Dr. Marques Sophie
Office 519
Linear algebra
Spring Semester 2016
marques@cims.nyu.edu
Problem Set # 8
The following equations are considered over the reals numbers. All the answer should
be justified unless mentioned differently.
Problem 1 :
Find a basis for the set of vector in R3 on the plane x ´ 3y ` 2z “ 0.
Solution : Note that the plane¨in the
˛ question that we denote P is defined by x “
x
3y ´ 2z. So that any point x “ ˝ y ‚ on this plane is of the form
z
˛
¨
˛
¨ ˛
¨ ˛ ¨
´2
3
3y ´ 2z
x
‚ “ y ˝ 1 ‚` z ˝ 0 ‚
y
x “ ˝ y ‚“ ˝
1
0
z
z
So that
˛
˛ ¨
´2
3
P “ Spant˝ 1 ‚, ˝ 0 ‚u
1
0
¨
˛
˛ ¨
´2
3
t˝ 1 ‚, ˝ 0 ‚u is a linearly independent set since the vectors are not multiples. So
1
0
˛
¨ ˛ ¨
´2
3
that t˝ 1 ‚, ˝ 0 ‚u is not a basis.
1
0
¨
Problem 2 :
Find bases for NulpAq and ColpAq where
¨
˛
´2 4 ´2 ´4
A “ ˝ 2 ´6 ´3 1 ‚
´3 8
2 ´3
Solution : We row reduce A
¨
˛
1 0 6 5
A „ Row reduce „ ˝ 0 2 5 3 ‚
0 0 0 0
We know that a basis for column space is formed by the set of the pivot columns of
A. We see via the echelon form of A, B that the pivot columns of A are the column 1
1
˛
˛ ¨
4
´2
and 2. Thus, t˝ 2 ‚, ˝ ´6 ‚u is a basis for ColpAq. NulpAq “ tx P R4 |Ax “ 0u.
´8
´3
Since¨A and
˛ B are row equivalent, we know that NulpAq “ NulpBq. From Bx “ 0 where
x1
˚ x2 ‹
‹
x“˚
˝ x3 ‚ we get the system
x4
"
x1 ` 6x3 ` 5x4 “ 0
2x2 ` 5x3 ` 3x4 “ 0
¨
which is equivalent to
"
¨
x1 “
´6x3 ´ 5x4
x2 “ ´5{2x3 ´ 3{2x4
˛
x1
˚ x2 ‹
‹
So that any element x “ ˚
˝ x3 ‚ P NulpAq is of the
x4
˛
¨
¨
´6
´6x3 ´ 5x4
˚ ´5{2
˚ ´5{2x3 ´ 3{2x4 ‹
‹ “ x3 ˚
˚
‚
˝ 1
˝
x3
0
x4
where x3 and x4 can be any scalar.
form
˛
´5
‹
‹
˚
‹ ` x4 ˚ ´3{2 ‹
‚
˝ 0 ‚
1
˛
¨
˛ ¨
´5
´6
˚ ´5{2 ‹ ˚ ´3{2
‹ ˚
So that we know that NulpAq “ Spant˚
˝ 1 ‚, ˝ 0
1
0
dimpNulpAqq
¨ “ 2 number
˛ ¨ of free
˛ variable.
´6
´5
˚ ´5{2 ‹ ˚ ´3{2 ‹
‹ ˚
‹
So that t˚
˝ 1 ‚, ˝ 0 ‚u is a basis for NulpAq.
0
1
¨
˛
‹
‹u and we also know that
‚
Problem 3 :
In the vector space of all real valued functions, find a basis for the subspace spanned
by tsinptq, sinp2tq, sinptqcosptqu.
Solution : Note that sinp2tq “ 2sinptqcosptq, by the spannin set theorem we know
that
Spantsinptq, sinp2tq, sinptqcosptqu “ Spantsinptq, sinp2tqu
Note also that sinptq and sinp2tq are not multiples. So tsinptq, sinp2tqu is also a linearly
independent set. As a consequence, by the definition of a basis, tsinptq, sinp2tqu is a
2
basis for Spantsinptq, sinp2tq, sinptqcosptqu.
Problem 4 :
Is the set of polynomial tp1 , p2 u where p1 ptq “ 1 ` t2 and p2 ptq “ 1 ´ t2 linearly independent or not ?
