**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 8 **

Anton and Rorres

Section 5.4

1.(a)
Since R^{2} has dimension 2 a basis will only have 2
vectors.

(b)
Since R^{3} has dimension 3 a basis must have 3 vectors.

(c)
Since P^{2} has dimension 3 a basis must have 3 vectors.

(d)
Since M_{22} has dimension 4 a basis will only have 4
vectors.

3.
Three vestors in R^{3} will be a basis if they are linearly
independent. Thus we must solve the system k_{1}**u**_{1}
+ k_{2}**u**_{2} + k_{3}**u**_{3 }=
**0**.

(a) The coefficient matrix for this system is

which has determinant 6 and hence has only the trivial solution. Thus the vectors are linearly independent and hence a basis.

(b) The coefficient matrix for this system is

which has determinant 26 and hence has only the trivial solution. Thus the vectors are linearly independent and hence a basis.

(c) The coefficient matrix for this system is

which has determinant 0 and hence has a non-trivial solution. Thus the vectors are linearly dependent and hence not a basis.

(d) The coefficient matrix for this system is

which has determinant 0 and hence has a non-trivial solution. Thus the vectors are linearly dependent and hence not a basis.

7.
We must find constants p and q such that p**u**_{1} + q**u**_{2}
= **w**. Then (p, q) are the co-ordinates with respect to the
given basis.

(a) This is the standard basis and the co-ordinates are simply (3, -7).

(b) We must solve the system of equations

2p + 3q = 1

-4p + 8q = 1

This has solution p = 5/28, q = 3/14.

(c) We must solve the system of equations

p = a

p + 2q = b

This has solution p = a, q = (b - a)/2.

9.
We must find constants r, s and t such that r**p**_{1} +
s**p**^{2} + t**p**_{3} = **p**. Then (r,
s, t) are the coordinates.

(a) In this case we have the standard basis so the coordinates are (4, -3, 1).

(b) We must solve the system

r + s = 2

r + t = -1

s + t = 1

which has solution r = 0, s = 2, t = -1.

15. The row echelon form for this system is

x_{1}
+ 5x_{3} = 0

x_{2} - 7x_{3} = 0

x_{3} = 0

which has only the trivial solution so the dimension of the solution space is zero and a basis consists only of the zero vector (0, 0, 0).

Section 5.5

4. See solution in text.

5.(b) The row echelon form for this system is

x_{1}
+ x_{2} + 2x_{3} = 5

x_{2} + x_{3} = 7

Thus
the solution can be written as x_{3} = s, x_{2} = 7 -
s, x_{1} = -2 - s or in vector form

**x**
= (-2, 7, 0) + s(-1, -1, 1)

From
this we can extract the solution of the homogeneous system as **x**
= s(-1, -1, 1).

(c)
The row echelon form of this system is x_{1} - 2x_{2}
+ x_{3} + 2x_{4} = -1 which has solution

x_{2}
= r, x_{3} = s, x_{4} = t and x_{1} = -1 + 2r
- s - 2t or in vector form

**x**
= (-1, 0, 0, 0) + r(2, 1, 0, 0) + s(-1, 0, 1, 0) + t(-2, 0, 0, 1).

From this we can extract the solution of the homogeneous system as

**x**
= r(2, 1, 0, 0) + s(-1, 0, 1, 0) + t(-2, 0, 0, 1).

6.(c)
The reduced row echelon form for this matrix is
so
that the solution of the homogeneous system A**x** = **0** is
x_{3} = s, x_{4} = t, x_{1} = -s+(2/7)t, x_{2}
=-s-(4/7)t

Thus a basis for the null space is (-1, -1, 1, 0) and (2, -4, 0, 7). Note that in the last case we have multiplied the vector in the solution by 7 to get the vector with integer components.

Section 5.6

4. The rank of a matrix = dimension of the row space = dimension of the column space.

The rank of a matrix plus the nullity (the dimension of the null space) = the number of columns n.

