A basis is a spanning set with no wasted vectors. Every basis of a space has the same size โ that number is the dimension, $\dim V$.
A spanning set lets us reach every vector of a space; a linearly independent set contains no vector that the others can already build. A set that does both jobs is the most economical description of a vector space there is: nothing is missing, and nothing is wasted. Such a set is called a basis, and it turns abstract vectors into lists of numbers we can compute with โ exactly as the familiar $x$-, $y$- and $z$-directions do for ordinary space.
Bases also answer a question that sounds almost philosophical: how big is a vector space? A plane through the origin in $\mathbb{R}^3$ and the whole of $\mathbb{R}^3$ both contain infinitely many vectors, yet the plane is clearly "smaller". The way to measure this is to count the vectors in a basis. The surprise โ and the main theorem of this section โ is that every basis of a given space has the same number of vectors. That number is the dimension, and it tells an engineer how many independent controls a system has, a data scientist how many genuinely different measurements a data set contains, and a physicist how many numbers are needed to describe a state.
What Is a Basis?
๐ Definition 5.5 โ Basis
Let $V$ be a vector space. An indexed set of vectors $\mathcal B=\{\mathbf{b}_1,\dots,\mathbf{b}_p\}$ in $V$ is a basis for $V$ if
$\mathcal B$ is linearly independent, and
$\mathcal B$ spans $V$, that is, $V=\operatorname{span}\{\mathbf{b}_1,\dots,\mathbf{b}_p\}$.
The zero space $\{\mathbf{0}\}$ has the empty set as its basis.
Both conditions matter. A spanning set may contain redundant vectors (it is "too big"); an independent set may fail to reach everything (it is "too small"). A basis is exactly the right size.
Standard bases
The spaces we use most often each come with a natural basis.
$\mathbb{R}^n$The columns of $I_n$: $\mathbf{e}_1=\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix},\ \dots,\ \mathbf{e}_n=\begin{bmatrix}0\\\vdots\\0\\1\end{bmatrix}$. Any $\mathbf{x}=(x_1,\dots,x_n)$ equals $x_1\mathbf{e}_1+\dots+x_n\mathbf{e}_n$, and the $\mathbf{e}_i$ are independent because $I_n$ has a pivot in every column.
$P_n$The monomials $\{1,x,x^2,\dots,x^n\}$. Every polynomial of degree at most $n$ is $a_0+a_1x+\dots+a_nx^n$, and $a_0+a_1x+\dots+a_nx^n$ is the zero polynomial only when every $a_i=0$.
$M_{m\times n}$The matrices $E_{ij}$ with a $1$ in row $i$, column $j$ and $0$ elsewhere. For $M_{2\times2}$: $E_{11}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $E_{12}$, $E_{21}$, $E_{22}$, and $\begin{bmatrix}a&b\\c&d\end{bmatrix}=aE_{11}+bE_{12}+cE_{21}+dE_{22}$.
Standard bases are convenient, but they are far from the only ones. A space has infinitely many bases, and choosing a good one is often the key to a problem.
๐งญ Problem-solving strategy โ is this set a basis of $\mathbb{R}^n$?
Put the vectors in as the columns of a matrix $A$.
Row reduce. The set is independent exactly when every column has a pivot, and it spans $\mathbb{R}^n$ exactly when every row has a pivot.
So a basis of $\mathbb{R}^n$ must have exactly $n$ vectors, and then $A$ is square: the set is a basis $\iff$ $A$ is invertible $\iff$ $\det A\ne0$.
For $P_n$ or $M_{m\times n}$, first replace each polynomial or matrix by its list of coefficients or entries, then use the same test.
โ๏ธ Example 1 โ Testing a basis of $\mathbb{R}^3$
Let $\mathbf{v}_1=\begin{bmatrix}1\\2\\1\end{bmatrix}$, $\mathbf{v}_2=\begin{bmatrix}0\\1\\3\end{bmatrix}$, $\mathbf{v}_3=\begin{bmatrix}2\\3\\0\end{bmatrix}$, $\mathbf{w}=\begin{bmatrix}2\\5\\5\end{bmatrix}$.
