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A basis as the set that spans and wastes nothing, why every basis of a space has the same size, and the dimension that follows from it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say what a basis is and why both halves of the definition are needed, find the coordinate column of a vector in a given basis, extract a basis from a spanning set by taking the pivot columns of the original matrix, extend an independent set to a basis by appending the standard vectors, give the dimension of the standard spaces of columns, polynomials and matrices, and use the fact that any set of the right size with either property is a basis.
You can test a set of vectors for independence and decide what it spans. Those two properties pull against each other: adding vectors makes spanning easier and independence harder, removing them does the reverse. This lesson is about the sets that manage both at once, and about the surprising fact that all of them are the same size.
A basis of a vector space $V$ is a set of vectors that is linearly independent and spans $V$. Given a basis $\mathcal{B} = \{b_1, \dots, b_n\}$, every $v \in V$ is $c_1b_1 + \dots + c_nb_n$ for exactly one list of scalars, and that list is the coordinate vector $[v]_{\mathcal{B}}$. The dimension of $V$ is the number of vectors in any basis of it. The standard basis of $\mathbb{R}^n$ is $e_1, \dots, e_n$, the columns of the identity matrix.
A spanning set reaches everything and may be wasteful. An independent set wastes nothing and may not reach everything. A basis does both, and that combination is exactly what makes coordinates work:
Spanning gives existence — every $v$ can be written as a combination of the basis. Independence gives uniqueness — it can be written that way in only one manner. If $v = \sum c_ib_i$ and also $v = \sum d_ib_i$, subtracting gives $\sum (c_i - d_i)b_i = 0$, and independence forces $c_i = d_i$ for every $i$.
So a basis is precisely what is needed to turn an abstract vector into a definite list of numbers, and neither half of the definition is decoration.
Every basis of a space has the same size. Suppose $\mathcal{B}$ has $n$ vectors and $\mathcal{C}$ has $m$. Each vector of $\mathcal{C}$ is a combination of $\mathcal{B}$, so writing everything in $\mathcal{B}$-coordinates puts $\mathcal{C}$ inside $\mathbb{R}^n$ as $m$ independent vectors — and more than $n$ vectors in $\mathbb{R}^n$ are dependent, so $m \le n$. Swapping the roles gives $n \le m$. Hence $m = n$, and the common value deserves a name: the dimension.
That theorem is what makes dimension a property of the space rather than of the basis you happened to choose, and every count in the rest of this course rests on it.
| Space | A basis | Dimension |
|---|---|---|
| $\mathbb{R}^n$ | $e_1, \dots, e_n$ | $n$ |
| $P_n$ | $1, x, x^2, \dots, x^n$ | $n + 1$ |
| $M_{2 \times 2}$ | the four single-entry matrices | $4$ |
| a plane through $0$ in $\mathbb{R}^3$ | any two independent vectors in it | $2$ |
| $\{0\}$ | the empty set | $0$ |
Another way: picture
A basis of $\mathbb{R}^2$ is a pair of arrows that are not parallel — any such pair. They set up a grid of parallelograms over the plane, and a vector's coordinates say how far to go along each of the two grid directions. Change the basis and the grid tilts and stretches; every point still has coordinates, and they are different numbers for the same point. The standard basis is the one grid that happens to be square, which is the only thing special about it.
Another way: steps
Two procedures, and they are the same procedure run from opposite ends.
Too many vectors: shrink. A spanning set that is dependent has a redundant vector, and dropping it leaves the span unchanged — because the dropped vector was a combination of the others, so anything reached through it is reached without it. Repeat until independent. Mechanically: put the vectors in as columns, reduce, and take the pivot columns of the original matrix.
Why the original? Because row operations change the columns. They preserve which combinations of columns vanish, which is why the pivot positions are trustworthy, but the column vectors themselves are altered, and the reduced columns generally span a different subspace. Pivot positions come from the reduced form; the vectors come from the original.
Too few vectors: grow. An independent set that does not span misses some vector $w$; adjoining it keeps the set independent, because $w$ was not a combination of what was there. Repeat until spanning. Mechanically in $\mathbb{R}^n$: write your independent vectors as the first columns, append $e_1, \dots, e_n$ after them, reduce, and keep the pivot columns. The pivots take your vectors first and then whichever standard vectors are needed to finish the job.
Both halves terminate for the same reason — the dimension bounds an independent set above and a spanning set below — and both end at the same size, which is the theorem of this lesson.
Two facts get used constantly once dimension exists.
In a space of dimension $n$, any $n$ independent vectors are a basis, and any $n$ vectors that span are a basis. One property plus the right count gives the other free. So to check that three given vectors are a basis of $\mathbb{R}^3$ you need only reduce and see three pivots; there is no separate spanning check to run. This shortcut is used far more often than the definition.
