MA-305 Homework 5

Due at 4:05pm, Thursday, October 23, 1997
Extended to 1:00pm, Friday, October 24, 1997

Solution

You may do the calculations necessary for these problems either by hand or with Maple. Please submit a handwritten solution, or email an ASCII/Postscript/html document to the TA. (You may also submit through email which attached your Maple worksheet.)

1. *Determine whether the given set V is a vector space under the operation +¨ and ר:

   V is the set of all ordered triples of real numbers of the form ( 0, y, z);

   ( 0, y, z) +¨ ( 0, y', z') = ( 0, y + y', z + z')

   and

   c ר ( 0, y, z) = ( 0, 0, cz).

   Solution:

   Not a vector space. No multiplicative identity exists for ( 0, y, z) if y is nonzero.

2. *Determine whether the given set V is a vector space under the operation +¨ and ר:

   V is the set of all polynomials of the form at² + bt + c , where a, b, and c are real numbers with b = a + 1 ;

   ( a₁t² + b₁t + c₁ ) +¨ ( a₂t² + b₂t + c₂ ) = ( a₁ + a₂ )t² + ( b₁ + b₂ )t + ( c₁ + c₂ )

   and

   r ר ( at² + bt + c ) = (ra)t² + (rb)t + c.

   Solution:

   Not a vector space. The operation +¨ is not closed in V.
   Because ( a₁t² + b₁t + c₁ ) +¨ ( a₂t² + b₂t + c₂ ) = ( a₁t² + (a₁ + 1)t + c₁ ) +¨ ( a₂t² + (a₂ + 1)t + c₂ ) = ( a₁ + a₂ )t² + ( a₁ + a₂ + 2)t + ( c₁ + c₂ ) .

3. *M₂₃ is the vector space of all 2 × 3 matrices under the usual operations of matrix addition and scalar multiplication. Which of the following subset of the vector space M₂₃ are subspaces? The set of all matrices of the form

   (a)	[	a	b	c	]	, where b = a + c.
   	[	d	0	0	]

   (b)	[	a	b	c	]	, where c > 0.
   	[	d	0	0	]

   (c)	[	a	b	c	]	, where a = -2c and f = 2e +d.
   	[	d	e	f	]

   Solution:

   (a) A vector space.
   (b) Not a vector space
   

   The scalar multiplication is not closed. Because for c > 0 ,

   (-1)

   [

   a

   b

   c

   ]

   =

   [

   -a

   -b

   -c

   ]

   [

   d

   0

   0

   ]

   [

   -d

   0

   0

   ]

   (c) A vector space.

4. *Find a set of vectors spanning the solution space of Ax=0, where

   	[	1	0	1	0	]
   A =	[	1	2	3	1	]
   	[	2	1	3	1	]
   	[	1	1	2	1	]

   Solution:

   	[	-1	]
   {	[	-1	]	}
   	[	1	]
   	[	0	]

5. *Which of the following sets of vectors in R³ are linearly dependent? For those that are, express one vector as a linear combination of the rest.

   (a) { ( 1, 2, -1 ), ( 3, 2, 5 ) }.
   (b) { ( 4, 2, 1 ), ( 2, 6, -5 ), ( 1, -2, 3 ) }.
   (c) { ( 1, 1, 0 ), ( 0, 2, 3 ), ( 1, 2, 3 ), ( 3, 6, 6 ) }.
   

   Solution:

   (a) Not linearly dependent.
   (b) Linearly dependent: ( 1, -2, 3 ) = (1/2)*( 4, 2, 1 ) - (1/2)*( 2, 6, -5 )
   (c) Linearly dependent: ( 3, 6, 6) = 2*( 1, 1, 0) + ( 0, 2, 3) + ( 1, 2, 3)

6. *Suppose that S = {v₁ , v₂ , v₃} is a linearly independent set of vectors in a vector space V. Show that T = {w₁ , w₂ , w₃} is also linearly independent, where w₁ = v₁ + v₂ + v₃ , w₂ = v₂ + v₃ , and w₃ = v₃ .

   Solution:

   S is a basis of Span(S).
   Span(S) = Span(T) because:
   v₁ = w₁ - w₂ , v₂ = w₂ - w₃ , v₃ = w₃.
   Since both S and T have the same number of vectors, T must be another basis of Span(S). Therefore, T = {w₁ , w₂ , w₃} is also linearly independent.

7. *Which fo the following sets of vectors are bases for R³ ?

   (a) { ( 1, 2, -0 ), ( 0, 1, -1 ) }.
   (b) { ( 1, 1, -1 ), ( 2, 3, 4 ), ( 4, 1, -1 ), ( 0, 1, -1 ) }.
   (c) { ( 3, 2, 2 ), ( -1, 2, 1 ), ( 0, 1, 0 ) }.
   (d) { ( 1, 0, 0 ), ( 0, 2, -1 ), ( 3, 4, 1 ), ( 0, 1, 0 ) }.
   

   Solution:

   (a) Not a basis for R³.
   (b) Not a basis for R³.
   (c) A basis for R³.
   (d) Not a basis for R³.

8. Find a basis for the null space of the given matrix A.

   	[	1	2	1	0	]
   A =	[	2	3	4	7	]
   	[	3	5	5	7	]
   	[	3	4	7	14	]

   Solution:

   	[	-5	]		[	-14	]
   {	[	2	]	,	[	7	]	}
   	[	1	]		[	0	]
   	[	0	]		[	1	]

*: problems from ``Introductory Linear Algebra with Applications'' by Kolman and Hill