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 notions of +, , and just as for
regular real numbers. Fie
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 the circumference
of the circle to its diameter. Well,
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 expressions you find. If
all has gone well, you have found
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 M is an invertible matrix. i. Show
that 0 is not an eig
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 any matrix polynomial we
have A n + c1A n1 + cn1A + cnI
MX = V , MY1 = 0 = MY2 . where M = 1 0
1 2 1 1 1 1 0 1 2 3 5 2 1 4 and V =
1 2 3 1 . 325 326 Sample First
Midterm (e) This amounts to explicitly
performing the matrix manipulations MX V,
MY1, and MY2 to verify that they are all zero
vectors. 3. As usual
+ 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 matrix are not independent.
Indeed, 2R2 R1 = R3 in 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 military vehicles, so the
exact units, coordinate system and
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 Mk v =
Mk1 v = . . . = k v . So v is an eigenvector of
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 solved (either using an
augmented matrix and row reducing
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) + + d qL(uq) = L(d 1u1 + + d quq).
But since the u j are li
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 + w W. Hence u + w
U W. So closure holds in U W and this
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: If I choose a real
vector at random, what is the proba
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 clever tricks you use. 6.
For which values of a does U
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) : B3 B by (L1 + L2)(v)
= L1(v) + L2(v). Is this map a lin
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 be diagonalized. Next, let us make a
simplifying assumpt
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 origin1 at O, 1This is a spy satellite. The
exact location o
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 h 6= 3. 339
340 Sample Second Midterm 340 F Sample
Fina
, 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 the column vectors of O and O0 into
change of basis matri
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 with determinant 0. (d) Use one of
the above examples to
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 solution set
for the system in the form S = cfw_X0 + X i i
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 invertible. 5. Write out the singular
value decomposition the
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 other where
it is . For CauchySchwartz, |X (MX)| |X|
|MX|
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 bicycle in terms of time.
Suppose M is the matrix for the li
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 z z2 = det 1 x x2
0 y x y2 x 2 0 z x z2 x 2 = (y x)(z
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 ! . Compute M12. (Hint: 212 =
4096.) 8. The Cayley Hamilto