Final Exam Solutions, Math. 221 (s. 02), 11 May 2004
[This may now be the finished product -- more details will be coming if required.]
[For best reading, widen your window so that this entire line is visible, and no horizontal scroll bar shows up.]1. The system of equations
x + y + 2z = a
x + z = b
2x + y + 3z = c
reduces, straightforwardly, to (among other things) the system
x + z = b
y + z = a − b
0 = c − (a + b).
This system is inconsistent -- i.e., has NO solutions, answering (iv) -- when
c − (a + b) ≠ 0, i.e., when
c ≠ a + b.
But when the system IS consistent, i.e., when c = a + b,
there are infinitely many solutions (answering (i)), all with
z arbitrary,
y = (a − b) − z ( = a − (b + z)), and
x = b − z.
As for (iii) and (ii), there are no values of c left
for which the system can have exactly one solution;
and it's never the case that a system
of linear equations can (as some exams hoped) have
more than one solution,
but not infinitely many -- cf. page 4 (reddish
display announcement) or page 59 (Theorem 1.6.1).
2. Don't let the reciprocals of the variables throw you: just treat
1/x, 1/y, and 1/z, as unknowns in their own right,
and solve
3/x + 2/y − 1/z = −15
5/x + 3/y + 2/z = 0
3/x + 1/y + 3/z = 11
for 1/x, 1/y, and 1/z. Then invert once
more, to find x, y, and z. Standard techniques
yield
1/x = −4, 1/y = 2, and 1/z = 7,
whence
x = −1/4, y = 1/2, and z = 1/7.
3. (i) Trace(A +/− B) = Trace(A) +/− Trace(B): always true, cf. Ex. 1.3.23.
(ii) Trace(AB) = Trace(A)Trace(B): not true in general.
Fails, for example, when A = B = the 2×2 matrix with
top row (0, 1) and bottom row (1, 0). Here AB = I2×2,
which has trace 2; but A and B each have trace 0.
(iii) Trace(AB) = Trace(BA): always true. Here's why.
Both those traces turn out to be equal to the sum of ALL the entries in
the matrix C mentioned in the suggestion.
For example, Trace(AB) is the sum of
all the entries on the principal diagonal
of AB. Now the first such entry is
the sum of all possible products
a1j bj1
(right? top row of A "zipped" against
the first column of B), i.e.,
the sum of the entries in the top row of the
matrix C mentioned in the suggestion;
the second such entry is the sum of all the products
a2j bj2,
i.e., the sum of all the entries in the second row of that matrix C;
and so on, down the line.
Thus the trace Trace(AB) of AB turns out to be the sum got by
adding together all the various sums of all the entries in any given
row of C, i.e., it's just the sum of ALL the entries (in all the
rows) in C.
Trace(BA) admits an analogous reinterpretation: each entry
along the diagonal of BA is equal to the sum of all the entries
in the corresponding column of C; and so Trace(BA) = the sum
of all those column-sums, i.e., again, the sum of all the entries in C. QED
4. Matrices of type A will (i) always have determinant zero, and hence
(ii) never be invertible. To understand why, notice for example that the row space
of such an A can never have dimension more than 4, because elementary
row manipulations just among the first three rows alone will surely already
lead to a row of zeros, so that rank(A) < 4.
Or notice that each of the 5! = 120 different products of 5 selected entries
(selecting one from each row and from each column) will have to include
as a factor one of the later 0 's from at least one of the first three rows.
Or, yet again, try evaluating the determinant by blocks: subdivide the
matrix by separating column 2 from column 3 and row 2 from row 3:
because of the 2×3 block of zeroes at the upper right, the
determinant of a type A matrix will be the product of the
determinant of the upper left 2×2 block with the
determinant of the lower right 3×3 block; but the latter block,
because its top row is all-zero, has determinant zero, hence so does A.
Matrices of type B will likewise (i) always have determinant zero,
and hence (ii) never be invertible. To understand why, notice for example that
the only way one of the 5! = 120 different products of 5 selected entries
(selecting one from each row and from each column) can try to be non-zero
is by selecting a non-zero entry from each row. But once the only non-zero
entry in the second row has been selected, it's no longer possible to use
a non-zero entry from the fourth row, as that lies in the same column as
the second-row choice. So all the 5! products contributing to the determinant
are zero, and hence so is the determinant itself.
Or again, by means of a suitable row operation, involving only
subtracting from one of rows 2 or 4 an appropriate multiple
of the other, a matrix of type B can be reduced
to one having an all-zero row, hence has determinant zero.
