v}_1+2\mathbf{v}_2 .$$ (Column 3 of the reduced matrix, $(-1,2,0)$, shows these weights directly.)

Check $(1,2,3)-2(4,5,6)+(7,8,9)=(1-8+7,\ 2-10+8,\ 3-12+9)=(0,0,0)$. ✓

Geometry The three vectors lie in one plane through the origin. Any two of them, e.g. $\mathbf{v}_1,\mathbf{v}_2$, are independent (columns 1 and 2 are pivot columns) and span that plane. Step through this in Interactive 2.

📐 Theorem 5.8 — Recognising dependence

Let $S=\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ be vectors in a vector space $V$.

If $\mathbf{0}\in S$, then $S$ is linearly dependent (take weight $1$ on $\mathbf{0}$ and $0$ on the rest).
Two vectors $\{\mathbf{u},\mathbf{v}\}$ are linearly dependent if and only if one is a scalar multiple of the other.
For $k\ge2$, $S$ is linearly dependent if and only if at least one of the vectors is a linear combination of the others.

Proof of 3. If $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{0}$ with some $c_j\ne0$, divide by $c_j$ and solve for $\mathbf{v}_j$. Conversely, $\mathbf{v}_j=\sum_{i\ne j}a_i\mathbf{v}_i$ rearranges to a relation in which $\mathbf{v}_j$ has weight $-1\ne0$.

📐 Theorem 5.9 — Too many vectors are dependent

If $k\gt n$, then any $k$ vectors in $\mathbb{R}^n$ are linearly dependent. Reason: $A\mathbf{c}=\mathbf{0}$ has $n$ equations and $k\gt n$ unknowns; there are at most $n$ pivots, so at least one variable is free and there is a nontrivial solution.

⚠️ Common mistakes with independence

1. "Dependent" does not mean every vector is a combination of the others. In $\{(1,0),(2,0),(0,1)\}$ the first two are dependent, but $(0,1)$ is not a combination of $(1,0)$ and $(2,0)$. Theorem 5.8 says at least one vector is.

2. Checking pairs is not enough. $(1,0,1)$, $(0,1,1)$, $(1,1,2)$: no vector is a multiple of another, yet the third is the sum of the first two.

3. The converse of Theorem 5.9 is false. Having $k\le n$ vectors does not make them independent — $(1,2,3)$ and $(2,4,6)$ are dependent in $\mathbb{R}^3$.

4. Pivot columns of $A$, not of the reduced matrix. Row operations change the column vectors; they keep the relations between columns. Use the reduced matrix to find the relation, then state it for the original vectors.

What independence looks like

In $\mathbb{R}^2$Two vectors are dependent exactly when they lie on one line through the origin (are collinear). Three or more vectors in $\mathbb{R}^2$ are always dependent.

In $\mathbb{R}^3$Two vectors are dependent when collinear; three vectors are dependent exactly when they lie in one plane through the origin (are coplanar). Four or more are always dependent.

Removing a redundant vectorIf $\mathbf{v}_k$ is a combination of $\mathbf{v}_1,\dots,\mathbf{v}_{k-1}$, then $\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\}=\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_{k-1}\}$ (Theorem 5.10).

📐 Theorem 5.10 — Discarding redundant vectors keeps the span

If $\mathbf{v}_k$ is a linear combination of $\mathbf{v}_1,\dots,\mathbf{v}_{k-1}$, then $\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_{k-1},\mathbf{v}_k\}=\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_{k-1}\}$. Reason: in any combination $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k$, replace $\mathbf{v}_k$ by its expression in the others; the result is a combination of $\mathbf{v}_1,\dots,\mathbf{v}_{k-1}$ alone. Repeating this, the pivot columns of $A$ always span the same space as all the columns.

The chosen vectors are the columns of $A$. Press Next step (or Play) to row reduce $A$ exactly, one operation at a time; the current pivot is ringed. When the reduction is finished the pivot columns are highlighted and the picture shows the span: the vectors in $\mathbb{R}^3$, the plane or line they span, and any free column's dependence relation.

Vectors:

step—

row operation—

pivot columns—

independent?—

For polynomials and matrices we test independence by comparing coefficients or entries, exactly as in Example 3.

✏️ Example 5 — Dependence in $P_2$

Are $p_1=1+2x-x^2$, $p_2=2+x$ and $p_3=3x-2x^2$ linearly independent in $P_2$?

