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Subspaces, Basis, Dimension, and Rank
Definition. A subspace of Rn^ is any collection S of vectors in Rn^ such that
- The zero vector ~ 0 is in S.
- If ~u and ~v are in S, then ~u+~v is in S (that is, S is closed under addition).
- If ~u is in S and c is a scalar, then c~u is in S (that is, S is closed under multiplication by scalars).
Remark. Property 1 is needed only to ensure that S is non-empty; for non-empty S property 1 follows from property 3, as 0 ~a = ~ 0.
Theorem 3.19. Let ~v 1 , ~v 2 ,... , ~vk be vectors in Rn. Then span (~v 1 , ~v 2 ,... , ~vk) is a subspace of Rn.
Subspaces Associated with Matrices
Definition. Let A be an m × n matrix.
- The row space of A is the subspace row(A) of Rn^ spanned by the rows of A.
- The column space of A is the subspace col(A) of Rm^ spanned by the columns of A.
If we need to determine if ~b belongs to col(A), this is actually the same problem as whether ~b ∈ span of the columns of A; see the method on p. 9.
If we need to determine if ~b belongs to row(A), then we can apply the same method as above to the columns ~bT^ and col(AT^ ). Another method for the same task is described in Example 3.41 in the Textbook.
Theorem 3.20. Let B be any matrix that is row equivalent to a matrix A. Then row(B) =row(A).
See the theorem on p. 13.
Theorem 3.21. Let A be an m × n matrix and let N be the set of solutions of the homogeneous linear system A~x = ~ 0. Then N is a subspace of Rn.
Definition. Let A be an m × n matrix. The null space of A is the subspace of Rn^ consisting of solutions of the homogeneous linear system A~x = ~ 0. It is denoted by null(A).
Theorem. Let B be any matrix that is row equivalent to a matrix A. Then null(B) =null(A).
This is the Fund. Th. on e.r.o.s, see p. 4.
E.g., the set {[x 1 , x 2 , x 3 ] | x 1 + x 2 + x 3 = 0} is automatically a subspace of R^3 — no need to verify those closedness properties 1, 2, 3, as this is the null space of the homogeneous system x 1 + x 2 + x 3 = 0 (consisting of one equation).
Basis
Definition. A basis for a subspace S of Rn^ is a set of vectors in S that
- spans S and
- is linearly independent.
Remark. It can be shown that this definition is equivalent to each of the following two definitions:
Definition ′. A basis for a subspace S of Rn^ is a set of vectors in S that spans S and is minimal with this property (that is, any proper subset does not span S).
Definition ′′. A basis for a subspace S of Rn^ is a set of vectors in S that is linearly independent and is maximal with this property (that is, adding any other vector in S to this subset makes the resulting set linearly dependent).
Method for finding a basis of row (A). Reduce A to r.r.e.f. R by e.r.o.s. (We know row(A) = row(R).) The non-zero rows of R, say, ~b 1 ,... ,~br , form a basis of row(R) = row(A). Indeed, they clearly span row(R), as zero rows contribute nothing. The fact that the non-zero rows are linearly indepen- dent can be seen from columns with leading 1s: in a linear combination∑ ci~bi the coordinate in the column of the 1st leading 1 is c 1 , since there are only zeros above and below this leading 1; also the coordinate in the column of the 2nd leading 1 is c 2 , since there are only zeros above and below this leading 1; and so on. If
ci~bi = ~ 0 , then we must have all ci = 0. Moreover, the same is true for any r.e.f. Q (not necessarily reduced r.e.f.): The non-zero rows of Q form a basis of row(Q)=row(A).
1st Method for finding a basis of col (A). Use the previous method applied to AT^.
For the examination, no need to have proof. But, for the com- pleteness of exposition, I give a proof of existence of basis, Theo- rem 3.23+(a), here.
The existence of basis. Let S be a non-zero subspace (that is, S does not consist of zero vector only) of Rn. Then S has a basis.
Proof. Consider all linearly independent systems of vectors in S. Since S contains a non-zero vector ~v 6 = ~ 0 , there is at least one such system: ~v. Now, if ~v 1 ,... , ~vk is a system of linearly independent vectors in S, we have k 6 n by Theorem 2.8.
We come to a crucial step of the proof: choose a system of linearly in- dependent vectors ~v 1 ,... , ~vk in such way that k is maximal possible and consider U = span(v 1 ,... , vk).
Observe that U ⊆ S. If U = S, then ~v 1 ,... , ~vk is a basis of S by definition of the basis, and our theorem is proven. Therefore we can assume that U 6 = S and chose a vector ~v ∈ S \ U (in S but not in U ).
The rest of proof of Theorem 3.23 can be taken from the text- book.
Definition. If S is a subspace of Rn, then the number of vectors in a basis for S is called the dimension of S, denoted dim S.
Remark. The zero vector ~ 0 by itself is always a subspace of Rn. (Why?) Yet any set containing the zero vector (and, in particular, {~ 0 }) is linearly dependent, so {~ 0 } cannot have a basis. We define dim{~ 0 } to be 0.
Examples. 1) As we know, the n standard unit vectors form a basis of Rn; thus, dim Rn^ = n.
- If ~v 1 ,... , ~vk are linearly independent vectors, then they form a basis of span(~v 1 ,... , ~vk), so then dim span(~v 1 ,... , ~vk) = k.
We shall need a slightly more general result:
Theorem 3.23++. (a) If v 1 ,... , vk are linearly independent vectors in a subspace S, then they can be included in (complemented to) a basis of S; in particular, k ≤ dim S.
(b) If one subspace is contained in another, S ⊆ T , then dim S ≤ dim T. If both S ⊆ T and dim S = dim T , then S = T.
Example. If we have some n linearly independent vectors ~v 1 ,... , ~vn in Rn, they must also form a basis of Rn, as the dimension of their span is n and we can apply Theorem 3.23++(b).
Theorem 3.24. The row and column spaces of a matrix A have the same dimension.
Definition The rank of a matrix A is the dimension of its row and column spaces and is denoted by rank(A).
Theorem 3.25. For any matrix A,
rank (AT^ ) = rank (A)
Definition The nullity of a matrix A is the dimension of its null space and is denoted by nullity(A).
Theorem 3.26. The Rank–Nullity Theorem
If A is an m × n matrix, then
rank (A) + nullity (A) = n
Theorem 3.27. The Fundamental Theorem of Invertible Matrices
Let A be an n × n matrix. The following statements are equivalent:
a. A is invertible.
b. A~x = ~b has a unique solution for every ~b in Rn.
c. A~x = ~ 0 has only the trivial solution.
d. The reduced row echelon form of A is In.
e. A is a product of elementary matrices.
f. rank(A)= n.
g. nullity(A)= 0.
h. The column vectors of A are linearly independent.
i. The column vectors of A span Rn.
j. The column vectors of A form a basis for Rn.
k. The row vectors of A are linearly independent.
l. The row vectors of A span Rn.
m. The row vectors of A form a basis for Rn.
