on the plane x+2y3zt = 0 in R 4 . Then find three
independent vectors. Why not four? This plane is the
nullspace of what matrix? Problems 1118 are about the
space spanned by a set of vectors. Take all linear
combinations of the vectors 11. Describe the su

linear algebra. We never use the terms basis of a
matrix or rank of a space or dimension of a basis.
These phrases have no meaning. It is the dimension of
the column space that equals the rank of the matrix, as
we prove in the coming section. Problem Set

P: Projection onto -line P = " c 2 cs cs s2 # . This matrix
has no inverse, because the transformation has no
inverse. Points on the perpendicular line are projected
onto the origin; that line is the nullspace of P. Points on
the -line are projected to th

the same three rotations, but each one through 180?
150 Chapter 2 Vector Spaces 7. On the space P3 of cubic
polynomials, what matrix represents d 2/dt2 ? Construct
the 4 by 4 matrix from the standard basis 1, t, t 2 , t 3 .
Find its nullspace and column s

they also produce linear transformations. Having gone
that far, there is no reason to stop. The operations in the
linearity condition (1) are addition and scalar
multiplication, but x and y need not be column vectors in
R n . Those are not the only spaces

fundamental theorem of linear algebra connects the
dimensions of the subspaces: Nullspace: dimension 1,
contains x = (1,.,1). Column space: dimension r = n1,
any n1 columns are independent. Row space: dimension
r = n1, independent rows from any spanning t

= a11v1 +am1vm, this is the first column of a matrix
multiplication VA: W = h w1 w2 wn i = h v1 vm i
a11 . . . am1 = VA. We dont know each ai j, but we
know the shape of A (it is m by n). The second vector w2
is also a combination of the vs. The coeffici

function that satisfies dy dx = 3. (c) Find all functions that
satisfy dy dx = 3. 39. The cosine space F3 contains all
combinations y(x) = Acos x + Bcos2x +Ccos3x. Find a basis
for the subspace that has y(0) = 0. 40. Find a basis for the
space of function

On edge i, the conductance is ci and the resistance is
1/ci . Ohms Law says that the current yi through the
resistor is proportional to the voltage drop ei : Ohms Law
yi = ciei (current) = (conductance)(voltage drop). This is
also written E = IR, voltage

column space have the same dimension r! This is one of
the most important theorems in linear algebra. It is often
abbreviated as row rank = column rank. It expresses a
result that, for a random 10 by 12 matrix, is not at all 2.4
The Four Fundamental Subsp

plane V. In this case, with subspaces of dimension 2 and 1
in R 4 , there is room for a third subspace. The line L
through z = (0,0,5,4) is perpendicular to V and W. Then
the dimensions add to 2+1+1 = 4. What space is
perpendicular to all of V, W, and L?

(0,0,2), w4 = (0,0,0), and any b? Problems 1937 are
about the requirements for a basis. 19. If v1,.,vn are
linearly independent, the space they span has
dimension . These vectors are a for that space. If the
vectors are the columns of an m by n matrix, th

coordinate vectors e1,e2,e3 into three given vectors
v1,v2,v3? When will that matrix be invertible? 1.24 If
e1,e2,e3 are in the column space of a 3 by 5 matrix, does
it have a left-inverse? Does it have a right-inverse? 1.25
Suppose T is the linear transf

(0,x1,x2,x3). Find also the left shift matrix B from R 4 back
to R 3 , transforming (x1,x2,x3,x4) to (x2,x3,x4). What are
the products AB and BA? 2.6 Linear Transformations 151
18. In the vector space P3 of all p(x) = a0 + a1x + a2x 2 +
a3x 3 , let S be t

many free variables are there in the solution to Ax = b?
How many free variables are there in the solution to A T y
= f ? How many edges must be removed to leave a
spanning tree? 13. In the graph above with 4 nodes and 6
edges, find all 16 spanning trees.