(a) Is $\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$ a basis of $\mathbb{R}^3$? If so, write $\mathbf{x}=(3,7,7)$ in terms of it. (b) Is $\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{w}\}$ a basis?
๐ก Show solutionโถ
(a) Determinant Three vectors in $\mathbb{R}^3$ give a square matrix, so compute its determinant (expanding along row 1): $$\det\begin{bmatrix}1&0&2\\2&1&3\\1&3&0\end{bmatrix}=1(0-9)-0+2(6-1)=-9+10=1\ne0 .$$ The matrix is invertible, so its columns are independent and span $\mathbb{R}^3$: they form a basis.
Check $1(1,2,1)+2(0,1,3)+1(2,3,0)=(3,7,7)$. โ Because $\mathcal B$ is a basis, these weights are the only ones that work (Theorem 5.11 below).
(b) Notice $\mathbf{w}=2\mathbf{v}_1+\mathbf{v}_2=(2,4,2)+(0,1,3)=(2,5,5)$. The set is linearly dependent, so it is not a basis โ its span is only the plane spanned by $\mathbf{v}_1$ and $\mathbf{v}_2$.
โ๏ธ Example 2 โ A basis of $P_2$ that is not the standard one
Show that $\mathcal B=\{1+x,\ x+x^2,\ 1+x^2\}$ is a basis of $P_2$, and express $p(x)=3+x+2x^2$ in terms of it.
๐ก Show solutionโถ
Translate Record each polynomial $a_0+a_1x+a_2x^2$ as the column $(a_0,a_1,a_2)$: $$1+x\mapsto\begin{bmatrix}1\\1\\0\end{bmatrix},\quad x+x^2\mapsto\begin{bmatrix}0\\1\\1\end{bmatrix},\quad 1+x^2\mapsto\begin{bmatrix}1\\0\\1\end{bmatrix}.$$
Test $$\det\begin{bmatrix}1&0&1\\1&1&0\\0&1&1\end{bmatrix}=1(1-0)-0+1(1-0)=2\ne0,$$ so the three columns form a basis of $\mathbb{R}^3$, and the three polynomials form a basis of $P_2$ ($\dim P_2=3$).
Weights $a(1+x)+b(x+x^2)+c(1+x^2)=3+x+2x^2$ gives $a+c=3$, $a+b=1$, $b+c=2$. Adding all three: $2(a+b+c)=6$, so $a+b+c=3$; hence $b=0$, $c=2$, $a=1$.
Check $1(1+x)+0(x+x^2)+2(1+x^2)=3+x+2x^2$. โ
Why a basis gives unique "addresses"
๐ Theorem 5.11 โ Unique representation
Let $\mathcal B=\{\mathbf{b}_1,\dots,\mathbf{b}_n\}$ be a basis of a vector space $V$. Then every $\mathbf{x}$ in $V$ can be written in exactly one way as $$\mathbf{x}=c_1\mathbf{b}_1+c_2\mathbf{b}_2+\dots+c_n\mathbf{b}_n .$$
Why. Such weights exist because $\mathcal B$ spans $V$. If also $\mathbf{x}=d_1\mathbf{b}_1+\dots+d_n\mathbf{b}_n$, subtracting gives $\mathbf{0}=(c_1-d_1)\mathbf{b}_1+\dots+(c_n-d_n)\mathbf{b}_n$, and independence forces every $c_i-d_i=0$. The weights $c_1,\dots,c_n$ are called the coordinates of $\mathbf{x}$ relative to $\mathcal B$; Section 5.5 is devoted to them. Interactive 1 below lets you watch them on a slanted grid.
Interactive 1 โ Coordinates on a Slanted Grid
The two arrows $\mathbf{b}_1$ (red) and $\mathbf{b}_2$ (blue) generate a grid of parallelograms. The point $\mathbf{x}$ is reached by walking $c_1$ steps along $\mathbf{b}_1$ and then $c_2$ steps along $\mathbf{b}_2$ (dashed path). When the vectors form a basis the grid covers the whole plane and every point gets one pair $(c_1,c_2)$. Choose "nearly parallel" or "parallel" to watch the grid squash and the coordinates misbehave. Drag the point, or press Play.
