Math 312: Linear Algebra

Homework 1
Lawrence Tyler Rush
<me@tylerlogic.com>

July 11, 2012

1


For this problem, let us assume that a,b,c,d,e,f F for some field F. The system of equations
ax + cy  =   e
bx+ dy  =   f
has the matrix form of Au = v where
    (  a  c )     ( x  )         ( e )
A =    b  d  , u =   y  , and v =  f

or alternatively

 (    )    (   )   (   )
x   a   +y   c   =    e
    b        d       f

Given this form, we see that in order for a solution to exist for any e,f F then a,b,c,d need to be such that the vectors

(    )     (   )
   a   and   c
   b         d

spans the entire space Fcols2, where Fcols2 denotes the vector space of two dimensional column-vectors.

2


Let v1,,vn Fn for Fn over F and A Mn×m(F) such that vi = (a1i,,ani) with each aij F and Aij = aij.

Given this definition of A, we see that

A = (vt1|v2t|⋅⋅⋅|vtm )

that is, v1t,,vnt are the columns of A.

(a)


Let Span(v1,,vm) = Fn. Then for any b Fn there exist c1,,cm F such that c1v1 + ⋅⋅⋅ + cmvm = b. Since v1,,vm,b are all row vectors, this is akin to saying that
c1vt1 + ⋅⋅⋅+ cmvtm = bt

but this is just Act = bt where ct = (c1,,cm)t. Hence for any vector bt in the column-space of Fn, Ax = bt will have a solution, i.e. the row-reduced echelon form of A will have a pivot in each row, so it will have n pivots.

(b)


Let the row-reduced echelon form of A have n pivots, that is, every row of A has a pivot since it has n rows. This means that no matter the b Fcolsn, Ax = b will have a solution. In other words, for any b there exist c1,,cm F such that x1v1t + ⋅⋅⋅ + xmvmt = b. This is equivalent to saying that for any b′∈ Fn there exist c1,,cm F such that
c1v1 + ⋅⋅⋅+ cmvm = b′

Hence v1,,vm span Fn.

(c)


Given our solution in part (a) if Span(v1,,vm) = Fn, then the matrix A with the ith column being vi, will have n pivots in its RREF. Because A is an n×m matrix and because each pivot in a RREF matrix must be in a different row and column than any other pivot, then m n.

3


For the following solutions let F be a field and fix A Mm×n(F).

(a)


Let S be the subset of Mm×k(F) defined by {AB : B Mn×k(F )}. For b,b′∈ F and C,C′∈ S there exist B,B′∈ Mm×k(F ) such that both C = AB and C= AB. Making use of this and Lemma A.1 we attain
      ′ ′          ′   ′              ′ ′          ′ ′
bC + bC  = b(AB )+ b (AB  ) = A (bB )+ A(bB ) = A(bB + bB )

and thus bC + bC′∈ S. Given this and that 0M m×k(F) are both in S, so S is a subspace of Mn×k(F).

(b)


Let S be the subset of Mk×n(F) defined by {B A : B Mk×m(F )}. For b,b′∈ F and C,C′∈ S there exists B,B′∈ Mk×m(F) such that C = BA and C= BA. Making use of this we obtain the following.
      ′ ′          ′  ′             ′ ′           ′ ′
bC + bC  = b(BA )+ b (B  A) = (bB)A + (bB )A = (bB + bB )A

Therefore bC + bC′∈ S. Since this is true and 0M k×n(F) S, then S is a subspace of Mk×n(F).

(c)


Let S be the subset of Mn×k(F) defined by {B Mn×k(F) : A B = 0}. For b,b′∈ F and B,B′∈ S we have
        ′ ′             ′ ′            ′   ′        ′
A (bB + bB  ) = A(bB)+ A (bB ) = b(AB )+ b(AB ) = b0+ b0 = 0

since AB = 0 and AB= 0. This implies bB + bB′∈ S, so S is a subspace of Mn×k(F).

