. Then we have

(1)

1.2 Matrices and Linear Operators 17

An important and less trivial consequence of the rank formula is the following.

1.28 Theorem (Rank of the transpose). Let A ∈ M

m,n

. Then we have

(i) the maximum number of linearly independent columns and the maximum number of linearly independent rows are equal, i.e.,

Rank A = Rank A

^T

,

(ii) let p := Rank A. Then there exists a nonsingular p × p square submatrix of A.

Proof. (i) Let A = [a

ⁱ_j

], let a

1

, a

2

, . . . , a

n

be the columns of A and let p := Rank A. We assume without loss of generality that the first p columns of A are linearly independent and we define B as the m×p submatrix formed by these columns, B := [a

1

| a

2

| . . . | a

p

].

Since the remaining columns of A depend linearly on the columns of B, we have

a

^k_j

=

p

X

i=1

r

ⁱ_j

a

^k_i

∀k = 1, . . . , m, ∀j = p + 1, . . . , n

for some R = [r

ⁱ_j

] ∈ M

_p,n−p

(K). In terms of matrices, h

a

_p+1

˛

˛ a

_p+2

˛

˛ . . .

˛

˛ a

_n

i

= h a

₁

˛

˛ . . .

˛

˛ a

_p

i

R = BR, hence

A = h B ˛

˛

˛ BR i

= B h Id

p

˛

˛ R i .

Taking the transposes, we have A

^T

∈ M

n,m

(K), B

^T

∈ M

p,m

(K) and

A

^T

= 0 B B

@ Id

p

R

^T

1 C C A

B

^T

. (1.6)

Since [ Id

p

| R]

^T

is trivially injective, we infer that ker A

^T

= ker B

^T

, hence by the rank formula

Rank A

^T

= m − dim ker A

^T

= m − dim ker B

^T

= Rank B

^T

, and we conclude that

Rank A

^T

= Rank B

^T

≤ min(m, p) = p = Rank A.

Finally, by applying the above to the matrix A

^T

, we get the opposite inequality Rank A = Rank (A

^T

)

^T

≤ Rank A

^T

, hence the conclusion.

(ii) With the previous notation, we have Rank B

^T

= Rank B = p. Thus B has a set of p independent rows. The submatrix S of B made by these rows is a square p × p matrix

with Rank S = Rank S

^T

= p, hence nonsingular. ⊓ ⊔

1.29 ¶. Let A ∈ M

m,n

(K), let A(x) := Ax and let (v

1

, v

2

, . . . , v

n

) be a basis of K

ⁿ

. Show the following:

(i) A is injective if and only if the vectors A(v

1

), A(v

2

), . . . , A(v

n

) of K

^m

are linearly independent,

(ii) A is surjective if and only if {A(v

1

), A(v

2

), . . . , A(v

_n

)} spans K

^m

,

(iii) A is bijective iff {A(v

1

), A(v

2

), . . . , A(v

n

)} is a basis of K

^m

.