Matrices of type C, on the other hand, being upper-triangular,
will have as determinant whatever the number is at the intersection
of row 3 with column 3, and will be invertible whenever that number ≠ 0.
5. To say the columns of A are all mutually perpendicular
and of length one is to say that AT A = I.
But for square matrices like A, any "left inverse"
(like AT) must be a "right inverse" as well -- A AT = I -- and that equation, spelled out
entry by entry, lets you read off that the various rows of A
all have length 1 and are mutually perpendicular.
6. (i) If λ is an eigenvalue for A,
with corresponding eigenvector x,
so that Ax = λx,
then AAx = A·λx = λ·Ax =
λ·λx = λ2x.
On the other hand, if A is idempotent,
then AAx = Ax = λx.
Combining both relations, we obtain λ2x
= λx ,
from which,
because x ≠ 0, we may infer λ2 = λ.
And from this it follows that λ must = 0 or = 1.
(ii) (I − A)(I − A) =
I(I − A) − A(I − A) =
(I I − I A) −
(A I − A A) = I − A − A + A =
I − A .
(iii) If A is idempotent and not 0, then A has at least one
non-zero column. Claim: such a column -- call it c -- must satisfy
Ac = c. Reason: if column # j, for example, is the non-zero
column we're calling c, and if we write (as usual)
ej for the column vector with
a 1 in position # j and 0 's in all remaining positions, then
c = Aej, and so
Ac = A Aej =
AA ej = Aej =
c, as claimed.
But then c is an eigenvector
for A linked with eigenvalue 1.
If instead A is not I,
then I − A is a non-zero idempotent and, by the same reasoning
as for A,
also has eigenvector/eigenvalue pair (v, 1), i.e.,
(I − A) v = v.
But this means v − Av = v,
or Av = 0, i.e., v is an eigenvector for A
linked with eigenvalue zero.
(iv) Where x = Az and y = (I − A)z, we have:
Ax = AAz = Az = x.
Ay = A(I − A)z =
Az − AAz = Az − Az = 0.
Similarly, (I − A)x = 0 and
(I − A)y = y , while x + y = z .
Send e-mail if you can't discover
why x and y, if
both non-zero, must then be linearly independent. (Or see Theorem 7.2.2.)
(v) The column space of A, the eigenspace for A corresponding
to eigenvalue 1, the null space of I − A,
the eigenspace for I − A corresponding to
eigenvalue 0, and even the range space of A, are all the same.
The column space of I − A, the eigenspace for I − A corresponding
to eigenvalue 1, the null space of A, the eigenspace for A
corresponding to eigenvalue 0, and even the range space of
I − A, are likewise all the same.
(vi) For idempotent n×n matrix A,
either A = I, or A = 0, or every vector in
Rn can (as in (iv)) be expressed as the sum of a vector
in the eigenspace for 0 and a vector in the eigenspace for 1. In the first
two (special) cases, A is already diagonal. In the general case,
the eigenvectors for A clearly span Rn (it's (iv)
that tells you this), and so
A can be diagonalized by means of any matrix P whose
columns are all the vectors in a basis for the eigenspace for eigenvalue 1,
followed by all the vectors in a basis for the eigenspace for eigenvalue 0.
[The diagonalized form of A is then a matrix with a 1 in each of the first rank(A) positions along the main diagonal, and a 0
everywhere else.]
Feedback. Too many thought that identity matrices and all-zero
matrices were the only idempotent matrices possible. Just to illustrate how many and varied idempotents can be, here are a few (add your own matrix brackets, please):
1 0 0 1 0 0
1 17 0 1/3 1/3 1/3
0 1/2 1/2 0 1 0
0 0 0 1/3 1/3 1/3
0 1/2 1/2 0 0 0
0 0 1 1/3 1/3 1/3
7. (i) If λ is an eigenvalue for M, with (non-zero)
eigenvector x, we have Mx = λx, whence also
M2 x = M Mx = M(λx) =
λ Mx = λλx = λ2x,
and, continuing the same reasoning,
Mmx = λmx.
But since Mm = 0, this means
λmx = 0. But since x is non-zero,
this in turn means λm = 0, whence λ = 0.
(ii) Remember that the column space ( = range space) of
a product matrix AB is always a subspace of
the column space ( = range space) of the first factor A.
Consequently, the column space of Mk+1 =
Mk M
is always a subspace of the column space of the first factor
Mk.
(iii) Suppose Mk = Mk+1.
Of course we don't know how k compares with m, so we have to
take all possibilities into account. The possibility
k > m is an easy case, for then k − m > 0
and Mk = Mm
Mk−m = 0 Mk−m =
0,
as desired.