(d)


Let S be the subset of Mk×m(F) defined by {B Mk×m(F) : B A = 0}. For b,b′∈ F and B,B′∈ S we have
(bB + b′B ′)A = (bB )A+ (b′B ′)A = b(BA )+ b′(B ′A) = b0+ b′0 = 0

by Lemma A.1 and because BA = 0 and BA = 0. This indicates that bB + bB′∈ S. Thus S is a subspace of Mk×m(F)

4


(a)


This subset W1 is a subspace of 3 since for a,b and (3x,x,x),(3y,y,y) 3 we have
a(3x,x,− x)+ b(3y,y,− y) = (3ax,ax,− ax)+ (3by,by,− by) = (3(ax + by),ax +by,− (ax + by))

which is an element of W1.

(b)


This subset W2 is not a subspace of 3 because it does not contain the zero vector since the zero vector does not fit the form of vectors in W2, i.e. there are no a2,a3such that
(0,0,0) = (a3 + 2,a2,a3)

(c)


Given (x,y,z),(x,y,z) W3 we have that
2x − 7y+ z = 0 and 2x′ − 7y′ + z′ = 0

So for a,b F we have the following, given that a(x,y,z) + b(x,y,z) = (ax + bx,ay + by,az + bz)

        ′         ′         ′                      ′     ′   ′                    ′    ′  ′
2(ax + bx )− 7(ay+ by )+ (az+ bz ) = (2ax− 7ay+ az)+ (2bx − 7by +bz ) = a(2x− 7y+ z)+ b(2x − 7y +z ) = 0

then a(x,y,z) + b(x,y,z) W3, and thus W3 a subspace of 3.

(d)


Given (x,y,z),(x,y,z) W4 we have that
                  ′    ′   ′
x − 4y− z = 0 and x − 4y − z = 0

So for a,b F we have the following, given that a(x,y,z) + b(x,y,z) = (ax + bx,ay + by,az + bz)

(ax+ bx′)− 4(ay+ by′)− (az + bz′) = (ax− 4ay− az) +(bx′ − 4by′ − bz′) = a(x− 4y − z)+ b(x′ − 4y′ − z′) = 0

then a(x,y,z) + b(x,y,z) W4, and hence W4 is a subspace of 3

(e)


This subset, W5, is not a subspace of 3 as it does not contain the zero vector since 0 + 2(0) 3(0) = 01.

(f)


The subset W6 requires that all elements, (x,y,z), in it have that 5x2 3y2 + 6z2 = 0, that is to say that
     ∘ ----------
y = ±  5∕3x2 + 2z2

So (3,√ --
  17,1) and (3,√ --
  23,2) are both in W6. They’re sum, however, is (6,√--
 17 + √ --
  23,3) which since

  2    √ --  √-- 2    2              ∘ ------                    ∘ ------           ∘ ------
5(6 )− 3( 17+  23) +6(3 ) = 180− 3(17+2  17(23)+23 )+54 = 234− (51+6  17(23)+69 ) = 252+6 17(23) ⁄= 0

then W6 is not closed under addition, and so its not a subspace of 3.

5


(a)


Let S be the set of all matrices in A Mm×n(F) for which the entries of A are all zero except for any one entry. There will only be mn of these matrices in Mm×n(F), and they will span the space.

More formally, we can define S as the set {Ak : Ak Mm×n(F ),k ∈{1,,m},ℓ ∈{1,,n}} where each entry of the matrix Ak is defined by

(Ak ) = δ (i)δ (j)
  ℓij   k   ℓ

where δ is the Kronecker-delta according to the field F, i.e. we use the field’s multiplicative and additive identities rather than the integers 1 and 0.

With this more specific definition, we can write any matrix B Mm×n(F) with the following linear combination of elements of S.

    m∑  (    i      i           i)
B =     Bi1A 1 + Bi2A2 +⋅⋅⋅+ BinAn
    i=1

(b)


Here we can use our ideas from the previous problem. We’ll continue our trend of using 1 and 0 in order to be able to generate the whole space. We define S Mn×n(F) in much the same manner as we did before, as the set {Ak : Ak Mn×n(F),k ∈{1,,n},ℓ ∈{1,,n},k }, however, we need to redefine Ak to get the symmetry we desire. Putting it to δk(i)δ(j) + δk(j)δ(i) gets us our symmetry, but has the annoying side effect of resulting in a value of one being added to itself whenever = k. Hence, we need to “neglect” one of the terms whenever equals k. To do this, simply multiply one term by (1 δk())1 to get the following entry-wise definition.
(Akℓ)ij = (1− δk(ℓ))δk(i)δℓ(j)+ δk(j)δℓ(i)

