We want to show that the recursively defined determinant is a "multilinear“ and "alternating“ mapping, where we identify
Mat
n
(
K
)
≅
(
K
n
)
n
,
{\displaystyle {}\operatorname {Mat} _{n}(K)\cong (K^{n})^{n}\,,}
so a matrix is identified with the
n
{\displaystyle {}n}
-tuple of the rows of the matrix. We consider a matrix as a tuple of columns
(
v
1
⋮
v
n
)
{\displaystyle {\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}}
where the entries
v
i
{\displaystyle {}v_{i}}
are row vectors of length
n
{\displaystyle {}n}
.
Let
K
{\displaystyle {}K}
be a field, and
n
∈
N
+
{\displaystyle {}n\in \mathbb {N} _{+}}
. Then the
determinant
Mat
n
(
K
)
=
(
K
n
)
n
⟶
K
,
M
⟼
det
M
,
{\displaystyle \operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K,M\longmapsto \det M,}
is
multilinear. This means that for every
k
∈
{
1
,
…
,
n
}
{\displaystyle {}k\in \{1,\ldots ,n\}}
,
and for every choice of
n
−
1
{\displaystyle {}n-1}
vectors
v
1
,
…
,
v
k
−
1
,
v
k
+
1
,
…
,
v
n
∈
K
n
{\displaystyle {}v_{1},\ldots ,v_{k-1},v_{k+1},\ldots ,v_{n}\in K^{n}}
,
and for any
u
,
w
∈
K
n
{\displaystyle {}u,w\in K^{n}}
,
the identity
det
(
v
1
⋮
v
k
−
1
u
+
w
v
k
+
1
⋮
v
n
)
=
det
(
v
1
⋮
v
k
−
1
u
v
k
+
1
⋮
v
n
)
+
det
(
v
1
⋮
v
k
−
1
w
v
k
+
1
⋮
v
n
)
{\displaystyle {}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}+\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,}
holds, and for
s
∈
K
{\displaystyle {}s\in K}
,
the identity
det
(
v
1
⋮
v
k
−
1
s
u
v
k
+
1
⋮
v
n
)
=
s
det
(
v
1
⋮
v
k
−
1
u
v
k
+
1
⋮
v
n
)
{\displaystyle {}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\su\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=s\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,}
holds.
Proof
This proof was not presented in the lecture.
◻
{\displaystyle \Box }
Proof
This proof was not presented in the lecture.
◻
{\displaystyle \Box }
The relation between rank, invertibility and linear independence was proven in
fact .
Suppose now that the rows are
linearly dependent .
After exchanging rows, we may assume that
v
n
=
∑
i
=
1
n
−
1
s
i
v
i
{\displaystyle {}v_{n}=\sum _{i=1}^{n-1}s_{i}v_{i}}
.
Then, due to
fact
and
fact ,
we get
det
M
=
det
(
v
1
⋮
v
n
−
1
∑
i
=
1
n
−
1
s
i
v
i
)
=
∑
i
=
1
n
−
1
s
i
det
(
v
1
⋮
v
n
−
1
v
i
)
=
0
.
{\displaystyle {}\det M=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\\sum _{i=1}^{n-1}s_{i}v_{i}\end{pmatrix}}=\sum _{i=1}^{n-1}s_{i}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\v_{i}\end{pmatrix}}=0\,.}
Now suppose that the rows are linearly independent. Then, by exchanging of rows, scaling and addition of a row to another row, we can transform the matrix successively into the identity matrix. During these manipulations, the determinant is multiplied with some factor
≠
0
{\displaystyle {}\neq 0}
. Since the determinant of the identity matrix is
1
{\displaystyle {}1}
, the determinant of the initial matrix is
≠
0
{\displaystyle {}\neq 0}
.
◻
{\displaystyle \Box }