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

rectangular matrix cannot have both existence and
uniqueness. If m is different from n, we cannot have r = m
and r = n. A square matrix is the opposite. If m = n, we
cannot have one property without the other. A square
matrix has a left-inverse if and onl

are dependent, so are the rows. (b) The column space of
a 2 by 2 matrix is the same as its row space. (c) The
column space of a 2 by 2 matrix has the same dimension
as its row space. (d) The columns of a matrix are a basis
for the column space. 29. For wh

integration are inverse operations. Or at least integration
followed by differentiation brings back the original
function. To make that happen for matrices, we need the
differentiation matrix from quartics down to cubics,
which is 4 by 5: Adiff = 0 1 0 0

120121002200000001000000
. 156 Chapter 2 Vector Spaces (b) The first 3
rows of U are a basis for the row space of Atrue or
false? Columns 1, 3, 6 of U are a basis for the column
space of Atrue or false? The four rows of A are a basis
for the row space of

kernel contains only v = 0. Why are these transformations
not invertible? (a) T(v1,v2) = (v2,v2) W = R 2 . (b) T(v1,v2)
= (v1,v2,v1 +v2) W = R 3 . (c) T(v1,v2) = v1 W = R 1 . 30.
Suppose a linear T transforms (1,1) to (2,2) and (2,0) to
(0,0). Find T(v) w