With this we can see that when k = ,

(Akℓ)ij = δk(j)δℓ(i) = δℓ(j)δk(i) = δk(i)δℓ(j) = (Akℓ)ji

and when k,

   k                                                k
(A ℓ)ij = δk(i)δℓ(j)+ δk(j)δℓ(i) = δk(j)δℓ(i)+ δk(i)δℓ(j) = (A ℓ)ji

so we confirm the symmetry of each matrix in S.

Since the number of elements of S is governed by the possible values for k and , then we look to those for the size of S. The value of k ranges from 1 to n and will always be greater than or equal to k, meaning that when k = 1,k = 2,,k = n the are n,n 1,,1 values of . Summing these we get

                    ( n)         n(n-+1)-
n + (n− 1)+ ⋅⋅⋅+ 1 =  2 (n + 1) =   2

This is the size of S, as needed.

Now, any B Symn can be written as a linear combination of the elements of S in the following manner.

       (         )
    ∑n ( ∑n     i)
B =         BijAj
    i=1  j=i

(c)


Let S Mn×n(F) be the set {Ak : Ak Mn×n(F ),k ∈{1,,n},ℓ ∈{1,,n},k < ℓ}, but define Ak, entry-wise, in the following way.
  k
(A ℓ)ij = δk(i)δℓ(j) − δk(j)δℓ(i)

We can see this results in skew-symmetry,

  k                                                     k
(Aℓ)ij = δk(i)δℓ(j)− δk(j)δℓ(i) = − (δk(j)δℓ(i)− δk(i)δℓ(j)) = − (A ℓ)ji

as well as for when i = j (the diagonals)

(Ak ) = δ (i)δ (j)− δ(j)δ(i) = δ (i)δ (i)− δ (i)δ (i) = 0
  ℓ ij   k   ℓ      k   ℓ     k   ℓ     k   ℓ

which is required in Skewn.

Given that k ranges over 1,,n, k < ℓ, and ranges over k + 1,,n then the size of S is the following.

                          (     )
                           n-−-1    n(n−-1)-
(n − 1)+ (n− 2)+ ⋅⋅⋅+ 1 = n   2    =    2

Now, any B Skewn can be written as a linear combination of the elements of S in the following manner.

     n  (  n       )
B = ∑   ( ∑   BijAi)
     i=1  j=i+1    j

6


Let V be a vector space over a field F and there exist some subspaces W1 and W2.

(a) W1 W2


Let u and v each be vectors in W1 W2 and a,b F. Then u and v are also each individually in W1 and W2, but being subspaces of V , W1 and W2 then both also contain the sum au + bv and hence au + bv W1 W2.

(b) W1 W2


It is not always true that the union of two subspaces is a subspace. To derive a counter-example, we can apply what we learned about some subspaces of Mm×n(F) in problem 3.

Letting A1 and A2 be

(  1  0 )     ( 0  0 )
   0  0   and   0  1

respectively, we know that W1 = {A1B : B M2×2(F)} and W2 = {A2B : B M2×2(F)} are subspaces of M2×2(F). Elements of W1 have thus form

   (      )   (      ) (      )   (      )
     a  b       1  0     a  b       a  b
A1   c  d   =   0  0     c  d   =   0  0
(6.1)

and likewise elements of W2 are

   ( a  b )   ( 0  0 ) ( a  b )   ( 0  0 )
A2   c  d   =   0  1     c  d   =    c d
(6.2)

Since the identity matrix is in M2×2(F) then A1 W1 and A2 W2, and therefore they are both in W1 W2. If this union were a subspace, then

          ( 1  0 )
A1 + A2 =   0  1

would need to be in it, but this is neither in the form of equation 6.1 nor equation 6.2. So W1 W2 is not a subspace.