Solution : There is no non zero c scalar such that cp1 ptq “ p2 ptq or p2 ptq “ cp1 ptq if so
then one would have that 1 “ ´1 which is impossible. So they are not multiples. So,
the set tp1 , p2 u is linearly independent.
Problem 5 :
Given vectors u1 , ¨ ¨ ¨ , up and w in V, show that w is a linear combination of u1 , ¨ ¨ ¨ , up
if and only if rwsB is a linear combination of the coordinate vectors ru1 sB , ¨ ¨ ¨ , rup sB .
Solution : Let u1 , ¨ ¨ ¨ , up and w in V. Suppose that w is a linear combination of
u2 , ¨ ¨ ¨ , up . That is there is scalar c1 , ¨ ¨ ¨ , cp , such that w “ c1 u1 ` ¨ ¨ ¨ ` cp up .
Thus rwsB “ rc1 u1 ` ¨ ¨ ¨ ` cp up sB , since the coordinate map is linear this implies that
rwsB “ c1 ru1 sB ` ¨ ¨ ¨ ` cp rup sB
Thus, rwsB is a linear combination of the coordinate vectors ru1 sB , ¨ ¨ ¨ , rup sB .
For the converse suppose rwsB is a linear combination of the coordinate vectors ru1 sB , ¨ ¨ ¨ , rup sB .
That is there is scalar c1 , ¨ ¨ ¨ , cp , such that
rwsB “ c1 ru1 sB ` ¨ ¨ ¨ ` cp rup sB
since the coordinate map is linear this implies that rwsB “ rc1 u1 ` ¨ ¨ ¨ ` cp up sB . The
coordinate map being one to one implies that then.
w “ c1 u1 ` ¨ ¨ ¨ ` cp up
Problem 6 :
Let p1 ptq “ 1 ` t2 , p2 ptq “ t ´ 3t2 and p3 ptq “ 1 ` t ´ 3t2 .
1. Use coordinate vectors to show that these polynomials form a basis for P2 .
2. Consider the basis B “ tp1 , p2 , p3 u for P2 . Find q in P2 , given that
¨
˛
´1
rqsB “ ˝ 1 ‚
2
Solution :
1. We consider the standard basis for P2 which is S “ t1, t, t2 u. The coordinates
vectors of these polynomial for the standard basis are then :
¨ ˛
¨
˛
¨
˛
1
0
1
rp1 sS “ ˝ 0 ‚, rp2 sS “ ˝ 1 ‚, rp3 sS “ ˝ 1 ‚
1
´3
´3
3
In order to prove that that tp1 , p2 , p3 u form a basis one can prove that the matrix
A “ rrp1 sS , rp2 sS , rp3 sS s whose column are the coordinate vectors of the polynomials is invertible. For this, one can row reduce the matrix A and see if there is
a pivot in each row.
˛
¨
1 0 1
A „ Row reduce „ ˝ 0 1 1 ‚
0 0 ´1
Since we have a pivot in each row by the invertible matrix theorem we know that
A is invertible, , thus the column of A form a basis of R3 . Since the coordinate
map is an isomorphism form R3 to P2 , the corresponding polynomials tp1 , p2 , p3 u
form a basis for P2 .
2. By definition if
¨
˛
´1
rqsB “ ˝ 1 ‚
2
Then one has q “ ´p1 ` p2 ` 2p3 . Thus qptq “ 1 ` 3t ´ 10t2 .
Problem 7 :
The first four Laguerre polynomials are 1, 1 ´ t, 2 ´ 4t ` t2 and 6 ´ 18t ` 9t2 ´ t3 . Show
that these polynomials form a basis of P3 .
Solution : Let p1 ptq “ 1, p2 ptq “ 1 ´ t, p3 ptq “ 2 ´ 4t ` t2 , p4 ptq “ 6 ´ 18t ` 9t2 ´ t3 .