particular solution of Ax = b. The complete solution has
this arbitrary constant c (like the +C when we integrate in
calculus). This has a meaning if we think of x1, x2, x3, x4
as the potentials (the voltages) at the nodes. The five
components of Ax give

independent. Then A has an n by m left-inverse B such
that BA = In. This is possible only if m n. 122 Chapter 2
Vector Spaces In the existence case, one possible solution
is x = Cb, since then Ax = ACb = b. But there will be other
solutions if there are o

that is projected to zero. 2.6 Linear Transformations 149
(ii) For reflections, that same basis gives H = 1 0 0 1 .
The second basis vector is re- flected onto its negative, to
produce this second column. The matrix H is still 2PI
when the same basis is u

and column, which are deleted when column 4 is
removed from A (making A TCA invertible). The 4 by 4
matrix would have all rows and columns adding to zero,
and (1,1,1,1) would be in its nullspace. Notice that A TCA
is symmetric. It has positive pivots and

9. Find a 1 by 3 matrix whose nullspace consists of all
vectors in R 3 such that x1 + 2x2 +4x3 = 0. Find a 3 by 3
matrix with that same nullspace. 10. If Ax = b always has
at least one solution, show that the only solution to A T y
= 0 is y = 0. Hint: Wha

first column, which is zero. Ap2 is the second column,
which is p1. Ap3 is 2p2 and Ap4 is 3p3. The nullspace
contains p1 (the derivative of a constant is zero). The
column space contains p1, p2, p3 (the derivative of a
cubic is a quadratic). The derivativ

row space. A similar rule applies to every echelon matrix
U or R, with r pivots and r nonzero rows: The nonzero
rows are a basis, and the row space has dimension r. That
makes it easy to deal with the original matrix A. 2M The
row space of A has the same

0 3 3 3 0 0 0 0 0 1 0 1 and B = 1 1 4 4 5 5 .
23. Suppose the 3 by 3 matrix A is invertible. Write bases
for the four subspaces for A, and also for the 3 by 6
matrix B = [A A]. 24. What are the dimensions of the four
subspaces for A, B, and C, if I is the

multiply the rows of A to produce the zero row: y TA = h
y1 ym ih A i = h 0 0 i . The dimension of this
nullspace N(A T ) is easy to find, For any matrix, the
number of pivot variables plus the number of free
variables must match the total number of colum

)b b b b b b b Purdue Ohio State MIT Michigan USC Texas
Notre Dame Georgia Tech Figure 2.7: Part of the graph for
college football. If football were perfectly consistent, we
could assign a potential x j to every team. Then if
visiting team v played home t

unsolvable if Harvard beats Yale, Yale beats Princeton,
and Princeton beats Harvard. More than that, the score
differences in that loop of games have to add to zero:
Kirchhoffs law for score differences bHY +bYP +bPH = 0.
This is also a law of linear alge

transforms (1,0) and (0,1) to (1,4) and (1,5) is M = . The
combination a(1,4) + b(1,5) that equals (1,0) has (a,b) = ( ,
). How are those new coordinates of (1,0) related to M or
M1 ? 41. What are the three equations for A, B, C if the
parabola Y = A+Bx +

complicated, 124 Chapter 2 Vector Spaces to show how it
can be broken into simple pieces. For linear algebra, the
simple pieces are matrices of rank 1: Rank 1 A =
2 1 1 4 2 2 8 4 4 2 1 1 has r = 1. Every row is a
multiple of the first row, so the row spa

kxk 2 = 1 2 +2 2 +3 2 and kxk = q x 2 1 +x 2 2 +x 2 3 . The
extension to x = (x1,.,xn) in n dimensions is immediate.
By Pythagoras n1 times, the length kxk in R n is the
positive square root of x T x: Length squared kxk 2 = x 2 1
+x 2 2 +x 2 n = x T x. (1

does that mean when the fs are currents into the nodes?
4. Compute the 3 by 3 matrix A TA, and show that it is
symmetric but singularwhat vectors are in its
nullspace? Removing the last column of A (and last row
of A T ) leaves the 2 by 2 matrix in the up