Kenneth Kuttler

8.6. FREDHOLM ALTERNATIVE 173

which says that b is a vector in span(a1, · · · ,an) . This shows that there exists a solutionto the system, 8.4 if and only if b is contained in span(a1, · · · ,an) . In words, there is asolution to 8.4 if and only if b is in the column space of A. In terms of rank, the followingproposition describes the situation.

Proposition 8.5.28 Let A be an m× n matrix and let b be an m× 1 column vector. Thenthere exists a solution to 8.4 if and only if

rank(

A | b)= rank(A) . (8.5)

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 8.4.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 8.5is the same. ■

8.6 Fredholm AlternativeBefore reading this, it would be helpful to read about the dot product earlier in Section 3.1.However, what you need to know is this: For x ∈ Rp meaning

x =(

x1 x2 · · · xp

)and y another such thing in Rp,x ·y is defined as ∑ j x jy j ≡ x1y1 + x2y2 + · · ·+ xpyp. Thisis called the dot product.

There is a very useful version of Proposition 8.5.28 known as the Fredholm alterna-tive. I will only present this for the case of real matrices here. Later a much more elegantand general approach is presented which allows for the general case of complex matrices.

The following definition is used to state the Fredholm alternative.

Definition 8.6.1 Let S⊆Rm. Then S⊥ ≡{z ∈ Rm : z · s = 0 for every s ∈ S} . The funny ex-ponent, ⊥ is called “perp”.

Now note

ker(AT )≡ {z : AT z = 0

{z :

∑k=1

zkak = 0

}Here the ak are the rows of A because they are the columns of AT .

Lemma 8.6.2 Let A be a real m×n matrix, let x ∈ Rn and y ∈ Rm. Then

(Ax ·y) =(x·AT y

)

8.6. FREDHOLM ALTERNATIVE 173which says that b is a vector in span(aj,--- ,a,). This shows that there exists a solutionto the system, 8.4 if and only if b is contained in span (aj,---,a,). In words, there is asolution to 8.4 if and only if b is in the column space of A. In terms of rank, the followingproposition describes the situation.Proposition 8.5.28 Let A be an m x n matrix and let b be an m x 1 column vector. Thenthere exists a solution to 8.4 if and only ifrank ( A | b ) =rank (4). (8.5)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 m )where Mf + 0 in B. Therefore, there will exist a solution to the system 8.4.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 8.5is the same.8.6 Fredholm AlternativeBefore reading this, it would be helpful to read about the dot product earlier in Section 3.1.However, what you need to know is this: For x € R? meaningx=( x xX2 °'° xp )and y another such thing in R’,x-y is defined as ))jxjyj =x1y1 +x2y2 +++: +Xpyp. Thisis called the dot product.There is a very useful version of Proposition 8.5.28 known as the Fredholm alterna-tive. I will only present this for the case of real matrices here. Later a much more elegantand general approach is presented which allows for the general case of complex matrices.The following definition is used to state the Fredholm alternative.Definition 8.6.1 Let SCR”. Then S+ = {2 € R” :z-s =0 for every s € S}. The funny ex-ponent, | is called “perp”.Now notemker (A‘) = {z:A'z =0}= {* yea = ohk=1Here the a, are the rows of A because they are the columns of A’.Lemma 8.6.2 Let A be a real m x n matrix, let x € R" and y € R”. Then(Ax-y) = (x-A’y)