2Linear maps

IB Linear Algebra

2.5 Elementary matrix operations

We are now going to re-prove our corollary that we can find

P, Q

such that

Q

−1

AP

=

I

r

0

0 0

in a way that involves matrices only. This will give a

concrete way to find P and Q, but is less elegant.

To do so, we need to introduce elementary matrices.

Definition

(Elementary matrices)

.

We call the following matrices of

GL

n

(

F

)

elementary matrices:

S

n

ij

=

1

.

.

.

1

0 1

1

.

.

.

1

1 0

1

.

.

.

1

This is called a reflection, where the rows we changed are the ith and jth row.

E

n

ij

(λ) =

1

.

.

.

1 λ

.

.

.

1

.

.

.

1

This is called a shear, where λ appears at the i, jth entry.

T

n

i

(λ) =

1

.

.

.

1

λ

1

.

.

.

1

This is called a shear, where λ 6= 0 appears at the ith column.

Observe that if A is a m × n matrix, then

(i) AS

n

ij

is obtained from A by swapping the i and j columns.

(ii) AE

n

Ij

(λ) is obtained by adding λ× column i to column j.

(iii) AT

n

i

(λ) is obtained from A by rescaling the ith column by λ.

Multiplying on the left instead of the right would result in the same operations

performed on the rows instead of the columns.

Proposition.

If

A ∈ Mat

n,m

(

F

), then there exists invertible matrices

P ∈

GL

m

(F), Q ∈ GL

n

(F) so that

Q

−1

AP =

I

r

0

0 0

for some 0 ≤ r ≤ min(m, n).

We are going to start with

A

, and then apply these operations to get it into

this form.

Proof.

We claim that there are elementary matrices

E

m

1

, ··· , E

m

a

and

F

n

1

, ··· , F

n

b

(these E are not necessarily the shears, but any elementary matrix) such that

E

m

1

···E

m

a

AF

n

1

···F

n

b

=

I

r

0

0 0

This suffices since the

E

m

i

∈ GL

M

(

F

) and

F

n

j

∈ GL

n

(

F

). Moreover, to prove the

claim, it suffices to find a sequence of elementary row and column operations

reducing A to this form.

If

A

= 0, then done. If not, there is some

i, j

such that

A

ij

6

= 0. By swapping

row 1 and row

i

; and then column 1 and column

j

, we can assume

A

11

6

= 0. By

rescaling row 1 by

1

A

11

, we can further assume A

11

= 1.

Now we can add

−A

1j

times column 1 to column

j

for each

j 6

= 1, and then

add −A

i1

times row 1 to row i 6= 1. Then we now have

A =

1 0 ··· 0

0

.

.

. B

0

Now

B

is smaller than

A

. So by induction on the size of

A

, we can reduce

B

to

a matrix of the required form, so done.

It is an exercise to show that the row and column operations do not change

the row rank or column rank, and deduce that they are equal.