The
rank of A^{T} = rank of A (the row space and column space are
interchanged on transposition). Thus if A is an m by n matrix, A^{T}
is an n by m matrix so the rank plus the nullity of A^{T} = m

Thus we have

(a) rank = 3 = dimension of row space = dimension of column space

Since
m = n = 3 the nullity of A and A^{T} are both 0.

(b) rank = 2 = dimension of row space = dimension of column space

Since
m = n = 3 the nullity of A and A^{T} are both 1.

(c) rank = 1 = dimension of row space = dimension of column space

Since
m = n = 3 the nullity of A and A^{T} are both 2.

(d) rank = 2 = dimension of row space = dimension of column space

Since
m = 5 and n = 9 the nullity of A is 7 and the nullity of A^{T}
is 3.

(e) rank = 2 = dimension of row space = dimension of column space

Since
m = 9 and n = 5 the nullity of A is 3 and the nullity of A^{T}
is 7.

(f) rank = 0 = dimension of row space = dimension of column space

Since
m = n = 4 the nullity of A and A^{T} are both 4.

(g) rank = 2 = dimension of row space = dimension of column space

Since
m = 6 and n = 2 the nullity of A is 0 and the nullity of A^{T}
is 4.

5. The largest possible rank of an m by n matrix is the minimum value of m and n.

The largest possible rank gives the smallest possible nullity.

(a) Since A is 4 by 4 the largest possible rank is 4, the smallest possible nullity is 0.

(b) Since A is 3 by 5 the largest possible rank is 3, the smallest possible nullity is 2.

(c) Since A is 5 by 3 the largest possible rank is 3, the smallest possible nullity is 0.

6. From question 5, the largest possible rank of an m by n matrix is min(m, n) and the smallest possible nullity is n - min(m, n).

Section 6.3

3. (a) the second and third vectors are not orthogonal;

(b) all vector are orthogonal;

(c) the second and third vectors are not orhtogonal

(d) all vectors are othogonal

4. All the vectors in question 3 have unit length but only sets (b) and (d) are orthogonal and hence orthonormal.

9.
Since the vectors **v**_{1}, **v**_{2} and **v**_{3}
are an orthonormal set and there are 3 of them they form an
orthonormal basis for R^{3}. Thus any vector can be
expressed as **v** = a**v**_{1} +b**v**_{2}
+c**v**_{3} where a = **v**.**v**_{1}, b =
**v**.**v _{2}**, c =

(a) a = -7/5, b = 1/5, c = 2.

(b) a = -37/5, b = -9/5, c = 4.

(c) a = -3/7, b = -1/7, c = 5/7.

17.
(a) Since **u**_{1} and **u**_{2} are
orthogonal we can take **v**_{1} = **u**_{1}
and **v**_{2} = **u**_{2}.
Thus

**v**_{3}
= **u**_{3} - (**u**_{3}.**v**_{1}/**v**_{1}.**v**_{1})
**v**_{1} - (**u**_{3}.**v**_{2}/**v**_{2}.**v**_{2})
**v**_{2}** **=
(1, 2, 1) - (4/3)(1, 1, 1) - (1/2)(-1, 1, 0) = (1/6, 1/6, -1/3).

Now
normalize the vectors to get **
**

(b)
Take **v**_{1} = **u**_{1} then **v**_{2}
= **u**_{2}** **-
(**u**_{2}.**v**_{1}/**v**_{1}.**v**_{1})
**v**_{1} = (3, 7, -2) - 3(1, 0, 0) = (0, 7, -2) and

**v**_{3}
= **u**_{3} - (**u**_{3}.**v**_{1}/**v**_{1}.**v**_{1})
**v**_{1} - (**u**_{3}.**v**_{2}/**v**_{2}.**v**_{2})
**v**_{2}** **=
(0, 4, 1) - 0(1, 0, 0) - (26/53)(0, 7, -2) =

(0, 30/53, 105/53).

Now
normalize the vectors to get **
**