The Dimension of a Vector Space

Home , Coordinate vector, Dimension (vector space)

340 Coordinate Vectors Deﬁnition ~ ~ Let β = {b1,..., bn} be an ordered basis for a vector space V and let ~x ∈ V .

The coordinates of ~x relative to the basis β (or the β-coordinates of ~x) are the weights c1,..., cn such that ~ ~ ~x = c1b1 + ··· + cnbn.

In this case, the vector in Rn given by   c1 ~  .  [x]β =  .  cn

is called the coordinate vector of x relative to β (or the

β-coordinate vector of ~x). 341 Example 3 0 Let β = {~b , ~b } where ~b = and ~b = and 1 2 1 1 2 1

1 0 let E = {e~ , e~ } where e~ = and e~ = . 1 2 1 0 2 1

2 Given [~x] = , what is ~x ? β 3

6 Given [~x] = , what is ~x ? E 5

342 Linearly Independent Sets Cannot Have Too Many Vectors Theorem ~ ~ If a vector space V has a basis β = {b1,..., bn}, then any set in V containing more than n vectors must be linearly dependent.

Proof Idea Suppose {u~1,..., u~p} is a set of vectors in V where p > n. n Then the coordinate vectors {[u~1]β,..., [u~p]β} are in R . Since p > n, an earlier result tells us that this set of coordinate vectors must be linearly dependent in Rn, and it follows that {u~1,..., u~p} is linearly dependent in V .

343 All Bases of V Have the Same Number of Vectors Theorem If a vector space V has a basis of n vectors, then every basis of V must consist of n vectors.

Proof Idea Let β1 be a basis for V consisting of exactly n vectors.

Suppose that β2 is any other basis for V . By the deﬁnition of basis, we know that β1 and β2 are both linearly independent sets.

Suppose for the sake of contradiction that β2 does not have n vectors. Then either β1 has more vectors than β2 or β2 has more vectors than β1. In either case, by the Theorem we proved on the prior slide we obtain a contradiction to the linear independence of β1 and β2. 344 Dimension of a Vector Space Definition If V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V , written as dim(V ), is the number of vectors in a basis for V .

The dimension of the zero vector space {~0} is deﬁned to be 0.

If V is not spanned by a ﬁnite set, then V is said to be inﬁnite-dimensional.

345 Example

2 3 The standard basis for P3(R) is {1, t, t , t }.

Thus dim(P3(R)) = 4.

In general, dim(Pn(R)) = n + 1.

346 Example

n The standard basis for R is {e~1,..., e~n}, where e~1,..., e~n are the columns of the n × n identity matrix.

In general, dim(Rn) = n.

347 Example  a + 2b + 2c     2a + 2b + 4c + d  Let W =   : a, b, c, d ∈ R .  b + c + d    3a + 3c + d 

Find a basis for W and determine dim(W ).

348 Dimensions of Subspaces of R3

I 0-dimensional subspace: {~0}

I 1-dimensional subspaces: Span{u~} where u~ 6= ~0 lines passing through the origin

I 2-dimensional subspaces: Span{u~, ~v} where u~ and ~v are nonzero and not multiples of each other planes passing through the origin

I 3-dimensional subspace: R3

349 Dimensions of Subspaces of V Theorem Let H be a subspace of a ﬁnite-dimensional vector space V .

Any basis for H can be extended, if necessary, to a basis for V .

Hence H is ﬁnite-dimensional and dim(H) ≤ dim(V ).

350 Example      1 1  Let H = Span 0 , 1 .  0 0 

Then H is a subspace of R3 and dim(H) < dim(R3).

351 The Basis Theorem Theorem Let V be a p-dimensional vector space with p ≥ 1.

Any linearly independent set of exactly p vectors in V is automatically a basis for V .

Any set of exactly p vectors that spans V is automatically a basis for V .

352 Example

2 Show that {t, 1 − t, 1 + t − t } is a basis for P2(R).

353 Dimensions of Col(A) and Nul(A) 1 2 3 4 Let A = . 2 4 7 8

Find dim(Col(A)) and dim(Nul(A)).

354 Dimensions of Col(A) and Nul(A)

I dim(Col(A)) = number of pivot columns of A

I dim(Nul(A)) = number of free variables of A

355 Acknowledgement Statements of results and some examples follow the notation and wording of Lay’s Linear Algebra and Its Applications, 5th edition.