basis-free definition of determinant

basis-free definition of determinant

The

definition

of

determinant

as a multilinear mapping on rows can be
modified to provide a

basis-free definition of determinant

. In order
to make it clear that we are not using bases. we shall speak in terms
of an

endomorphism

of a

vector space

over

k

rather than speaking of
a matrix whose entries belong to

k

. We start by recalling some
preliminary facts.

Suppose

V

is a

finite-dimensional

vector space of

dimension

n

over
a field

k

. Recall that a multilinear map

f

:

V

n

→

k

is

alternating

if

f

⁢

(

x

)

=

0

whenever there exist distinct indices

i

,

j

∈

[

n

]

=

{

1

,

…

,

n

}

such that

x

i

=

x

j

. Every alternating
map

f

:

V

n

→

k

is

skew-symmetric

, that is, for each

permutation

π

∈

𝔖

n

, we have that

f

⁢

(

x

)

=

sgn

⁡

(

π

)

⁢

f

⁢

(

x

π

)

, where

x

π

denotes

(

x

π

⁢

(

i

)

)

i

∈

[

n

]

, the result of

π

permuting the entries of

x

.

Since the

trivial

map

0

:

V

n

→

k

is alternating and any

linear
combination

of alternating maps is alternating, it follows that
alternating maps form a

subspace

of the space of multilinear maps. In
the following

proposition

we show that this subspace is
one-dimensional.

Theorem

.

Suppose

V

is a finite-dimensional vector space of dimension

n

over
a field

k

. Then the space of alternating maps from

V

n

to

k

is
one-dimensional.

Proof.

We use a basis here, but we will throw it away later. We need the
basis here because each map we will consider has exactly as many

elements

as a basis of

V

. So let

B

=

{

b

i

:

i

∈

[

n

]

}

be a
basis of

V

.

Suppose

f

and

g

are nontrivial alternating maps from

V

n

to

k

.
We claim that

f

and

g

are

linearly dependent

. Let

x

∈

V

n

. We
may assume that the entries of

x

are basis vectors, that is, that

X

=

{

x

i

:

i

∈

[

n

]

}

⊂

{

b

i

:

i

∈

[

n

]

}

.
If

X

⊊

B

, then there exist distinct indices

i

,

j

∈

[

n

]

such
that

x

i

=

x

j

. Since

f

and

g

are alternating, it follows that

f

⁢

(

x

)

=

g

⁢

(

x

)

=

0

, which implies that

f

⁢

(

b

)

⁢

g

⁢

(

x

)

=

g

⁢

(

b

)

⁢

f

⁢

(

x

)

.
On the other hand, if

X

=

B

, then there is a permutation

π

∈

𝔖

n

such that

x

=

b

π

. Since

f

and

g

are
skew-symmetric, it follows that

f

⁢

(

b

)

⁢

g

⁢

(

x

)

=

sgn

⁡

(

π

)

⁢

f

⁢

(

b

)

⁢

g

⁢

(

b

)

=

g

⁢

(

b

)

⁢

f

⁢

(

x

)

.

In either case we find that

f

⁢

(

b

)

⁢

g

⁢

(

x

)

=

g

⁢

(

b

)

⁢

f

⁢

(

x

)

. Since

f

⁢

(

b

)

and

g

⁢

(

b

)

are fixed scalars, it follows that

f

and

g

are linearly
dependent.

So far we have shown only that the dimension of the space of
alternating maps is less than or equal to one. In order to show that
the space is one-dimensional we simply need to find a nontrivial
alternating form. To do this, let

{

b

i

*

:

i

∈

[

n

]

}

be the
natural basis of

V

*

, so that

b

i

*

⁢

(

b

j

)

is the

Kronecker delta

of

i

and

j

for any

i

,

j

∈

[

n

]

. Define a map

f

:

V

n

→

k

by

f

⁢

(

x

)

=

∑

π

∈

𝔖

n

sgn

⁡

(

π

)

⁢

∏

i

∈

[

n

]

b

i

*

⁢

(

x

π

⁢

(

i

)

)

.

One can check that

f

is

multilinear

and alternating. Moreover,

f

⁢

(

b

)

=

1

, so it is nontrivial. Hence the space of alternating maps
is one-dimensional.
∎

For an alternate view of the above results, we could look instead at

linear maps

from the

exterior product

⋀

n

V

into

k

. The
proposition above can be viewed as saying that the dimension of

⋀

n

V

is

(

n

n

)

=

1

.

We define the determinant of an endomorphism in terms of the action of
the endomorphism on alternating maps. Recall that if

M

:

V

→

V

is an endomorphism, its

pullback

M

*

is the unique

operator

such
that

(

M

*

⁢

f

)

⁢

(

x

i

)

i

∈

[

n

]

=

f

⁢

(

M

⁢

(

x

i

)

)

i

∈

[

n

]

.

Since the space of alternating maps is one-dimensional and
endomorphisms of a one-dimensional space reduce to

scalar
multiplication

, it follows that

M

*

⁢

f

is a

scalar multiple

of

f

.
We call this scalar the determinant. It is

well-defined

because the
scalar depends on

M

but not on

f

.

Definition

.

Suppose

V

is a finite-dimensional vector space of dimension

n

over
a field

k

, and let

M

:

V

→

V

be an endomorphism. Then the

determinant

of

M

is the unique scalar

det

⁡

(

M

)

such that

M

*

⁢

f

=

det

⁡

(

M

)

⁢

f

for all alternating maps

f

:

V

n

→

k

.

Title

basis-free definition of determinant

Canonical

name

BasisfreeDefinitionOfDeterminant

Date of creation

2013-03-22 16:51:39

Last modified on

2013-03-22 16:51:39

Owner

mps (409)

Last modified by

mps (409)

Numerical id

9

Author

mps (409)

Entry type

Definition

Classification

msc 15A15

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