Assignment 2: Solutions, comments, marking scheme

A marking scheme will appear at the end of this file some time after it is posted. For now, I just want to post some solutions and say some stuff.

First some general comment: This is an honours course, and one is supposed to do things carefully in such a course. This assignment is the first place we are proving things involving properties of matrix algebra and related facts, so it seems to me we should do that carefully. Thus, below, I will check things like sizes, and check whether sums or products are defined, etc.

First, in problem 1, let an nxn matrix  A  be given.

1.(a) (18.(a), 2.1)

A - A^T  is skew-symmetric.

Pf:  A^T  is again  nxn,  so  A - A^T  is defined. 

(A - A^T)^T  =  (A + (-1)A^T)^T  =  A^T + (-1)A^TT  =

A^T + (-1)A  =  A^T - A  =  (easy but long using stuff
from p. 39 or the defn. of the negative of a matrix ) 

- (A - A^T). //

1.(b) (18.(b), 2.1)

First, some comments:

I noticed that part (c) had an answer in the back, but 
did not give myself the extra few seconds to realize 
that that answer would spoil part (b) if people looked 
at it. The purpose of doing 17 in class and assigning
18.(a) was to get people to do (b) with NO further
"help". Even so, I essentially spoiled (b) with the
hint given in class. 

Now... I have the impression that people looked at
the answer to (c) in the back of the text and somehow
decided that it answers (b). It doesn't. (b) asks us
to show that THERE ARE matrices S, W with certain
properties. (c) asks us to show that there is no 
choice about the S and the W in (b), i.e., that 
if we have ALREADY SHOWN that there is such an S and
such a W, then we have found the only ones having
those properties.

Answer to (b):

From class, we know that  A + A^T  is symmetric
(problem 17 of 2.1). Call it  C.

Let  S  be  (1/2)C.  Then  S^T  =  (Theorem 2, p. 41)
(1/2) C^T  =  (1/2)C  =  S.  So S is symmetric.

From (a),  D = A - A^T  is skew-symmetric. Let  W
denote  (1/2)D.  Then

W^T = (1/2)D^T  =  (1/2)(-D)  =  (1/2)((-1)D) =

(Theorem 1, property 7)  ((1/2)(-1)) D  =

((-1)(1/2)) D  = -((1/2)D) = - W.  So W is

And  S + W = (1/2) (A + A^T) + (1/2) (A - A^T) =

(use several properties of matrix algebra --

5, 1, 6, 8, 3 of Theorem 1, at least) ... = A.

2. (20.(a) of 2.2)

Suppose  AX = 0  for every nx1  fellow  X.
Fix an  i  between 1 and n inclusive.  Let X have a
1 in row i and 0's in all other rows. Then

AX = column i of A.  So  column i of A is 0.  Since
this holds for all such i's, A is 0.

2. (37.(a) of 2.2)

Let  A, B  be  nxn.  Then  A+B is defined and is nxn.
One checks that all the sums and products written below
are defined.

(A + B)^2  =  (A+B)(A+B) =  (Theorem 1, part 4)  A(A+B) + B(A+B) =

(Theorem 1, part 3, twice)  AA + AB + BA + BB = 

A^2 + AB + BA + B^2.

Now, suppose  AB = BA.  Then

(A+B)^2 = A^2 + AB + AB + B^2 = A^2 + (1)AB + (1)AB + B^2 =

A^2 + (1+1)AB + B^2 = A^2 + 2AB + B^2.

For the implication in the other direction, we basically
need a cancellation law for matrix addition:

If U, V, W are matrices of the same size, and
U + W = V + W,  then  U = V.
Proof: Suppose the first equation holds for such 
matrices. Then
U = U + 0 = U + (W + (-W)) = (U + W) + (-W) =
(V + W) + (-W) = ... = V. //

So we can cancel matrices in sums, on the right. By
commutativity of matrix addition, we can cancel them
on the left too.

Now suppose  (A+B)^2 = A^2 + 2 AB + B^2.  Then from
the expression for the left side given at the start
of this solution, we can cancel  A^2  on the left,
and then  B^2 on the right.  And then write  2 AB
as  (1+1) AB = 1AB + 1AB = AB + AB, and finally 
cancel an  AB on either left or right of what is
left after the two cancellations in the expression
referred to above, to get that  AB = BA. //

The Chief