💡 Show solution▶

Set up $ap_1+bp_2+cp_3=(a+2b)+(2a+b+3c)x+(-a-2c)x^2$. For this to be the zero polynomial, every coefficient must vanish: $$a+2b=0,\qquad 2a+b+3c=0,\qquad -a-2c=0 .$$

Reduce $$\begin{bmatrix}1&2&0\\2&1&3\\-1&0&-2\end{bmatrix}\xrightarrow[R_3\to R_3+R_1]{R_2\to R_2-2R_1}\begin{bmatrix}1&2&0\\0&-3&3\\0&2&-2\end{bmatrix}\xrightarrow[R_3\to R_3-2R_2]{R_2\to-\frac13R_2}\begin{bmatrix}1&2&0\\0&1&-1\\0&0&0\end{bmatrix}.$$ $c$ is free, so the polynomials are dependent.

Relation Take $c=1$: $b=1$ and $a=-2b=-2$. So $-2p_1+p_2+p_3=0$, i.e. $p_2+p_3=2p_1$.

Check $p_2+p_3=(2+x)+(3x-2x^2)=2+4x-2x^2=2(1+2x-x^2)=2p_1$. ✓

✏️ Example 6 — Independence in $M_{2\times2}$

Show that $A_1=\begin{bmatrix}1&1\\0&0\end{bmatrix}$, $A_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $A_3=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ are linearly independent in $M_{2\times2}$. Do they span $M_{2\times2}$?

💡 Show solution▶

Set up $$aA_1+bA_2+cA_3=\begin{bmatrix}a+c&a+b\\b&c\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}.$$

Solve The bottom row gives $b=0$ and $c=0$; then the top-left entry gives $a=0$. Only the trivial solution, so $A_1,A_2,A_3$ are independent.

Span? No. Every combination has the form $\begin{bmatrix}a+c&a+b\\b&c\end{bmatrix}$, whose top-left entry equals (top-right) $-$ (bottom-left) $+$ (bottom-right). For example $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ fails this ($1\ne0-0+0$), so it is not in the span. Independence and spanning are different properties: in Section 5.3 we shall see that $M_{2\times2}$ needs four independent matrices.

Exercises

Linear combinations and spanning sets.

1 Write $\mathbf{b}=(5,3)$ as a linear combination of $(1,2)$ and $(3,-1)$.
2 Is $(2,5,3)$ in $\operatorname{span}\{(1,1,0),(0,1,1)\}$? If so, give the weights.
3 Is $(1,2,3)$ in $\operatorname{span}\{(1,1,0),(0,1,1)\}$?
4 For which value of $h$ is $(1,h,7)$ in $\operatorname{span}\{(1,2,1),(0,1,3)\}$?
5 Do $(1,2)$ and $(3,6)$ span $\mathbb{R}^2$? Describe their span.
6 Do $(1,2)$, $(2,1)$ and $(1,1)$ span $\mathbb{R}^2$?
7 Do $(1,0,1)$, $(2,1,0)$ and $(0,1,-2)$ span $\mathbb{R}^3$? If not, find an equation of their span.
8 Is $1+x^2$ in $\operatorname{span}\{1+x,\ x-x^2\}$ in $P_2$?
9 Is $\begin{bmatrix}1&2\\3&4\end{bmatrix}$ in the span of $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, $\begin{bmatrix}0&1\\1&0\end{bmatrix}$, $\begin{bmatrix}1&1\\0&0\end{bmatrix}$?
10 Show that $\operatorname{span}\{(1,-1,0)\}=\operatorname{span}\{(1,-1,0),(2,-2,0)\}$ and describe this set.

Check: reduce the augmented matrix [ v₁ … vₖ | b ] exactly and read off consistency

Part:

💡 Show solutions▶

1 $a(1,2)+c(3,-1)=(5,3)$: $a+3c=5$, $2a-c=3$. Then $c=2a-3$, so $a+6a-9=5$, $a=2$, $c=1$: $\ (5,3)=2(1,2)+(3,-1)$.

2 $a(1,1,0)+b(0,1,1)=(a,\ a+b,\ b)$. Matching: $a=2$, $b=3$, and $a+b=5$ ✓. Yes: $(2,5,3)=2(1,1,0)+3(0,1,1)$.

3 $(a,\ a+b,\ b)=(1,2,3)$ forces $a=1$, $b=3$, but then $a+b=4\ne2$. Inconsistent: not in the span.

4 $a(1,2,1)+b(0,1,3)=(a,\ 2a+b,\ a+3b)$. First and third entries: $a=1$, $1+3b=7$, $b=2$. Then $h=2a+b=4$.