A subspace $W \subseteq V$ has $\dim W \le \dim V$, with equality only when $W = V$. A basis of $W$ is an independent set in $V$, so it has at most $\dim V$ vectors. And if it has exactly $\dim V$ of them, it is an independent set of full size in $V$ and therefore a basis of $V$ — so $W$, which contains its span, is all of $V$.
That second fact is worth more than it looks. It means a proper subspace is strictly smaller in dimension, so the subspaces of $\mathbb{R}^3$ are exactly the objects of dimension $0, 1, 2, 3$, and there is nothing in between a plane and the whole space. It is also how many later proofs finish: show that one subspace sits inside another and that the dimensions agree, and the two are equal.
Taking the pivot columns of the reduced matrix. The pivot positions come from the reduced form and the vectors come from the original. Row operations change the columns, so the reduced columns usually span something else.
Counting the vectors handed over instead of the basis. Three vectors can span a two-dimensional space. Dimension is what survives the pruning.
Thinking the standard basis is the basis. Every space of dimension $n$ has infinitely many bases, and the standard one is special only in $\mathbb{R}^n$ and only because it is convenient. A vector's coordinates depend on the basis; the vector does not.
Off by one in $P_n$. The polynomials of degree at most $n$ have dimension $n + 1$, because the exponents run from $0$ to $n$. The polynomials of degree exactly $n$ are not even a subspace.
Expecting the empty basis to be a mistake. $\{0\}$ has dimension $0$ and the empty set as its basis. The empty combination is the zero vector, so the empty set does span it, vacuously and correctly.
Do $(1, 1, 0)$, $(2, 2, 0)$, $(0, 1, 1)$ span a plane or all of $\mathbb{R}^3$? Columns: $\begin{pmatrix} 1 & 2 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.
Vectors as columns.
Row-reducing gives pivots in columns 1 and 3 and none in column 2, since the second vector is twice the first.
Two pivots, so dimension two.
A basis is $\{(1, 1, 0), (0, 1, 1)\}$ — columns 1 and 3 of the original. The span is a plane, not the whole space, even though three vectors were offered.
Original columns, pivot positions.
Extend $\{(1, 2, 1)\}$ to a basis of $\mathbb{R}^3$. Append the standard vectors: columns $(1,2,1), e_1, e_2, e_3$.
Your vectors first, so the pivots prefer them.
Reducing, the pivots land in columns 1, 2 and 3 — that is $(1,2,1)$, $e_1$ and $e_2$ — and column 4 is free.
Three pivots in a three-dimensional space.
So $\{(1,2,1), e_1, e_2\}$ is a basis. A different ordering of the appended vectors would give a different basis, equally correct and equally of size three.
Many bases, one size.
Three vectors in a space of dimension three, so one check suffices: independence or spanning, either one.
Use the shortcut, not the definition.
As columns the matrix is upper triangular with ones on the diagonal, so it reduces to the identity: three pivots.
Triangular with non-zero diagonal is already the answer.
Independent, and three of them, so it is a basis. Its coordinate grid is a sheared version of the usual one, and every vector still has exactly one coordinate list in it.
In the basis $u = (1, 1)$, $v = (1, -1)$ of $\mathbb{R}^{2}$, write the coordinate column of the vector $(5, -1)$.
This task has no paper form; do it on a device.
You are given the spanning set $(3, 0, 0)$, $(0, 6, 0)$, $(3, 6, 0)$ and asked for a basis of its span. Put the steps in order.
Number the steps in order (write the number in the box):
What is the dimension of the span of $(5, 0, 0)$, $(0, 3, 0)$ and $(5, 3, 0)$ in $\mathbb{R}^{3}$?
Answer:
$V$ is a vector space of dimension $4$. Which of these is impossible?
Fill in the dimension of each space.
| Dimension | |
|---|---|
| $\mathbb{R}^{5}$ | |
| Polynomials of degree at most $3$ | |
| The $3 \times 3$ matrices | |
| The zero subspace |
The plane $x + 3y + 4z = 0$ in $\mathbb{R}^{3}$ is a subspace of dimension $2$. Take $y$ and $z$ as the free variables. Enter the $3 \times 2$ matrix whose first column is the basis vector with $y = 1, z = 0$ and whose second column is the one with $y = 0, z = 1$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
In the basis $u = (1, 1)$, $v = (1, -1)$ of $\mathbb{R}^{2}$, write the coordinate column of the vector $(7, -1)$.
This task has no paper form; do it on a device.
You can produce a basis from a spanning set or from an independent one, and give the dimension of a subspace. Say in your own words why every basis of a space has the same number of vectors. Next: the four subspaces every matrix carries, and how one reduction gives a basis for each.
10. Your turn: is $\{(1, 0, 0), (1, 1, 0), (1, 1, 1)\}$ a basis of $\mathbb{R}^3$?, step 3
Count plus one property is enough.