V I L L A N O V A U N I V E R S I T Y
Department of Electrical & Computer Engineering
ECE 8412 Introduction to Neural Networks
Homework 04 Due 18 October 2005
Chapters 5
Name: Santhosh Kumar Kokala
Complete the following and submit your work to me as a hard copy of a WORD document.
You can also fax your solutions to 610-519-4436.
1
1
1
 
 
 
For the problems below you given arbitrary vectors x  1 , y   1  , and z   1 .
1
 1
 1
 
 
 
01.
Represent the given vectors using the standard basis vectors.
A.
1
1  x  y  z
0
0
 0
1
02.
Are the given vectors linearly independent?
A.
Writing the above vectors in the matrix equation form
1 
1   x  y  z
0
0
0
 
 1
1 
1 
0 
0 
 1  x  y  z where x  0 y  1  z  0
0
0
0
0
 
  0   0  
 1
0
0
1 
1 1 1 a1  0
1 1 - 1 a   0

 2   
1 - 1 - 1 a3  0
If the matrix in the equation has an inverse, then the solution will require that all the coefficients be
zero; therefore the vectors are independent. If the matrix is singular (has no inverse), then a nonzero
set of coefficients will work and the vectors are dependent. Using the Laplace expansion on the first
column, the determinant of this matrix is
1
1
1
1 -1
1 -1 1 1
1 1 -1  1
 (1)
1
 2  0  2  4
-1 -1
1 -1 1 -1
1 -1 -1
Therefore the vectors are linearly independent.
03.
Do the given vectors span the space R3?
A.
Since any vector in the space R3 can be written as linear combination of x, y and z hence the given
vectors span the space R3.
04.
Determine the following inner products,
x, y 
x, z 
y, z 
05.
x T y  x1 y1  x 2 y 2  x3 y 3  1 * 1  1 *1  1 * 1  1  1  1  1
x T z  x1 z1  x 2 z 2  x3 z 3  1 *1  1 * 1  1 * 1  1  1  1  1
y T z  y1 z1  y 2 z 2  y 3 z 3  1 *1  1 * 1  1 * 1  1  1  1  1
Determine the following norms,
x 
y 
z 
x x
y y 
z z 
T
1
2
T
1
2
T
1
2
 x1  x 2  x3  1  1  1  3
2

2
2
y1  y 2  y 3  1  1  1  3
2
2
2
 z1  z 2  z 3  1  1  1  3
2
2
2
06.
Determine the angle between each pair of vectors.
A.
The angle between two vectors x and y us defined by
cos  
( x, y )
x y
Using the norms and inner products calculated in the previous problems we get
cos  x , y 
cos  x , z 
cos  y , z 
1
3 3
1
3 3
1
3 3
07.
Show by calculation that x  y  x  y .
A.
1 1  2
x  y  1  1   2
1  1 0

1
1
  x , y  cos 1  
3
3

1
 1
  x , z  cos 1  
3
 3 

1
1
  x , y  cos 1  
3
3
and x  y  2 2  2 2  0 2  8  2 2
and we know that x  3 y  3
so x  y  3  3  2 3 .
2 22 3
Hence proved that
x y  x  y
-2-
08.
Calculate the orthogonal basis vectors v1, v2, and v3 by applying the Gram-Schmidt Orthogonalization
procedure to the vectors x, y, and z.
A.
Step 1.
1
v1  x  1
1
Step 2.
1 
1 1 11 
 1  2 
1 
1 1   3   3 
T
 1       
v y
v 2  y  1T v 1  1  
1  1  1  2 
1      3   3 
v1 v1
 1
1  1
1 1 11      13   4 3 
1
Step 3.
1   1 3 1 3
v 3  z  T v1  T v 2   1   1 3  1 3
v1 v1
v2 v2
 1  1 3  2
T
v1 z
T
v2 z
 1 
   1
  
3 0 
09.
Calculate the vector expansion of w = [1 2 3]T using the orthogonal basis v1, v2, and v3 from the
previous problem.
A.
Calculating the components w
w1
w2
w3
=
=
=
w, v1  = 6  2
v1 , v1  3
w, v2  =  2   3
4
v2 , v2  8 3
w, v3  =  1   1
2
v3 , v3  2
Therefore the vector expansion of w is x = w1v1 + w2v2 + w3v3
2 
1
1  1 
 3 
3 2  1    


x  21 
 1  2
4 3  2   
1
0  3 
 4 
 3 
-3-
10.
Calculate the vector expansion of w = [1 2 3]T using the basis vectors x, y, and z which were given at
the beginning of this assignment.
A.
1
B  1
1
1
1
-1
1
- 1
- 1
B
1
0.5
0.5 0

 0
0.5 - 0.5
0.5 - 0.5 0 
Step 2:
 0 .5 
0 
0.5 




r1  0  r2  0.5  r3   0.5
0.5
 0.5
0 
Step 3:
a1 = (r1, w) = 0.5
a2 = (r2, w) = 0
a3 = (r3, w) = 0.5
0
0.5
- 0.5
1 
0.52  2
3 
1 
- 0.52  0.5
3 
1 
02  0.5
3 
Step 4:
Verification:
w = a1v1 + a2v2+ a3v3
where v1=x v2=y v3=z
1
1 
1  1 
w= 21  0.51   0.5 1  2
1
 1
 1 3 
-4-