(c) W1 + W2


Let u,v W1 + W2. Hence there exist vectors w1u,w1v W1 and w2u,w2v W2 such that u = w1u + w2u and v = w1v + w2v. By this we have that for a,b F
au + bv = a(w  + w  )+ b(w  + w  ) = aw  + aw  + bw   + bw   = (aw  + bw  )+ (aw  + bw  )
            1u   2u      1v    2v      1u    2u     1v     2v      1u    1v      2u    2v

and therefore au + bv W1 + W2. Thus W1 + W2 is a subspace of V .

7


(a)


A matrix A Mn×n(ℝ ) that is both symmetric and anti-symmetric must maintain the condition Aij = Aji = Aji. Only the zero matrix of Mn×n(ℝ) does this, so Symn Skewn = {0}.

Let A Mn×n(ℝ ) and define matrices B and C by the following.

     1      t         1      t
B  = 2(A +A  ) and C = 2(A − A )

With these definitions, B,C Mn×n(ℝ). We then also have that

     ( 1       )     1             1            1             1              ( 1       )
Bij =  -(A +At )   = -(Aij+ (At)ij) =- (Aij+Aji) = -((At)ji+Aji) =- (Aji+ (At )ji) =  -(A + At)   = Bji
       2         ij   2             2            2             2                2        ji

and

     ( 1       )     1             1             1             1               ( 1       )
Cij =  -(A− At)   =  -(Aij− (At)ij) =-(Aij− Aji) = −-(Aji− Aij) = −-(Aji− (At)ji) = −-(A − At)  = − Cji
       2         ij   2             2             2             2                 2         ji

which indicate that B Symn and C Skewn. Looking at the sum of these two matrices

        1          1          1                   1
B + C = -(A + At)+ -(A − At) =- (A + At + A− At) = -(2A) = A
        2          2          2                   2

we see that it’s simply A, implying that A Symn Skewn.
 
With the above two results, we have that Mn×n(ℝ ) = Symn Skewn.

(b)


Let both f 𝔒 and f 𝔈. Then for every x , f(x) = f(x) = f(x), and thus f must be the constant zero function. This then implies that 𝔒 𝔈 = {f} = {0} since the constant zero function is the zero vector of C0(, ).

Now let h C0(, ) and define the functions f and g by the following.

       1                       1
f(x) = 2(h (x) +h (− x)) and g(x ) = 2(h(x)− h(− x))

With these definitions, f,g C0(, ). Given these we see that

f(− x) = 1(h(− x)+ h(− (− x))) = 1(h(− x)+ h(x)) = 1(h(x)+ h(− x )) = f(x )
        2                    2               2

and

        1                    1                 1
g(− x) = 2(h(− x) − h(− (− x))) = 2(h(− x) − h(x)) = − 2(h(x)− h(− x )) = − g(x)

which informs us that f 𝔈 and g 𝔒. Looking at the sum of f and g,

                     1              1               1                          1
(f+g)(x) = f(x)+g(x) = 2 (h(x)+h (− x))+ 2(h(x)− h(− x)) = 2(h(x)+h (− x)+h (x)− h(− x)) = 2 (2h(x)+h(x)) = h(x�

we see that it is simply h. So h 𝔒 𝔈.
 
By the results of the two paragraphs above, we have that C0(, ) = 𝔒 𝔈

A Extraneous Proofs

While there are some things in the solutions above that appear obvious to me, there are others that did not. For the latter, these are some of the proofs I generated in order to convince myself of certain properties.

Lemma A.1 (Commuting Scalars Among Matrices) Given a scalar c F, A Mm×(F), and B M×n(F) for some field F, c can commute about the matrix multiplication of A and B, that is

c(AB) = A(cB)
(A.3)

Proof. In this case, it suffices to show that the ijth entry on the left-hand-side of A.3 is equivalent to the right-hand-side. Given this we have the following due to the multiplicative commutativity of F.

                     (∑ℓ       )   ∑ℓ           ∑ℓ                      -|
(c(AB ))ij = c(AB )ij = c   AikBkj  =    AikcBkj =    Aik(cB )kj = (A (cB ))ij  --
                      k=1          k=1          k=1

References

[1]   Friedberg, S.H. and Insel, A.J. and Spence, L.E. Linear Algebra 4e. Upper Saddle River: Pearson Education, 2003.