M3. Multiplication is commutative a b = b a.
M4. There exists a multiplicative inverse a 1 if
a 6= 0. D. The distributive law holds a (b + c) =
ab + ac. Roughly, all of the above mean that
you have no
was not quite sure whether the statistician
was pulling his leg. How can you know that?
was his query. And what is this symbol here?
Oh, said the statistician, this is . And
what is that? The ratio of
your answer to part (a) in the form Y TMY = g.
Make sure you give formulas for the new
unknown column vector Y and constant g in
terms of X, M, C and f. You need not multiply
out any of the matrix exp
Suppose that L is bijective (i.e., one-to-one and
onto). i. Show that dim V = rankL = dim W. ii.
Show that 0 is not an eigenvalue of M. iii.
Show that M is an invertible matrix. (b) Now,
suppose that
that any 2 2 matrix is a zero of its own
characteristic polynomial (in fact this holds for
square matrices of any size). Now if A = P 1DP
then A2 = P 1DP P 1DP = P 1D2P. Similarly
Ak = P 1DkP. So for
+ z + w = 0, z + w = 0 w = (arbitrary), z = ,
y = 0, x = 1 . The solution set is
x y z y = 1 0 : R
5. First det 1 2 3 4 = 2 . All the other
determinants vanish because the first three
rows of each
for the proposed bridge? (d) Find an
orthogonal matrix P such that MP = P D where
D is a diagonal matrix. Be sure to also state
your result for D. 2The bridge is intended for
French and English milita
compute H2 = (I 2XXT )(I 2XXT ) = I 4XXT +
4XXT XXT = I 4XXT + 4X(XT X)XT = I 4XXT +
4X.1.XT = I . So, since HH = I, we have H1 = H.
6. We know Mv = v. Hence M2 v = MMv =
Mv = Mv = 2 v , and similarly
on the right and one going around the outside
of the circuit). Respectively, they give the
equations 60 I 80 3I = 0 80 + 2J V + 3J =
0 60 I + 2J V + 3J 3I = 0 . (F.1) The above
equations are easily so
spancfw_L(u1), . . . , L(uq). Now we show that
cfw_L(u1), . . . , L(uq) is linearly independent. We
argue by contradiction. Suppose there exist
constants d j (not all zero) such that 0 = d
1L(u1) + +
always yes. The proof is not difficult. Take a
vector u and w such that u U W 3 w. This
means that both u and w are in both U and W.
But, since U is a vector space, u + w is also
in U. Similarly, u +
What is the probability that the columns of M
form a basis for B3 ? (Hint: what is the
relationship between the kernel of M and its
eigenvalues?) Note: We could ask the same
question for real vectors:
18 19 20 21 22 23 24 25 . Now test
your skills on det 1 2 3 n n + 1 n +
2 n + 3 2n 2n + 1 2n + 2 2n + 3 3n . . . . . . . . .
n2n+1n2n+2n2n+3n2
. Make sure to jot down a few brief
notes explaining any
How many different linear transformations did
you find? Compare your answer to part (c). (g)
Suppose L1 : B3 B and L2 : B3 B are linear
transformations, and and are bits. Define a
new map (L1 + L2) :
Decomposition 309 Thus LL : W W and L
L : V V and both have eigenvalue
problems. Moreover, as is shown in Chapter
15, both L L and LL have orthonormal
bases of eigenvectors, and both MMT and
MTM can b
exactly to that same point after one orbit of
the earth. Unfortunately, if there is a small
mistake in the original location of the satellite,
which the engineers label by a vector X in R 3
with origi
that the matrix on the left hand side must be
invertible, so we examine its determinant det
1 4 10 4 5 7 3 0 h + 3 = 4 (4 (h +
3) 7 3) + 5 (1 (h + 3) 10 3) = 11(h 3)
Hence we obtain a basis whenever
, 0 1 0 , 1 2 0 1 2 .
The new matrix M0 of the linear
transformation given by M with respect to the
bases O and O0 is M0 = 1 0 0 2 0 0 , so
the singular values are 1, 2. Finally note that
arranging th
bc . (a) For which values of det M does M have
an inverse? (b) Write down all 22 bit matrices
with determinant 1. (Remember bits are either
0 or 1 and 1 + 1 = 0.) (c) Write down all 2 2
bit matrices w
x z + 2w = 1 x + y + z w = 2
y 2z + 3w = 3 5x + 2y z + 4w = 1 (a) Write
an augmented matrix for this system. (b) Use
elementary row operations to find its reduced
row echelon form. (c) Write the sol
Gram-Schmidt algorithm in terms of projection
matrices. 4. Show that if v1, . . . , vk are
linearly independent that the matrix M = (v1
vk) is not necessarily invertible but the matrix
MTM is invert
X and MX is given by X (MX) |X| |MX| =
(x 2 + y 2 ) cos p x 2 + y 2 p x 2 + y 2 = cos .
In other words, the angle is OR . You
should draw two pictures, one where the
angle between X and MX is , the ot
on a racetrack once every five seconds. Our
observations wont be exact, but so long as the
observations are right on average, we can
figure out a best-possible linear function of
position of the bicyc
for this value of a. Also we see that dimU = 2 in
this case. Your picture should be a plane in R 3
though the origin containing the vectors 1
2 3 and 2 1 0 . 7. det 1 x 1 y = y x
, det 1 x x2 1 y y2 1
integer)? If so, what would the associated
eigenvalue be? Now suppose that the matrix N
is nilpotent, i.e. N k = 0 for some integer k 2.
Show that 0 is the only eigenvalue of N. 7. Let
M = 3 5 1 3 ! .