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Linear Algebra
3/25/05
What’s on the Exam? #2
Material covered:
Text —
Chapter 3 — Sections 2, 3, 4, 6
Chapter 4 — Sections 1, 2
(not including the gray application pages).
Handout —
Results about bases, 3/2/05
Handwritten table, 3/18/05
The homework is still the best study guide.
Checklist (not guaranteed):
Given a vector space V and a subset S of V, decide whether S is a subspace (and
prove the answer).
— First check whether 0 is in S; if not, then S isn’t a subspace.
— Then decide whether S is closed under both addition and scalar
multiplication. If yes, then it’s a subspace; if no, then it isn’t.
The nullspace of an m-by-n matrix A, called N(A), is the set of vectors x in Rn
such that Ax = 0.
The “nullity” of A is the dimension of the nullspace of A.
Given A, find the dimension of its nullspace and a basis for its nullspace.
— To find the dimension of the nullspace, reduce A to REF. The
dimension of the nullspace is the number of free variables.
— To find a basis for the nullspace, reduce further to RREF, then
write the general solution to Ax = 0. Extract the basis from
the general solution; see example at the last line of page 137.
The kernel of a linear transformation L:V→W, ker(L), is the set of vectors x in W
such that L(x) = 0.
Note that if L:Rn→Rm and L is represented by the matrix A, then L(x) = Ax and
the ker(L) is the same as N(A).
The span of a sequence of vectors x1, …, xk is the set of all vectors of the form
x = ax1 + … + axk.
Find the dimension of span(x1, …, xk) and a basis for span(x1, …, xk).
— If the vectors x1, …, xk are linearly independent, then they are a basis
for their span. Otherwise, find one that is linearly dependent on the others
and throw it out; and keep doing this until the remaining vectors are
linearly independent. The vectors you have left are a basis, and the
number of vectors in the basis is the dimension of the span.
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The vectors x1, …, xk is a spanning set for V if their span is all of V.
Note that x1, …, xk are always a spanning set for their own span.
Given a set x1, …, xk of vectors in Rn, decide…
(a) whether they span Rn,
(b) whether they are linearly independent,
(c) whether they are a basis for Rn, and
(d) what is a basis for their span.
— Write the vectors as rows of a matrix. Do row operations until the
matrix is in REF. Then…
— If there is a row of zeros, then the original vectors were NOT
linearly independent. If there is no row of zeros, they were
linearly independent.
— The rows with pivots, if written as vectors, are a basis for
the span of the original vectors.
— If there are n rows with pivots, then the original vectors span
Rn; otherwise they don’t.
— They are a basis for Rn if they are linearly independent AND
they span Rn.
— Or: The vectors form a basis for Rn if, when formed into a matrix
the matrix is both SQUARE and NON-SINGULAR (det ≠ 0).
For an m-by-n matrix A:
Row space = space in Rn spanned by its row vectors
Column space = space in Rm spanned by its column vectors
Row rank = dimension of the row space = Column rank = dimension of
the column space = “rank” of A
Given A, find…
its rank;
a basis for its row space;
a basis for its column space.
— Do row operations on A until it is REF. Then a basis for its row space
is the set of row vectors with pivots in the REF. A basis for the
column space is the ORIGINAL column vectors in A in columns
that end up with pivots in the REF (that is, columns corresponding
to lead variables).
The Image of a linear transformation L from V to W is the subspace of W
formed by all vectors of the form L(x) (for any x in V). (Also called the
image of V under L, or in the text, the range of L.)
n
If L:R →Rm is represented by a matrix A, then the image of L is the same as
the column space of L.
Know that:
Rank of A
+ Dimension of N(A)
=
Number of columns in A.
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Be able to recognize a spanning set, linearly independent set, or a basis in the
polynomial spaces (P3, P4, etc.)
Given a function L from a vector space V to a vector space W, decide whether it
is a linear transformation.
— It is a linear transformation if the following are always true:
L(av) = aL(v) if a is a number and v is in V
L(v+w) = L(v)+L(w) if v and w are in V.
Given a linear transformation L:Rn→Rm find a matrix A such that
L(x) = Ax for every vector x in Rn.
(This is the matrix that represents L with respect to the standard bases
of Rn and Rm.)
— The j-th column of A is the vector L(ej).
Given a linear transformation L:V→W, and bases
v1,…,vn for V and
w1,…,wm for W, find a matrix A that represents the linear transformation
with respect to these bases.
— The matrix A is m-by-n. For each j = 1..n, write L(vj) as a combination
of the wi’s:
L(vj) = a1j w1 + … + amj wm.
Then the j-th column of A is
( a1j )
(…)
( amj ).
(end)
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