5 No: $(3,6)=3(1,2)$, so the span is only the line $y=2x$ through the origin.

6 Yes. $\begin{bmatrix}1&2&1\\2&1&1\end{bmatrix}\xrightarrow{R_2\to R_2-2R_1}\begin{bmatrix}1&2&1\\0&-3&-1\end{bmatrix}$ has a pivot in both rows. (Already $(1,2)$ and $(2,1)$ span $\mathbb{R}^2$; $(1,1)$ is redundant.)

7 No. $(0,1,-2)=(2,1,0)-2(1,0,1)$, so the matrix has only two pivots. The span is the plane through the origin with normal $(1,0,1)\times(2,1,0)=(-1,2,1)$, i.e. $-x+2y+z=0$. (Check: $0+2-2=0$ ✓.)

8 $a(1+x)+b(x-x^2)=a+(a+b)x-bx^2$. Matching $1+0x+1x^2$: $a=1$, $b=-1$, and $a+b=0$ ✓. Yes: $1+x^2=(1+x)-(x-x^2)$.

9 $a\begin{bmatrix}1&0\\0&1\end{bmatrix}+b\begin{bmatrix}0&1\\1&0\end{bmatrix}+c\begin{bmatrix}1&1\\0&0\end{bmatrix}=\begin{bmatrix}a+c&b+c\\b&a\end{bmatrix}$. Matching: $b=3$, $a=4$, $c=1-a=-3$, but then $b+c=0\ne2$. Not in the span.

10 $(2,-2,0)=2(1,-1,0)$ is redundant, so by Theorem 5.10 the spans agree. Both are the line through the origin in the direction $(1,-1,0)$, lying in the $xy$-plane along $y=-x$.

Linear independence. Decide whether the vectors are linearly independent; if not, give a dependence relation.

1 $(1,2)$, $(-2,-4)$
2 $(1,2)$, $(2,1)$
3 $(1,0,0)$, $(1,1,0)$, $(1,1,1)$
4 $(1,2,-1)$, $(0,1,4)$, $(2,5,2)$
5 $(1,2,3)$, $(0,0,0)$, $(4,1,2)$
6 $(1,1)$, $(1,-1)$, $(3,5)$
7 $(1,0,2,1)$, $(0,1,1,-1)$, $(1,1,3,0)$
8 $(1,1,0,0)$, $(0,1,1,0)$, $(0,0,1,1)$, $(1,0,0,1)$
9 For which values of $k$ are $(1,k)$ and $(k,4)$ linearly dependent?
10 $1+x$, $1-x$, $x^2$ in $P_2$
11 $1+x$, $x+x^2$, $1-x^2$ in $P_2$
12 $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $\begin{bmatrix}1&1\\0&0\end{bmatrix}$, $\begin{bmatrix}1&1\\1&0\end{bmatrix}$, $\begin{bmatrix}1&1\\1&1\end{bmatrix}$ in $M_{2\times2}$

💡 Show solutions▶

1 Dependent: $(-2,-4)=-2(1,2)$, so $2(1,2)+(-2,-4)=\mathbf{0}$.

2 Independent: neither is a multiple of the other (for two vectors, Theorem 5.8(2) settles it).

3 Independent: $\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$ is already in echelon form with three pivot columns.

4 Dependent. The reduced matrix is $\begin{bmatrix}1&0&2\\0&1&1\\0&0&0\end{bmatrix}$; column 3 gives $\mathbf{v}_3=2\mathbf{v}_1+\mathbf{v}_2$, i.e. $2\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}$. Check: $2(1,2,-1)+(0,1,4)=(2,5,2)$ ✓.

5 Dependent — the set contains $\mathbf{0}$: $0\mathbf{v}_1+1\mathbf{v}_2+0\mathbf{v}_3=\mathbf{0}$.

6 Dependent — three vectors in $\mathbb{R}^2$ (Theorem 5.9). Solving $a(1,1)+b(1,-1)=(3,5)$: $a+b=3$, $a-b=5$, so $a=4$, $b=-1$, and $4\mathbf{v}_1-\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}$.

7 Dependent: $\mathbf{v}_1+\mathbf{v}_2=(1,1,3,0)=\mathbf{v}_3$, so $\mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3=\mathbf{0}$.

8 Dependent, although no vector is a multiple of another. Row reducing gives $\begin{bmatrix}1&0&0&1\\0&1&0&-1\\0&0&1&1\\0&0&0&0\end{bmatrix}$, so with $c_4=1$: $c_1=-1$, $c_2=1$, $c_3=-1$. Hence $\mathbf{v}_1-\mathbf{v}_2+\mathbf{v}_3-\mathbf{v}_4=\mathbf{0}$; check: $(1,1,0,0)-(0,1,1,0)+(0,0,1,1)-(1,0,0,1)=(0,0,0,0)$ ✓.