Different bases of the same space can look completely different โ $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$ and $\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\}$ of Example 1, say โ yet both have three vectors. That is no accident.
๐ Theorem 5.12 โ Too many vectors are dependent
If a vector space $V$ has a basis with $n$ vectors, then any set in $V$ containing more than $n$ vectors is linearly dependent. Equivalently, a linearly independent set in $V$ has at most $n$ vectors.
Idea of the proof. Writing each of $m\gt n$ vectors in terms of the $n$ basis vectors turns the question "is there a nontrivial combination equal to $\mathbf{0}$?" into a homogeneous system of $n$ equations in $m$ unknowns. With more unknowns than equations there is always a free variable, hence a nontrivial solution.
๐ Theorem 5.13 โ All bases have the same size
If a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ has exactly $n$ vectors.
Why. If $\mathcal B$ has $n$ vectors and $\mathcal C$ is another basis, then $\mathcal C$ is independent, so by Theorem 5.12 it has at most $n$ vectors. Swapping the roles of $\mathcal B$ and $\mathcal C$, $\mathcal B$ has at most as many vectors as $\mathcal C$. So the two counts are equal.
๐ Definition 5.6 โ Dimension
If $V$ is spanned by a finite set, $V$ is called finite-dimensional, and the dimension of $V$, written $\dim V$, is the number of vectors in any basis of $V$. The dimension of the zero space $\{\mathbf{0}\}$ is $0$. If $V$ is not spanned by any finite set, $V$ is infinite-dimensional.
$\dim P_n=n+1$basis $1,x,\dots,x^n$ โ count the constant!
$\dim M_{m\times n}=mn$one $E_{ij}$ for each entry โ so $\dim M_{2\times2}=4$, and no three matrices can span $M_{2\times2}$
infinite-dimensionalthe space $P$ of all polynomials: any finite set has a highest degree $d$, and $x^{d+1}$ is not in its span
โ๏ธ Example 3 โ Counting parameters is not enough
Find a basis and the dimension of $H=\left\{\begin{bmatrix}a+2b-c\\2a+4b-2c\\b+c\\a+3b\end{bmatrix}:\ a,b,c\in\mathbb{R}\right\}$.
๐ก Show solutionโถ
Split by parameter $$\begin{bmatrix}a+2b-c\\2a+4b-2c\\b+c\\a+3b\end{bmatrix}=a\underbrace{\begin{bmatrix}1\\2\\0\\1\end{bmatrix}}_{\mathbf{u}_1}+b\underbrace{\begin{bmatrix}2\\4\\1\\3\end{bmatrix}}_{\mathbf{u}_2}+c\underbrace{\begin{bmatrix}-1\\-2\\1\\0\end{bmatrix}}_{\mathbf{u}_3},$$ so $H=\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3\}$ is a subspace of $\mathbb{R}^4$.
Look for redundancy $\mathbf{u}_2-3\mathbf{u}_1=(2-3,\ 4-6,\ 1-0,\ 3-3)=(-1,-2,1,0)=\mathbf{u}_3$. So $\mathbf{u}_3$ adds nothing new and may be dropped (Theorem 5.14 below).
Basis $\mathbf{u}_1$ and $\mathbf{u}_2$ are not multiples of each other (compare the third entries $0$ and $1$), so they are independent. A basis is $\{\mathbf{u}_1,\mathbf{u}_2\}$ and $\dim H=2$.
Moral Three parameters did not give a three-dimensional space. The dimension counts independent directions, not letters.