We consider the standard basis for P3 which is S “ t1, t, t2 , t3 u. The coordinates vectors
of these polynomial for the standard basis are then :
˛
˛
¨
˛
¨
¨
¨ ˛
6
2
1
1
‹
‹
˚
‹
˚
˚
˚ 0 ‹
‹ , rp2 sS “ ˚ ´1 ‹ , rp3 sS “ ˚ ´4 ‹ , rp4 sS “ ˚ ´18 ‹ ,
rp1 sS “ ˚
˝ 9 ‚
˝ 1 ‚
˝ 0 ‚
˝ 0 ‚
´1
0
0
0
In order to prove that that tp1 , p2 , p3 , p4 u form a basis one can prove that the matrix
A “ rrp1 sS , rp2 sS , rp3 sS , rp4 sS s whose column are the coordinate vectors of the polynomial
is invertible. But
¨
˛
1 1
2
6
˚ 0 ´1 ´4 ´18 ‹
‹
A“˚
˝ 0 0
1
9 ‚
0 0
0 ´1
The matrix has a pivots position in each row thus by the invertible matrix theorem
we know that A is invertible, thus the column of A form a basis of R4 and since the
coordinate map is an isomorphism from P3 onto R4 , thus tp1 , ¨ ¨ ¨ , p4 u form a basis for P3 .
Problem 8 :
Let H be an n-dimensional subspace of a n-dimensional vector space V. Show that
H “ V.
Solution : Let H be an n-dimensional subspace of a n-dimensional vector space V.
4
If dimpVq “ dimpHq “ 0 then H “ V “ t0u and the statement is true. Now suppose
dimpVq “ dimpHq “ n ą 0. By definition of the dimension, we know that H contains
a basis S consisting of n vectors, but this vectors being also vector of V since V is a
subspace of H, we have then n linearly independent vectors on a space V of dimension n, thus the basis theorem applied to V tells us that S is also a basis for V. Thus
H “ V “ SpanpSq.
Problem 9 :
Let V and W be vector spaces and T : V Ñ W be a linear transformation. Let H be
a subspace of V and let TpHq be the set off images of vectors in H. Then TpHq is a
subspace of W. Prove that dimpTpHqq ď dimpHq.
Solution : Since H is a nonzero subspace of a finite dimensional vector space V, H
is finite-dimensional and has a basis. Let tu1 , ¨ ¨ ¨ , up u be a basis for H. We show that
the set tTpu1 q, ¨ ¨ ¨ , Tpup qu span TpHq. Let y be in TpHq. Then there is a vector x in
H with Tpxq “ y. Since x is in H and tu1 , ¨ ¨ ¨ , up u is a basis for H, one can write
x “ c1 u1 ` ¨ ¨ ¨ ` cp up for some scalars c1 , ¨ ¨ ¨ , cp . Since the transformation T is linear,
one has
y “ Tpxq “ Tpc1 u1 ` ¨ ¨ ¨ ` cp up q “ c1 Tpu1 q ` ¨ ¨ ¨ ` cp Tpup q
Thus y is a linear combination of Tpu1 q, ¨ ¨ ¨ , Tpup q and tTpu1 q, ¨ ¨ ¨ , Tpup qu spans TpHq.
By the Spanning set theorem, this set contains a basis for TpHq. This basis then has
not more than p vectors and dimpTpHqq ď p “ dimpHq.
Problem 10 :
Let V and W be vector spaces and T : V Ñ W be a linear transformation. Let H be a
nonzero subspace of V, and suppose T is a one-to-one (linear) mapping of V into W.
Prove that dimpTpHqq “ dimpHq. If T happens to be one-to-one mapping of V onto
W, then dimpVq “ dimpWq. Isomorphic finite-dimensional vector spaces have the same
dimension.
Solution Since H is a nonzero subspace of a finite dimensional vector space V, H is
finite-dimensional and has a basis. Let tu1 , ¨ ¨ ¨ , up u be a basis for H, we have shown in
the previous exercise that tTpu1 q, ¨ ¨ ¨ , Tpup qu span TpHq. So we just need to show that
tTpu1 q, ¨ ¨ ¨ , Tpup qu are linearly independent. For this let c1 , ¨ ¨ ¨ cp be scalar such that
c1 Tpu1 q ` ¨ ¨ ¨ ` cp Tpup q “ 0
Since T is linear, then the previous equality is equivalent to
Tpc1 u1 ` ¨ ¨ ¨ ` cp up q “ 0
But since T is supposed to be one-to-one this implies that
c1 u1 ` ¨ ¨ ¨ ` cp up “ 0
But since tu1 , ¨ ¨ ¨ , up u form a basis for H by assumption there are in particular linearly
independent and the previous equality implies
c1 “ ¨ ¨ ¨ “ cp “ 0
5
Thus tTpu1 q, ¨ ¨ ¨ , Tpup qu is linearly independent and we knew it span TpHq thus form a
basis of TpHq. As a consequence,
dimpTpHqq “ dimpHq “ p
6