COMMENTS AND SOLUTIONS FOR EXAM 2

```1. [30]  True or false?

(a)  In a subspace of dimension 7,  any 8 vectors are
linearly dependent.  TRUE

(b)  The columns of an 8 times 5 matrix are independent if
and only if the matrix has rank 5. TRUE

(c) A subspace of R^n can fail to have an orthonormal
basis. FALSE

(d)  An n times n matrix A is orthogonal if and only if
A^T = A^(-1).  TRUE

(e) An n times n matrix is orthogonal if and only if
any two different columns are orthogonal. FALSE
(columns must have length one as well)

(f)  The orthogonal complement of the column space of
A is the same as the kernel of A^T.   TRUE

(g) The intersection of two subspaces of R^n is always
a subspace of R^n. TRUE

(h) The union of two subspaces of R^n is always
a subspace of R^n. FALSE

(i) The columns of an n by n matrix are independent
if and only if the rows are independent. TRUE

(j) If M is a 5 times 3 matrix, the columns of M form
an orthonormal set if and only if M^T M is the 3 by 3
identity matrix. TRUE

(k) The transpose of an orthogonal matrix can fail to
be orthogonal. FALSE

(l) If V is contained in W, and both are subspaces
of R^(11) of dimension 7, then V = W. TRUE

(m) If T is a linear transformation from R^8 to R^6
whose image has dimension 5, then the kernel of T has
dimension 1. FALSE (dim Ker T = 8 - 5 = 3)

(n) If V is a subspace of R^n,  then every vector
in R^n can be written uniquely in the form v+w,  where
v is in V and w  is in the orthogonal complement of V.
TRUE

(o) If an n by m matrix M has independent columns, then
it can be factored uniquely in the form QR,  where Q is
an n times m matrix whose columns form an orthonormal set,
and R is an m times m upper triangular matrix whose
diagonal entries are positive. TRUE

2. [18] When the matrix A =

1   2  1   0  -1
2   4  1  -1  -4
5  10  3  -2  -9

is put in reduced row echelon form the result is

1  2  0  -1  -3
0  0  1   1   2
0  0  0   0   0

(a) [8]  Give a basis for the image of A consisting of columns
of the matrix A. What is the dimension of the image of A?

Since the leading 1's of the RREF are in the first and third
columns, the first and third columns of  A  (the pivot columns)
are a basis for the column space or image of A,
i.e., the basis is

1       1
2  and  1
5       3

The dimension of the image is therefore 2.

(b) [8] Write the kernel of A in set notation,  and then
give a basis for the kernel of A.  What is the dimension
of the kernel of A?

Call the variables  v, w, x, y, z.  Use the RREF to solve for
the leading variables  v, x  in terms of  w, y, z.  This gives
the general solution:

-2w + y + z
w
{   -y - 2z   : w, y, z in  R} =
y
z

-2         1        1
1         0        0
{ w  0  +  y  -1 +  z  -2 : w, y, z in R}
0         1        0
0         0        1

The coefficient vectors of  w, y, z  give the required basis:

-2     1     1
1     0     0
0 ,  -1 ,  -2
0     1     0
0     0     1

The dimension of the kernel is 3.

(c) [2] Let B = A^T =   1   2   5
2   4  10
1   1   3
0  -1  -2
-1  -4  -9

Give a basis for the orthogonal complement of the image of B.

The orthogonal complement of im A^T  is the kernel of  A,  and
so one has the same basis as in part (b).

3. [18]
Which of the following are subspaces of R^4?  Explain your answers
briefly.

(a) The empty set.  NO, by definition subspaces are non-empty.

(b) The set consisting of the vector

0
0
0
0.

YES, this is non-empty and closed under addition and scalar
multiplication.

(c) The image of the linear transformation with matrix

1   2
3   4
0  -2
5  -3

(no calculations are needed).  YES,  the image of an m by n
matrix is always a subspace of  R^m. Here,  m = 4.

