2.1 VECTOR SPACES AND SUBSPACES
Elimination can simplify, one entry at a time, the linear system Ax=b. Fortunatelyit also simplifies the theory. The basic questions of existence and uniqueness-Is thereone solution, or no solution, or an infinity of solutions?-are much easier to answerafter elimination. We need to devote one more section to those questions, to find everysolution for an m by n system. Then that circle of ideas will be complete. - P69
The valuable thing for linear algebra is that the extension to n dimensions is so straight forward. For a vector in R⁷ we just need the seven components, even if the geometry is hard to visualize. Within all vector spaces, two operations are possible:
We can add any two vectors, and we can multiply all vectors by scalars.
In other words, we can take linear combinations.
Addition obeys the commutative law x + y = y + x; there is a "zero vector" satisfying0+x= x; and there is a vector"-x" satisfying -x + x = 0. Eight properties (includingthose three) are fundamental; the full list is given in Problem 6 at the end of this section.
A real vector space is a set of vectors together with rules for vector addition andmultiplication by real numbers. Addition and multiplication must produce vectors inthe space, and they must satisfy the eight conditions. - P69
DEFINITION
A subspace of a vector space is a nonempty subset that satisfies therequirements for a vector space: Linear combinations stay in the subspace.
(i) If we add any vectors x and y in the subspace, x + y is in the subspace.
(ii) If we multiply any vector x in the subspace by any scalar c, cx is in the subspace. - P70
We can describe all combinations of the two columns geometrically: Ax = b canbe solved if and only if b lies in the plane that is spanned by the two column vectors(Figure 2.1). This is the thin set of attainable b. If b lies off the plane, then it is not aolon bin combination of the two columns. In that case Ax = b has no solution. - P71
Figure 2.1 The column space C(A), a plane in three-dimensional space.
C(A). Requirements (i) and (ii) for a subspace of R" are easy to check:
(i) Suppose b and b‘ lie in the column space, so that Ax = b for some x and Ax‘ = b‘
for some x‘. Then A(x + x) = b+ b‘, so that b + b‘ is also a combinationof the columns. The column space of all attainable vectors b is closed underaddition.
(ii) If b is in the column space C(A), so is any multiple cb. If some combination ofcolumns produces b (say Ax = b), then multiplying that combination by c willproduce cb. In other words, A(cx) = cb. - P72
The nullspace of a matrix consists of all vectors x such that Ax = 0. It is denotedby N(A). It is a subspace of Rⁿ, just as the column space was a subspace of R^m - P73
2B For any m by n matrix A there is a permutation P, a lower triangular L withunit diagonal, and an m by n echelon matrix U, such that PA = LU. - P79
2C If Ax 0 has more unknowns than equations (n> m), it has at least onespecial solution: There are more solutions than the trivial x = 0. - P81
Problem Set 2.2
10. Which of these rules give a correct definition of the rank of A?
(a) The number of nonzero rows in R.
(b) The number of columns minus the total number of rows.
(c) The number of columns minus the number of free columns.
(d) The number of 1s in R. - P86
2.3 Linear indepandence, basic, and Dimension
The goal of this section is to explain and use four ideas:
1. Linear independence or dependence.
2. Spanning a subspace.
3. Basis for a subspace (a set of vectors).
4. Dimension of a subspace (a number). - P92
2.4 THE FOUR FUNDAMENTAL SUBSPACES
The previous section dealt with definitions rather than constructions. We know what abasis is, but not how to find one. Now, starting from an explicit description of a subspace,
we would like to compute an explicit basis.
Subspaces can be described in two ways. First, we may be given a set of vectorsthat span the space. (Example: The columns span the column space.) Second, we maybe told which conditions the vectors in the space must satisfy. (Example: The nullspaceconsists of all vectors that satisfy Ax = 0.) - P102
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