hw09
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moienMath 21b: Linear Algebra Spring 2016
Homework 9: Dimension
This homework is due on Monday, February 22, respectively on Tuesday, February 23, 2016.
1 Determine the rank and nullity of the following matrix and verify
that the rank nullity theorem holds in this case:
5
4 3 2 1
1 2 3 4 5
A= 1 1 1 1 1
1 2 3 4 5
5 4 3 2 1
2 a) In each of the 5 cases k = 1, 2, 3, 4, 5; give a 4 × 5 matrix A for
which the dimension of the kernel of A is k.
b) Is there a 4 × 5 matrix whose kernel has dimension 0? Explain
why or why not.
3 Find out whether the following set of vectors related to prime
numbers forms a basis of R4.
2 3 5 7
3 5 7 11
{ , , , }
5 7 11 13
7 11 13 17
4 The following
matrices
are meant
to look
like letters:
1 1 1 1 1 1 1 0 1
T = 0 1 0 , O = 1 0 1 , X = 0 1 0 ,
0 1 0 1 1 1 1 0 1
0 1 0 1 1 1 1 0 1
I = 0 1 0 , C = 1 0 0 , U = 1 0 1 ,
0 1 0 1 1 1 1 1 1
1 0 0
L = 1 0 0 .
1 1 1
a) Find four letters which have the same kernel.
b) Find three letters which have the same image.
4
5 Consider the subspace V consisting
of all vectors
in R which are
1 1
1 −1
perpendicular to both v = and w = .
−1 1
−1 −1
Find a basis for V .
Basis
If V is a linear space and {v1, .., vn} is a basis, then n is the
dimension of V . If ~v1, . . . , ~vp are linearly independent in
V and w ~ q span V then p ≤ q. The dimension of
~ 1, . . . , w
the image of A is called the rank of A. The dimension of
the kernel of A is called the nullity of A. The rank-nullity
theorem tells that the sum of the rank and the nullity is
equal to the number of columns of A. It’s easy to see why
this is true if you remember that the rank of A is the number
of leading 1s in rref(A) and the nullity of A is the number
of free variables.