The subspaces of $\mathbb{R}^2$ and $\mathbb{R}^3$
We can now prove that the list of subspaces in Section 5.1 is complete. By Theorem 5.15 below, every subspace $H$ of $\mathbb{R}^3$ is finite-dimensional with $\dim H\le\dim\mathbb{R}^3=3$, so $\dim H$ is $0$, $1$, $2$ or $3$. A basis of one vector spans a line through the origin, a basis of two independent (non-parallel) vectors spans a plane through the origin, and a subspace of dimension $3$ is all of $\mathbb{R}^3$ by Theorem 5.16. So each dimension has a clear picture:
dimension 0only the zero subspace $\{\mathbf{0}\}$
dimension 1lines through the origin, $\operatorname{span}\{\mathbf{v}\}$ with $\mathbf{v}\ne\mathbf{0}$
dimension 2planes through the origin, spanned by two non-parallel vectors
The same argument in $\mathbb{R}^2$, where the possible dimensions are $0$, $1$ and $2$, shows that its only subspaces are $\{\mathbf{0}\}$, the lines through the origin and $\mathbb{R}^2$ itself.
โ๏ธ Example 4 โ A plane and a line through the origin
(a) Find a basis and the dimension of the plane $W=\{(x,y,z):x-2y+z=0\}$. (b) Do the same for the line $L$ with parametric equations $x=2t,\ y=-t,\ z=3t$.
๐ก Show solutionโถ
(a) Solve The equation has one pivot ($x$) and two free variables $y,z$: $x=2y-z$. Hence $$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2y-z\\y\\z\end{bmatrix}=y\begin{bmatrix}2\\1\\0\end{bmatrix}+z\begin{bmatrix}-1\\0\\1\end{bmatrix}.$$
(a) Basis The vectors $(2,1,0)$ and $(-1,0,1)$ span $W$ and are independent (look at the free-variable entries $y,z$: one has $(1,0)$, the other $(0,1)$). So they form a basis and $\dim W=2$. Check: $2-2(1)+0=0$ and $-1-0+1=0$. โ
(b) Every point of $L$ is $t(2,-1,3)$, so $L=\operatorname{span}\{(2,-1,3)\}$. One nonzero vector is independent, so $\{(2,-1,3)\}$ is a basis and $\dim L=1$.
Building and Trimming Bases
In practice we are rarely handed a basis. Usually we have either a spanning set with too many vectors, or an independent set with too few. Two theorems say both situations can always be repaired.
๐ Theorem 5.14 โ The spanning set theorem (trimming)
Let $S=\{\mathbf{v}_1,\dots,\mathbf{v}_p\}$ be a set in $V$ and $H=\operatorname{span}S$.
If some $\mathbf{v}_k$ is a linear combination of the other vectors in $S$, then removing $\mathbf{v}_k$ leaves a set that still spans $H$.
If $H\ne\{\mathbf{0}\}$, some subset of $S$ is a basis of $H$.
In $\mathbb{R}^n$: put the vectors in as columns of $A$ and row reduce; the vectors in the pivot columns form a basis of $\operatorname{span}S$.
๐ Theorem 5.15 โ Extending an independent set; subspaces
Let $V$ be finite-dimensional. Every linearly independent set in $V$ can be extended to a basis of $V$ by adding suitable vectors. Consequently, if $H$ is a subspace of $V$, then $H$ is finite-dimensional and $\dim H\le\dim V$.
๐ Theorem 5.16 โ The basis theorem (the "$n$ vectors in $n$ dimensions" shortcut)
Let $V$ be a vector space with $\dim V=n\ge1$, and let $S$ be a set of exactly $n$ vectors in $V$.
If $S$ is linearly independent, then $S$ is a basis of $V$ (it automatically spans).
If $S$ spans $V$, then $S$ is a basis of $V$ (it is automatically independent).
In particular, if $H$ is a subspace of $V$ with $\dim H=\dim V$, then $H=V$.
The shortcut halves the work: when you already know you have the right number of vectors, you only need to check one of the two conditions. It is also why Example 1 could be settled by a single determinant.
Interactive 2 โ The Basis Builder in $\mathbb{R}^3$
Vectors are added one at a time. Each new vector either enlarges the span (point โ line โ plane โ all of $\mathbb{R}^3$) and is kept, or already lies in the span and is redundant (drawn dashed). The kept vectors are always a basis of the span, and the dimension is how many were kept. Use the slider to turn the picture.
vectors addedโ
$\dim$ of the spanโ
basis of the span (kept)โ
last vectorโ
โ๏ธ Example 5 โ Trimming a spanning set to a basis
Find a basis of $H=\operatorname{span}\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3,\mathbf{v}_4\}$ chosen from among the vectors $\mathbf{v}_1=(1,1,0)$, $\mathbf{v}_2=(2,2,0)$, $\mathbf{v}_3=(0,1,1)$, $\mathbf{v}_4=(1,3,2)$, and describe $H$ geometrically.
