122 CHAPTER 4. ROW OPERATIONS

4.4 Existence of Solutions to Linear Systems

Consider the linear system of equations,

Ax = b (4.4)

where A is an m× n matrix, x is a n× 1 column vector, and b is an m× 1 column vector.Suppose

A =(

a1 · · · an

)where the ak denote the columns of A. Then x = (x1, · · · , xn)T is a solution of the system4.4, if and only if

x1a1 + · · ·+ xnan = b

which says that b is a vector in span (a1, · · · ,an) . This shows that there exists a solutionto the system, 4.4 if and only if b is contained in span (a1, · · · ,an) . In words, there is asolution to 4.4 if and only if b is in the column space of A. In terms of rank, the followingproposition describes the situation.

Proposition 4.4.1 Let A be an m× n matrix and let b be an m× 1 column vector. Thenthere exists a solution to 4.4 if and only if

rank(A | b

)= rank (A) . (4.5)

Proof: Place(A | b

)and A in row reduced echelon form, respectively B and C. If

the above condition on rank is true, then both B and C have the same number of nonzerorows. In particular, you cannot have a row of the form(

0 · · · 0 ⋆)

where ⋆ ̸= 0 in B. Therefore, there will exist a solution to the system 4.4.Conversely, suppose there exists a solution. This means there cannot be such a row in

B described above. Therefore, B and C must have the same number of zero rows and sothey have the same number of nonzero rows. Therefore, the rank of the two matrices in 4.5is the same. ■

4.5 Fredholm Alternative

There is a very useful version of Proposition 4.4.1 known as the Fredholm alternative.I will only present this for the case of real matrices here. Later a much more elegant andgeneral approach is presented which allows for the general case of complex matrices.

The following definition is used to state the Fredholm alternative.

Definition 4.5.1 Let S ⊆ Rm. Then S⊥ ≡ {z ∈ Rm : z · s = 0 for every s ∈ S} . The funnyexponent, ⊥ is called “perp”.

Now note

ker(AT)≡{z : AT z = 0

}=

{z :

m∑k=1

zkak = 0

}

Lemma 4.5.2 Let A be a real m× n matrix, let x ∈ Rn and y ∈ Rm. Then

(Ax · y) =(x·ATy

)

122 CHAPTER 4. ROW OPERATIONS4.4 Existence of Solutions to Linear SystemsConsider the linear system of equations,Ax=b (4.4)where A is an m X n matrix, x is an X 1 column vector, and b is an m x 1 column vector.SupposeA= ( al oct: ay )where the a, denote the columns of A. Then x = (21,--- In) is a solution of the system4.4, if and only if%1a, +++: + apa, =bwhich says that b is a vector in span (aj,,--- ,a,). This shows that there exists a solutionto the system, 4.4 if and only if b is contained in span (aj,--- ,a,). In words, there is asolution to 4.4 if and only if b is in the column space of A. In terms of rank, the followingproposition describes the situation.Proposition 4.4.1 Let A be an m x n matrix and let b be an m x 1 column vector. Thenthere exists a solution to 4.4 if and only ifrank ( A | b ) = rank (A). (4.5)Proof: Place ( A | b ) and A in row reduced echelon form, respectively B and C. Ifthe above condition on rank is true, then both B and C have the same number of nonzerorows. In particular, you cannot have a row of the form(0 =. 0 *)where % #0 in B. Therefore, there will exist a solution to the system 4.4.Conversely, suppose there exists a solution. This means there cannot be such a row inB described above. Therefore, B and C' must have the same number of zero rows and sothey have the same number of nonzero rows. Therefore, the rank of the two matrices in 4.5is the same. ll4.5 Fredholm AlternativeThere is a very useful version of Proposition 4.4.1 known as the Fredholm alternative.I will only present this for the case of real matrices here. Later a much more elegant andgeneral approach is presented which allows for the general case of complex matrices.The following definition is used to state the Fredholm alternative.Definition 4.5.1 Let S CR”. Then St = {z € R™:2z-s=0 for everys € S}. The funnyexponent, L is called “perp”.Now notemker (A*) = {z -Alz = o} = f : So nak = ohk=1Lemma 4.5.2 Let A be a realm x n matriz, let x € R” andy € R™. Then(Ax-y) = (x-ATy)