Even easier are the cases k = m or k = m − 1:
in the former, we have Mk = Mm
= 0 directly, while in the latter, we have almost as simply:
Mk = Mk+1 =
M(m−1)+1 = Mm = 0.
There still remains only the possibility k < m − 1. Here, after
observing, for use in a moment, that (m − 1) − k > 0,
we notice that, after multiplying both sides of the hypothesis
Mk = Mk+1
first by M, then by M2, ..., and finally by M(m−1)−k, we can obtain the additional
equalities
Mk+1 = Mk+2,
Mk+2 = Mk+3, ...,
Mk+((m−1)−k) = Mk+((m−1)−k)+1 =
Mm = 0,
which, when combined, yield
Mk = Mk+1 = Mk+2 = Mk+3 = ... =
Mk+((m−1)−k) = Mk+((m−1)−k)+1 =
Mm = 0. This completes the proof.
(iv) We know from (ii) that, for each integer k > 0, the column space of
Mk+1
is always a subspace of the column space of
Mk, and that the column space of the matrix
Mm = 0 is the trivial space {0}.
Let, say, κ be the first of the integers k ≥ 1
for which the column space of Mk+1 coincides with the whole column space of Mk
(we know there are such integers because m is an example of one,
the column spaces both of Mm = 0 and of
Mm+1 being the trivial space {0}).
Thus the column spaces of Mκ and Mκ+1 are the same. Moreover,
the column space
of Mκ+2,
being the range space
of Mκ+2 = M Mκ+1,
is the collection of all (column) vectors Mx obtained
as x ranges over the column space
of Mκ+1, and this, of course,
is the same as the collection of all (column) vectors Mx obtained
as x ranges over the column space
of Mκ, i.e., is the same as
the range space (column space)
of Mκ+1.
Thus the column spaces not only of
Mκ and Mκ+1
but also of Mκ+2
all are the same. Continuing in the same way, we can see that
all the matrix powers Mκ+i
(i = 0, 1, 2, ..., m) have the same column space.
And since the last of these column spaces must be a subspace of the
trivial column space of Mm = 0,
hence itself trivial, we deduce that already the column space of
Mκ must have been trivial, as well, which
can only signify that Mκ = 0.
It remains only to explain why κ ≤ n. For this, notice that
each time we move from the column space of
Mi, for 1 ≤ i ≤ κ,
to that of Mi+1,
the dimension must be dropping by at least 1 (for otherwise those two
column spaces would have to coincide, which, by the way κ was
selected, does not happen yet for i < κ. Moreover, the
column space of M itself has dimension only < n (remember,
M has 0 as an eigenvalue, hence has non-trivial null-space,
hence has rank < n; but the rank of M is the dimension
of the column space of M; this dimension being < n is another reflection
of the fact that, with rank < n, M has range ≠ Rn).
Putting this together, we see that the number κ of steps needed
to reach the trivial column space of Mκ
cannot exceed n, i.e., κ ≤ n, as desired.
Feedback. Too many thought that the only way a matrix could be
nilpotent would be by having all its entries zero. Just to illustrate how many and varied (and non-zero) nilpotent matrices can be, here are a few (add your own matrix brackets, please):
0 0 0 0 0 0
2 0 −2
0 −1/2 1/2 1 0 0
0 0 0
0 −1/2 1/2 9 5 0
2 0 −2
8. (i) The lengths are (all of them): √(1 + 1 + 1) = √3 .
(ii) The dot product of any one of these
with any other is: −1 + 1 − 1 = −1,
and so the angle
between any two of them has cosine (−1)/(√3)2 =
−(1/3) .
(That's π/2 radians (or 90°) wider than the smaller acute angle in a
(1, √8, 3) right triangle,
which is to say, an obtuse angle of roughly 109° or about 1.9 radians.)
9. When z is a zero vector in the sense specified here, z + z' = z'.
Likewise, when z' is such a zero vector, z + z' = z.
Thus when both z and z' are zero vectors, z' = z + z' = z, as desired.
10. Some of the various topics regretted
as absent from the problems were:
• cross products in R3;
• geometric transformations in low dimensions and/or their relevance to computer graphic display methodology;
• linear transformations in general;
• linear (in)dependence, spanning, basis;
• least squares approximations;
• explicit diagonalizations;
• orthonormalizing a basis;
• complex numbers;
• finding inverse matrices; and
• explicit row-reductions.
Last revised: 17 May 2004.
Formulae breaking awkwardly across lines?
Let me know which, please, and I'll knit them together with " " 's, where needed.