Read off The pivots are in columns 1 and 3, so $\{\mathbf{v}_1,\mathbf{v}_3\}$ is a basis of $H$ and $\dim H=2$. The non-pivot columns show the redundancies: $\mathbf{v}_2=2\mathbf{v}_1$ and $\mathbf{v}_4=\mathbf{v}_1+2\mathbf{v}_3$.
Geometry $H$ is a plane through the origin. A normal vector is $\mathbf{v}_1\times\mathbf{v}_3=(1\cdot1-0\cdot1,\ 0\cdot0-1\cdot1,\ 1\cdot1-1\cdot0)=(1,-1,1)$, so $H$ is the plane $x-y+z=0$.
โ๏ธ Example 6 โ Extending an independent set to a basis
Extend $\{\mathbf{u}_1,\mathbf{u}_2\}$, with $\mathbf{u}_1=(1,2,0)$ and $\mathbf{u}_2=(0,1,1)$, to a basis of $\mathbb{R}^3$.
๐ก Show solutionโถ
Method Append a spanning set we know โ the standard basis โ and trim. Since $\mathbf{u}_1,\mathbf{u}_2$ come first and are independent, they will be pivot columns and survive the trimming.
Choose Pivots in columns 1, 2, 3: the set $\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{e}_1\}$ is a basis of $\mathbb{R}^3$.
Check By Theorem 5.16 we only need independence of three vectors in $\mathbb{R}^3$: $\det\begin{bmatrix}1&0&1\\2&1&0\\0&1&0\end{bmatrix}=1(0-0)-0+1(2-0)=2\ne0$. โ
โ ๏ธ Common mistakes with bases and dimension
1. "Three vectors in $\mathbb{R}^3$ are always a basis." Only if they are independent โ $\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{w}\}$ in Example 1 is not. The right number is necessary but not sufficient.
2. Counting entries instead of directions. A plane in $\mathbb{R}^3$ has vectors with three entries but dimension $2$. And $\dim P_2=3$, not $2$.
3. Trimming with the wrong vectors. When you row reduce to find pivot columns, take the basis vectors from the original list, not from the echelon form: row operations change the column vectors (Section 5.4 explains why).
4. A subspace that misses the origin. A line or plane that does not pass through $\mathbf{0}$ is not a subspace at all, so it has no dimension in this sense.
Exercises
I
Decide whether the set is a basis of the given vector space. Give a reason.
1 $\left\{\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}3\\4\end{bmatrix}\right\}$ in $\mathbb{R}^2$
2 $\left\{\begin{bmatrix}2\\-1\end{bmatrix},\begin{bmatrix}-6\\3\end{bmatrix}\right\}$ in $\mathbb{R}^2$
3 $\{(1,0,1),\ (0,1,1),\ (1,1,0)\}$ in $\mathbb{R}^3$
4 $\{(1,1,1),\ (1,2,3),\ (2,3,4)\}$ in $\mathbb{R}^3$
5 $\{(1,2,3),\ (4,5,6)\}$ in $\mathbb{R}^3$
6 $\{(1,0,0),\ (0,1,0),\ (0,0,1),\ (1,1,1)\}$ in $\mathbb{R}^3$
7 $\{1,\ 1+x,\ (1+x)^2\}$ in $P_2$
8 $\{1-x,\ x-x^2,\ 1-x^2\}$ in $P_2$
9 $\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}1&1\\1&0\end{bmatrix},\begin{bmatrix}1&1\\1&1\end{bmatrix}\right\}$ in $M_{2\times2}$
Check: three vectors in $\mathbb{R}^3$ โ type them in as columns
$\det$โ
rank (number of pivots)โ
๐ก Show solutionsโถ
1 Two vectors in $\mathbb{R}^2$: $\det\begin{bmatrix}1&3\\2&4\end{bmatrix}=4-6=-2\ne0$. Basis.
2 $(-6,3)=-3(2,-1)$, so the set is dependent ($\det=6-6=0$). Not a basis; it spans only a line.