9 Two vectors in $\mathbb{R}^2$ are dependent exactly when $\det\begin{bmatrix}1&k\\k&4\end{bmatrix}=4-k^2=0$, i.e. $k=2$ or $k=-2$. (For $k=2$: $(2,4)=2(1,2)$; for $k=-2$: $(-2,4)=-2(1,-2)$.)

10 Independent. $a(1+x)+b(1-x)+cx^2=(a+b)+(a-b)x+cx^2=0$ gives $a+b=0$, $a-b=0$, $c=0$, so $a=b=c=0$.

11 Dependent. $a(1+x)+b(x+x^2)+c(1-x^2)=(a+c)+(a+b)x+(b-c)x^2$; the equations $a+c=0$, $a+b=0$, $b-c=0$ have the solution $a=1$, $b=-1$, $c=-1$. So $(1+x)-(x+x^2)-(1-x^2)=0$ ✓.

12 Independent. $aM_1+bM_2+cM_3+dM_4=\begin{bmatrix}a+b+c+d&b+c+d\\c+d&d\end{bmatrix}=O$ gives, from the bottom right upwards, $d=0$, $c=0$, $b=0$, $a=0$.

III

Theory and applications.

1 True or false (explain): if $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is linearly dependent, then $\mathbf{w}$ is a linear combination of $\mathbf{u}$ and $\mathbf{v}$.
2 True or false: any four vectors in $\mathbb{R}^3$ are linearly dependent.
3 True or false: if $\{\mathbf{u},\mathbf{v}\}$ is independent and $\mathbf{w}\notin\operatorname{span}\{\mathbf{u},\mathbf{v}\}$, then $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is independent.
4 True or false: every nonempty subset of a linearly independent set is linearly independent.
5 Prove that if $\{\mathbf{u},\mathbf{v}\}$ is linearly independent, then so is $\{\mathbf{u}+\mathbf{v},\ \mathbf{u}-\mathbf{v}\}$.
6 Nutrition. One serving of food A contains $10$ g protein, $20$ g carbohydrate and $5$ g fat; food B contains $5$ g, $10$ g and $10$ g; food C contains $0$ g, $10$ g and $2$ g. How many servings of each give exactly $25$ g protein, $80$ g carbohydrate and $26$ g fat? Why is the answer unique?
7 Chemistry. Write each substance in $\mathrm{C_3H_8+O_2\to CO_2+H_2O}$ as a vector of (carbon, hydrogen, oxygen) atoms. Explain why balancing the equation means finding a dependence relation among four vectors in $\mathbb{R}^3$, and balance it with the smallest whole numbers.
8 Challenge. Let $\mathbf{u},\mathbf{v}\in\mathbb{R}^3$ be linearly independent and $\mathbf{w}=\mathbf{u}\times\mathbf{v}$. Prove that $\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ is linearly independent. (Hint: take the dot product of a relation with $\mathbf{w}$.)

💡 Show solutions▶

1 False. $\mathbf{u}=(1,0)$, $\mathbf{v}=(2,0)$, $\mathbf{w}=(0,1)$ are dependent ($2\mathbf{u}-\mathbf{v}=\mathbf{0}$), but $\mathbf{w}$ is not a combination of $\mathbf{u}$ and $\mathbf{v}$. Only some vector must be a combination of the others.

2 True, by Theorem 5.9 ($4\gt3$): the $3\times4$ homogeneous system has a free variable.

3 True. Suppose $a\mathbf{u}+b\mathbf{v}+c\mathbf{w}=\mathbf{0}$. If $c\ne0$ then $\mathbf{w}=-\tfrac ac\mathbf{u}-\tfrac bc\mathbf{v}\in\operatorname{span}\{\mathbf{u},\mathbf{v}\}$, which is false; so $c=0$, and then $a\mathbf{u}+b\mathbf{v}=\mathbf{0}$ forces $a=b=0$.

4 True. A dependence relation among the vectors of the subset, with weight $0$ on the other vectors, would be a dependence relation for the whole set.

5 If $a(\mathbf{u}+\mathbf{v})+b(\mathbf{u}-\mathbf{v})=\mathbf{0}$, then $(a+b)\mathbf{u}+(a-b)\mathbf{v}=\mathbf{0}$. Since $\mathbf{u},\mathbf{v}$ are independent, $a+b=0$ and $a-b=0$, so $a=b=0$.

