1. Prove that if A1, ..., An are pairwise orthogonal vectors in a Euclidean vector space, then: 
 
||A1 + A2 + ... + An||2 = ||A1||2 + ...+ ||An||2.     (1)

By definition, in a Euclidean vector space, ||A||2 = A*A. So, equation (1) can be re-written as 
 
(A1 + A2 + ... + An)*(A1 + A2 + ... + An) = A1*A1 + ... + An*An.  (2)

The left side of this equation can be expanded, by the properties of the dot product in a Euclidean vector space. So, equation (2) becomes 
 
 A1*A1 + A1*A2 + ... + A2*A1 + A2*A2 + ... +  An*An = A1*A1 + ... + An*An

But the vectors are pairwise orthogonal. Hence, every term that is not the dot product of a vector with itself will drop out on the left side (the dot product of two orthogonal vectors equals zero). The left side will become the right side.

The proof is complete.


2. A function T from C[0,1] to R2 is given by the following formula:

T(f(x)) = (f(0),f(1)).


Is it true that T is a linear transformation?

To determine if T is a linear transformaiton, we must test to see if the two properties of a linear transformation hold for T.

Property 1: 
 
If f(x) and g(x) are two functions from C[0,1] then 
 
T(f(x) + g(x)) = ((f(0)+g(0), f(1)+g(1)) 
                                   = (f(0),f(1)) + (g(0),g(1)) 
                                        (by the rules of addition in R2) 
                                   = T(f(x)) + T(g(x)).

So, the transformation T obeys the first property.



Property 2: 
 
If f(x) is a function from C[0,1] and k is any number then 
 
T(kf(x)) = (kf(0),kf(1)) 
                                     = k*(f(0),f(1)) 
                                    (by the rule of scalar multiplication in R2) 
                                     = k*T(f(x)).

So, the transformation T also obeys the second property. Hence, T is a linear transformation.

3. Find the standard matrix of the linear operator in R3 which first reflects a vector about the plane 2x+3y+z=0 and then rotates it 60 degrees counterclockwise about the z-axis.

In the homework we found the standard matrix of a reflection about the plane

ax + by + cz = 0


to be
A := [ (-a2 + b2 + c2)/%1 -2ab/%1 -2ac/%1 ]
[ -2ab/%1 (a2 - b2 + c2)/%1 -2bc/%1 ]
[ -2ac/%1 -2bc/%1 (a2 + b2 - c2)/%1 ] 

                          
%1 :=                            a2  + b2  + c2

If we plug in a=2, b=3, and c=1, we get the standard matrix
A := [ 3/7 -6/7 -2/7 ]
[ -6/7 -2/7 -3/7 ]
[ -2/7 -3/7 6/7 ]

The standard matrix for a counterclockwise rotation about the z-axis is similar to the rotation of an R2 vector in the xy-plane. We know during the rotation, the z-coordinate of the vector is left unchanged. Hence, the standard matrix is simply
B := [ cos(a) -sin(a) 0 ]
[ sin(a) cos(a) 0 ]
[ 0 0 1 ]

So, for a 60 degree rotation the standard matrix is
B := [ 1/2 -1/2 * 31/2 0 ]
[ 1/2 * 31/2 1/2 0 ]
[ 0 0 1 ]

Hence, the standard matrix of the linear operator that reflects a vector about the plane and then rotates it 60 degrees is BA.

> BA:= evalm(B&*A);
BA := [ 3/14 + 3/7 * 31/2 -3/7 + 1/7 * 31/2 -1/7 + 3/14 * 31/2 ]
[ 3/14 * 31/2 - 3/7 -3/7 * 31/2 - 1/7 -1/7 * 31/2 - 3/14 ]
[ -2/7 -3/7 6/7 ]

4. Find the vectors which span the null space of the following matrix:
A := [ 2 3 4 5 6 ]
[ 1 2 3 5 4 ]

The null space of this matrix consists of all vectors, v = (x,y,z,t,s), which satisfies the equation Av=0. This is a system of equations with standard matrix
B := [ 2 3 4 5 6 0 ]
[ 1 2 3 5 4 0 ]

with reduced row-echelon form
[ 1 0 -1 -5 0 0 ]
[ 0 1 2 5 2 0 ]

The general solution of the system is 
 
			                    x =   z + 5t 
			                    y = -2z - 5t - 2s 
			                    z = z 
			                    t = t 
			                    s = s,

or in vector notation

(x,y,z,t,s) = (z+5t, -2z-5t-2s, z, t, s) = z(1,-2,1,0,0) + t(5,-5,0,1,0) + s(0,-2,0,0,1).


So, the null space is spanned by the vectors (1,-2,1,0,0), (5,-5,0,1,0), and (0,-2,0,0,1).

5. What are the standard matrix and the range of the transformation T from R4 to R2 which takes every vector (x,y,z,t) to (2x+3y,x+4y+z+t)?

The transformation goes from R4 to R2 so the standard matrix must be a 2 by 4 matrix. It is simply the matrix with the coefficients of the terms in the result of the transformation as rows. Hence, the standard matrix, T, is
T := [ 2 3 0 0 ]
[ 1 4 1 1 ]

> v:= matrix([[x],[y],[z],[t]]);
v := [ x ]
[ y ]
[ z ]
[ t ]

> b:=evalm(T&*v);

			b :=       [     2x + 3y     ]
			           [  x + 4y + z + t ]

The range consists of all vectors, b, in R2, such that there is a vector v, in R4, that is a solution to the equation Tv = b. It is spanned by the column vectors of T. So, the range of the transformation is the set of all linear combinations of the column vectors of T.

We can guess that the range is all of R2. This would mean that the span of the column vectors is equal to the span of the basic vectors i and j in R2. By definition, if the spans of these two subsets are equal then the vectors of one subset are linear combinations of vectors from the other subset and vice-versa. It is obvious that the column vectors of T are linear combinations of i and j, so, all that is left is to prove that i and j are linear combinations of the column vectors of T. We have two systems of equations:
a *
[ 2 ]
[ 1 ]
+ b *
[ 3 ]
[ 4 ]
+ c *
[ 0 ]
[ 1 ]
+ d *
[ 0 ]
[ 1 ]
=
[ 1 ]
[ 0 ]

and
a *
[ 2 ]
[ 1 ]
+ b *
[ 3 ]
[ 4 ]
+ c *
[ 0 ]
[ 1 ]
+ d *
[ 0 ]
[ 1 ]
=
[ 0 ]
[ 1 ]

These systems have the augmented matrix
F := [ 2 3 0 0 1 0 ]
[ 1 4 1 1 0 1 ]

with reduced row-echelon form
[ 1 0 -3/5 -3/5 4/5 -3/5 ]
[ 0 1 2/5 2/5 -1/5 2/5 ]

It can be seen, from the reduced row-echelon form of the augmented matrice for the two systems, that both of these systems have solutions. In fact, they both have an infinite number of solutions. So, the span of the column vectors is equal to the span of the basic vectors i and j. Hence, the range of the transformation T is all of R2.