4 $(1,1,1)+(1,2,3)=(2,3,4)$: dependent ($\det=0$). Not a basis.
5 Only two vectors, and $\dim\mathbb{R}^3=3$: two vectors can span at most a plane. Not a basis (although the set is independent).
6 Four vectors in a $3$-dimensional space are dependent by Theorem 5.12 (indeed $(1,1,1)=\mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_3$). Not a basis.
7 Coefficient columns $(1,0,0)$, $(1,1,0)$, $(1,2,1)$ (since $(1+x)^2=1+2x+x^2$) form an upper triangular matrix with determinant $1\cdot1\cdot1=1\ne0$. Three independent vectors in the $3$-dimensional $P_2$: basis by Theorem 5.16.
8 $(1-x)+(x-x^2)=1-x^2$, so the set is dependent. Not a basis.
9 Flatten each matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ to $(a,b,c,d)$: the columns $(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)$ give an upper triangular $4\times4$ matrix with determinant $1$. Four independent vectors in the $4$-dimensional $M_{2\times2}$: basis.
II
Find a basis of the subspace and state its dimension.
1 $x=-y-z$: $(x,y,z)=y(-1,1,0)+z(-1,0,1)$. Basis $\{(-1,1,0),(-1,0,1)\}$, dimension $2$ (a plane).
2 $(x,y,z)=(3y,y,-2y)=y(3,1,-2)$. Basis $\{(3,1,-2)\}$, dimension $1$ (a line).
3 $a(1,2,1)+b(-1,0,1)$; the two vectors are not parallel. Basis $\{(1,2,1),(-1,0,1)\}$, dimension $2$.
4 $a(1,1,0,1)+b(1,-1,1,0)+c(0,2,-1,1)$, and $(0,2,-1,1)=(1,1,0,1)-(1,-1,1,0)$. Drop it: basis $\{(1,1,0,1),(1,-1,1,0)\}$, dimension $2$.
5 $p=a_0+a_1x+a_2x^2$ with $a_0+a_1+a_2=0$, so $a_0=-a_1-a_2$ and $p=a_1(x-1)+a_2(x^2-1)$. Basis $\{x-1,\ x^2-1\}$ (different degrees, so independent); dimension $2$.
6 $\begin{bmatrix}a&b\\b&d\end{bmatrix}=a\begin{bmatrix}1&0\\0&0\end{bmatrix}+b\begin{bmatrix}0&1\\1&0\end{bmatrix}+d\begin{bmatrix}0&0\\0&1\end{bmatrix}$. These three are independent (each has a $1$ where the others have $0$). Dimension $3$.
1 Remove vectors from $\{(1,2),(2,4),(1,3)\}$ to obtain a basis of $\mathbb{R}^2$.
2 Find a basis of $\operatorname{span}\{(1,0,2),(2,1,3),(0,1,-1),(3,1,5)\}$ chosen from among these vectors, and give its dimension.
3 Extend $\{(1,1,0)\}$ to a basis of $\mathbb{R}^3$ using standard basis vectors.
4 Extend $\{(1,0,1,0),(0,1,0,1)\}$ to a basis of $\mathbb{R}^4$.
5 Extend $\{1+x^2\}$ to a basis of $P_2$.
6 Find a basis of $P_2$ chosen from $\{1+x,\ 1-x,\ 2,\ x+x^2\}$.
๐ก Show solutionsโถ
1 $\begin{bmatrix}1&2&1\\2&4&3\end{bmatrix}\xrightarrow{R_2\to R_2-2R_1}\begin{bmatrix}1&2&1\\0&0&1\end{bmatrix}$: pivots in columns 1 and 3. Basis $\{(1,2),(1,3)\}$; $(2,4)=2(1,2)$ is the redundant vector.