6 Solve $a(10,20,5)+b(5,10,10)+c(0,10,2)=(25,80,26)$. From the protein equation $b=5-2a$; the carbohydrate equation $20a+10(5-2a)+10c=80$ gives $c=3$; the fat equation $5a+10(5-2a)+6=26$ gives $a=2$, so $b=1$. Answer: $2$ servings of A, $1$ of B, $3$ of C (check: $(20+5+0,\ 40+10+30,\ 10+10+6)=(25,80,26)$ g ✓). The three food vectors are independent ($\det=-750\ne0$), so the weights are unique.

7 $\mathrm{C_3H_8}=(3,8,0)$, $\mathrm{O_2}=(0,0,2)$, $\mathrm{CO_2}=(1,0,2)$, $\mathrm{H_2O}=(0,2,1)$. Atoms are conserved, so $x_1(3,8,0)+x_2(0,0,2)-x_3(1,0,2)-x_4(0,2,1)=\mathbf{0}$ — a dependence relation, which must exist because these are four vectors in $\mathbb{R}^3$. Carbon: $3x_1=x_3$; hydrogen: $8x_1=2x_4$; oxygen: $2x_2=2x_3+x_4$. With $x_1=1$: $x_3=3$, $x_4=4$, $x_2=5$. So $\mathrm{C_3H_8+5O_2\to3CO_2+4H_2O}$.

8 $\mathbf{w}\ne\mathbf{0}$ because $\mathbf{u},\mathbf{v}$ are not parallel. Suppose $a\mathbf{u}+b\mathbf{v}+c\mathbf{w}=\mathbf{0}$. Dot with $\mathbf{w}$, using $\mathbf{u}\cdot\mathbf{w}=\mathbf{v}\cdot\mathbf{w}=0$: $c\|\mathbf{w}\|^2=0$, so $c=0$. Then $a\mathbf{u}+b\mathbf{v}=\mathbf{0}$ gives $a=b=0$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. A farm supplier stocks four fertilisers. Is one of them redundant — can its nutrient content be mixed exactly from the other three?

Why it matters. Fertiliser is described by its N–P–K content (percent nitrogen, phosphate and potash by mass), so each product is a vector in $\mathbb{R}^3$. Four vectors in $\mathbb{R}^3$ are always dependent, so in principle one product can be blended from the others. Agronomists in India, Kenya, Brazil and Oman use exactly this reasoning to cut costs, reduce storage and plan blends when a product is out of stock.

What to do.

Sample case: urea $(46,0,0)$, diammonium phosphate (DAP) $(18,46,0)$, muriate of potash (MOP) $(0,0,60)$ and a compound "15–15–15" $(15,15,15)$ — each in kg of nutrient per 100 kg of product.
Find the weights $a,b,c$ with $a(46,0,0)+b(18,46,0)+c(0,0,60)=(15,15,15)$ by back substitution, and write the dependence relation among the four vectors.
Interpret: how many kg of urea, DAP and MOP supply the same nutrients as 1 kg of 15–15–15? How many kg of inert filler would make the mixture weigh 1 kg?
Look up current local prices for the four products and decide which is cheaper per kg of nutrient: buying the compound, or blending.

Tools: Calculator or spreadsheet; optionally Python/sympy for exact fractions.

Hand in: A half-page note with the weights, the dependence relation, the price comparison and one practical reason (granule size, mixing evenly, storage) why a farmer might still buy the compound.

Going further. Add a fifth product, e.g. single superphosphate $(0,16,0)$. Now there are many blends; which one is cheapest? This is the start of linear programming.

Instructor note: $c=\tfrac14$, $b=\tfrac{15}{46}\approx0.326$, $a=\tfrac{105}{529}\approx0.198$; filler $\approx0.225$ kg. The relation is $\tfrac{105}{529}\mathbf{u}+\tfrac{15}{46}\mathbf{d}+\tfrac14\mathbf{m}-\mathbf{c}=\mathbf{0}$. Students should notice that the three straight fertilisers are independent (a triangular matrix with nonzero diagonal), so the blend is unique.

🧪 Self-check quiz

0/5 correct

Q1. Vectors $\mathbf{v}_1,\dots,\mathbf{v}_k$ are linearly independent when $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k=\mathbf{0}$

Q2. The set $\{(1,2),(3,6)\}$ is

Q3. Any $5$ vectors in $\mathbb{R}^4$ are

Q4. The reduced row echelon form of $\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\mathbf{v}_3\end{bmatrix}$ is $\begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\end{bmatrix}$. Which is true?

Q5. Is $(1,2,3)$ in $\operatorname{span}\{(1,0,0),(0,1,0)\}$?

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