(d) The set of all solutions

x_1
x_2
x_3
x_4

of the linear equation
x_1 + x_2 +x_3 + x_4 = 6.
NO.  Does not contain 0,  is not closed under addition,
is not closed under scalar multiplication.

(e) The set of all vectors

x_1
x_2
x_3
x_4

such that x_1 is greater than or equal to 0.  NO
If x_1  > 0  and one multiplies the vector by a negative
scalar such  as -1, one does not stay in the set.  I.e.,
the set is not closed under scalar multiplication.
(It is closed under addition.)

(f) The kernel of the linear transformation with matrix

17  19  -29  11
67  17  -43  92

(no calculations are needed). YES,  the kernel of an m by n
matrix is a subspace of  R^n.  Here,  n = 4.

4. [17] (a) [10] Let v_1 =
1
1
1
1

and  v_2 =
5
3
3
5.

Use the Gram-Schmidt process to find an orthonormal basis
w_1, w_2 for the span of v_1 and v_2.

SOLUTION:  ||v_1|| = (1+1+1+1)^(1/2) = 4^(1/2) = 2 = r_{11}.
w_1 = (1/2) v_1 =

1/2
1/2
1/2
1/2

r_{12} = v_2 . w_1 = (5+3+3+5)(1/2) = 16/2 = 8.
The projection of v_2 on the line through w_1  is
8w_1 =

4
4
4
4.

v_2 minus the projection =

1
-1
-1
1

The length of this vector = 2 = r_{22}.  w_2 = (1/2)v_2 =

1/2
-1/2
-1/2
1/2

(b) [7] Give the QR factorization of the matrix M =

1  5
1  3
1  3
1  5.

(Note:  essentially all of the work needed was carried
out in part (a).)

M = QR  where  Q =

1/2  1/2
1/2 -1/2
1/2 -1/2
1/2  1/2

and  R =

2 8
0 2

5. [17] (a) [6] You are given that

-1/2^(1/2)  2/3  a
1/2^(1/2)  2/3  b
0      1/3  c

is an orthogonal matrix.  Determine the possible values of c.
(Notice that you do not need to determine a and b.)

First method.  The rows of the matrix are also an orthonormal
set (transpose of orthogonal is orthogonal) and so the third
row has length 1.  Thus,  0 + (1/9) + c^2 = 1  and c^2 = 8/9.
Therefore  c = 2(2^(1/2))/3  or  c = - 2(2^(1/2))/3.

Second method.  The third column is orthognal to the first
column and the second.  Taking the dot products and clearing
denominators one gets linear equations:

-a +  b     = 0
2a + 2b + c = 0

Row reduce to get the system:

a  - b     = 0
4b + c = 0

and then

a   + (1/4)c = 0
b + (1/4)c = 0

Thus,  a = b = - (1/4)c  and the general solution is spanned
by the vector

1/4
1/4
-1

So the third column is a multiple of this vector of length 1.
Divide by the length, which is (18/16)^(1/2) = 3(2)^(1/2)/4
to get one solution. The negative is the only other, so that
c  is plus or minus  4/3(2^(1/2)) = plus or minus 2(2^(1/2))/3.
(b) You are given that the columns of the matrix M =

-1/2  1/2 -1/2
-1/2 -1/2 -1/2
1/2 -1/2 -1/2
1/2  1/2 -1/2

form an orthonormal set.

(1) [5] Without any calculation, what can you say about M^T M?

It is the size 3 identity matrix

1 0 0
0 1 0
0 0 1

since the columns of  M  form an orthonormal set.

(2) [6] Give the matrix of orthogonal projection on the column
space of M  as the product of two explicit matrices.  You do not
need to calculate the product, but state what size the answer is.

The answer is  M M^T  =

-1/2  1/2 -1/2        -1/2 -1/2  1/2  1/2
-1/2 -1/2 -1/2         1/2 -1/2 -1/2  1/2
1/2 -1/2 -1/2        -1/2 -1/2 -1/2 -1/2
1/2  1/2 -1/2

which is 4 by 4.

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