2 $\begin{bmatrix}1&2&0&3\\0&1&1&1\\2&3&-1&5\end{bmatrix}\xrightarrow{R_3\to R_3-2R_1}\begin{bmatrix}1&2&0&3\\0&1&1&1\\0&-1&-1&-1\end{bmatrix}\xrightarrow{R_3\to R_3+R_2}\begin{bmatrix}1&2&0&3\\0&1&1&1\\0&0&0&0\end{bmatrix}$. Pivots in columns 1, 2: basis $\{(1,0,2),(2,1,3)\}$, dimension $2$. (Indeed $(0,1,-1)=(2,1,3)-2(1,0,2)$ and $(3,1,5)=3(1,0,2)+(0,1,-1)$.)
3 Row reduce $[\,(1,1,0)\ \mathbf{e}_1\ \mathbf{e}_2\ \mathbf{e}_3\,]$: $\begin{bmatrix}1&1&0&0\\1&0&1&0\\0&0&0&1\end{bmatrix}\xrightarrow{R_2\to R_2-R_1}\begin{bmatrix}1&1&0&0\\0&-1&1&0\\0&0&0&1\end{bmatrix}$. Pivots in columns 1, 2, 4: basis $\{(1,1,0),\mathbf{e}_1,\mathbf{e}_3\}$. ($\mathbf{e}_2=(1,1,0)-\mathbf{e}_1$ is not needed.)
4 Appending $\mathbf{e}_1,\dots,\mathbf{e}_4$ and row reducing gives pivots in columns 1โ4, so $\{(1,0,1,0),(0,1,0,1),\mathbf{e}_1,\mathbf{e}_2\}$ is a basis. Check: the $4\times4$ matrix with these columns has determinant $1\ne0$ (expand along row 3, then row 4).
5 In coefficients: $(1,0,1)$, then append $(1,0,0),(0,1,0),(0,0,1)$. Pivots in columns 1, 2, 3, so $\{1+x^2,\ 1,\ x\}$ is a basis of $P_2$. (Determinant of the columns $(1,0,1),(1,0,0),(0,1,0)$ is $1\ne0$.)
6 Coefficient columns $(1,1,0),(1,-1,0),(2,0,0),(0,1,1)$. Row reduction gives pivots in columns 1, 2, 4, so $\{1+x,\ 1-x,\ x+x^2\}$ is a basis. ($2=(1+x)+(1-x)$ is redundant.)
IV
Thinking with dimension, and an application.
1 For which values of $k$ is $\{(1,2,3),\ (2,5,7),\ (1,1,k)\}$ a basis of $\mathbb{R}^3$?
2 True or false: any four vectors in $\mathbb{R}^3$ are linearly dependent.
3 True or false: any three nonzero vectors in $\mathbb{R}^3$ form a basis of $\mathbb{R}^3$.
4 True or false: if $H$ is a subspace of $\mathbb{R}^3$ and $\dim H=3$, then $H=\mathbb{R}^3$.
5 True or false: $\dim P_3=3$.
6 Explain why the space $P$ of all polynomials is infinite-dimensional.
7Nutrition. One serving of each of three foods supplies (protein, carbohydrate, fat) in grams: $F_1=(10,20,5)$, $F_2=(5,30,0)$, $F_3=(20,0,10)$. Show that the three nutrient vectors form a basis of $\mathbb{R}^3$, and find the numbers of servings that give exactly $45$ g protein, $70$ g carbohydrate and $20$ g fat. What does the basis property say about other targets, and what is its practical limitation?
๐ก Show solutionsโถ
1 $\det\begin{bmatrix}1&2&1\\2&5&1\\3&7&k\end{bmatrix}=1(5k-7)-2(2k-3)+1(14-15)=k-2$. It is a basis exactly when $k\ne2$. (For $k=2$: $(1,1,2)=3(1,2,3)-(2,5,7)$.)
2True $\dim\mathbb{R}^3=3$, and Theorem 5.12 says more than $3$ vectors are dependent.
3False They must also be independent: $(1,0,0),(2,0,0),(0,1,0)$ are nonzero but all lie in the $xy$-plane.
4True A basis of $H$ is $3$ independent vectors in $\mathbb{R}^3$, hence a basis of $\mathbb{R}^3$ (Theorem 5.16); so $H=\mathbb{R}^3$.
5False $\{1,x,x^2,x^3\}$ is a basis, so $\dim P_3=4$.
6 If finitely many polynomials $p_1,\dots,p_k$ spanned $P$, let $d$ be the largest of their degrees. Every combination of them has degree at most $d$, so $x^{d+1}$ is not in their span โ a contradiction. No finite set spans $P$. (Equivalently, $1,x,x^2,\dots,x^m$ are independent for every $m$, which is impossible in a space of finite dimension $n$ once $m+1\gt n$.)
7 $\det\begin{bmatrix}10&5&20\\20&30&0\\5&0&10\end{bmatrix}=10(300)-5(200)+20(-150)=-1000\ne0$, so the vectors form a basis. Solving $s_1F_1+s_2F_2+s_3F_3=(45,70,20)$ gives $s_1=2$, $s_2=1$, $s_3=1$ servings. Check: $(20+5+20,\ 40+30+0,\ 10+0+10)=(45,70,20)$ โ. Because the vectors form a basis, every target $(p,c,f)$ is reached by exactly one choice of servings โ but some targets need a negative number of servings, which is impossible in a real diet. The basis guarantees a unique mathematical answer, not a practical one.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. Why can a phone or television screen show (almost) any colour with only three kinds of light, and which colours can it not show?
Why it matters. The human eye has three types of colour receptor (cones), so a perceived colour is described by three numbers: colour is, to a good approximation, a $3$-dimensional vector space. A screen's red, green and blue lights give three vectors in that space. If they form a basis, every colour is a unique combination of them โ but a screen can only use non-negative amounts of light. Display engineers, printing companies and camera makers all work with this "gamut" problem.
What to do.
Sample case (illustrative numbers in arbitrary units): suppose the three cone responses to the screen's lights are $\mathbf{r}=(8,3,0)$, $\mathbf{g}=(5,8,1)$, $\mathbf{b}=(1,2,9)$. Show that $\{\mathbf{r},\mathbf{g},\mathbf{b}\}$ is a basis of $\mathbb{R}^3$.
Find the light intensities that produce the target colours $(14,13,10)$, $(7,9,5)$ and $(2,6,7)$.
Which target needs a negative intensity? Explain why the screen cannot display it, even though the mathematics gives an answer.
Search for a CIE chromaticity diagram with a typical screen gamut triangle (for example sRGB) and describe where the "impossible" colours lie.
Tools: Calculator, a spreadsheet or Python (to solve $3\times3$ systems), the internet for the gamut diagram
Hand in: A half-page note with the three solutions, the colour that cannot be displayed, and one sentence linking "basis" to "every colour has a unique recipe".
Going further. Printers use cyan, magenta, yellow and black inks โ four "vectors" in a $3$-dimensional colour space. Why do they still use four, when three already span? (Think about cost, dark tones and non-negativity.)
Instructor note: $\det[\mathbf{r}\ \mathbf{g}\ \mathbf{b}]=428\ne0$. Intensities: $(14,13,10)\to(1,1,1)$; $(7,9,5)\to\left(\tfrac{26}{107},\tfrac{197}{214},\tfrac{97}{214}\right)\approx(0.243,0.921,0.453)$; $(2,6,7)\to\left(-\tfrac{55}{214},\tfrac{287}{428},\tfrac{301}{428}\right)\approx(-0.257,0.671,0.703)$ โ the last needs negative red, so it lies outside the gamut. Students should separate "spans" (mathematics) from "reachable with non-negative weights" (physics).
๐งช Self-check quiz
0/5 correct
Q1. What is $\dim P_3$?
Q2. What is $\dim M_{2\times3}$?
Q3. Which set is a basis of $\mathbb{R}^2$?
Q4. Three linearly independent vectors in $\mathbb{R}^3$ โฆ
Q5. The subspace $\{(x,y,z):x+y+z=0\}$ of $\mathbb{R}^3$ has dimension