# Determinants

Page Contents

## Determinant of a Matrix (definition)

The #{~{determinant}} of an ~n # ~n matrix A is defined as:

|A| _ = _ sum{&zeta.(&pi.)prod{~a_{~i,&pi.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ }

where P_{~n} is the symmetric group (the group of all permutations on ~n elements) and &zeta.(&pi.) is the sign of the permutation. [Briefly, any permutation is a product of a finite number of transpositions of two elements at a time. If the number of transpositions is odd the sign of the permutation is &minus.1, if it is even the sign is +1.]

## Determinant of Transpose

#{Lemma}: |A^T| = |A|

Proof:

|A| _ = _ sum{&zeta.(&pi.)prod{~a_{~i,&pi.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ }

_ _ _ _ = _ sum{&zeta.(&pi.)prod{~a^T_{&pi.(~i),~i},~i = 1,~n} , &pi. &in. P_{~n}, _ }

_ _ _ _ = _ sum{&zeta.(&pi.)prod{~a^T_{~j,&pi.^{&minus.1}( ~j)}, _ ~j = 1,~n} , &pi. &in. P_{~n}, _ } _ _ _ _ _ _ where ~j = &pi.(~i)

_ _ _ _ = _ sum{&zeta.(&sigma.)prod{~a^T_{~j ,&sigma.( ~j)}, _ ~j = 1,~n} , &sigma. &in. P_{~n}, _ } _ = _ |A^T|

where &sigma. = &pi.^{&minus.1}. _ Note that _ &zeta.(&sigma.) = &zeta.(&pi.).

## Elementary Row Operations

#{Lemma}: Let R_{*} represent the elementary row matrices , then

1. |R_{λ~s} A| _ = _ λ |A|
2. |R_{~s,~t} A| _ = _ &minus. |A|
3. |R_{~s+λ~t} A| _ = _ |A|

Proof:

1. If _ B = R_{λ~s} A _ then _ ~b_{~i ~j} = ~a_{~i ~j} _ except _ ~b_{~s ~j} = λ~a_{~s ~j}

|B| _ = _ sum{&zeta.(&pi.)prod{~b_{~i,&pi.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ }

_ _ _ _ = _ sum{&zeta.(&pi.) λprod{~a_{~i,&pi.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ } _ = _ λ |A|

(since one element is not equal, namely _ ~b_{~s,&pi.(~s)} = λ~a_{~s,&pi.(~s)} _ for any &pi..)

Corollary: If A has a zero row then |A| = 0
For suppose the ~s^{th} row of A is zero, then _ R_{λ~s} A = A _ for any λ so _ |A| = λ |A| _ &forall. λ _ &imply. _ |A| = 0.
_

2. If _ B = R_{~s,~t} A _ then _ ~b_{~i ~j} = ~a_{~i ~j} _ except _ ~b_{~s ~j} = ~a_{~t ~j} _ and _ ~b_{~t ~j} = ~a_{~s ~j}.
Let _ &sigma. = &pi. &tau._{~s,~t} _ (where &tau._{~s,~t} is the transpose of elements ~s and ~t), _ then _ ~b_{~i &pi.(~i)} = ~a_{~i &sigma.(~i)} _ &forall. ~i. _ So

|B| _ = _ sum{&zeta.(&pi.)prod{~a_{~i,&sigma.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ }

Now _ &zeta.(&pi.) = &minus.&zeta.(&sigma.) _ since &sigma. has one more transposition, so

|B| _ = _ &minus.sum{&zeta.(&sigma.)prod{~a_{~i,&sigma.(~i)},~i = 1,~n} , &sigma. &in. P_{~n}, _ } _ = _ &minus. |A|

_
3. If _ B = R_{~s+λ~t} A _ then _ ~b_{~i ~j} = ~a_{~i ~j} _ except _ ~b_{~s ~j} = ~a_{~s ~j} + λ ~a_{~t ~j}. _ So
_

prod{~b_{~i,&pi.(~i)},~i = 1,~n} _ = _ prod{~a_{~i,&pi.(~i)},~i = 1,~n} _ _ _ + _ _ _ λ ~a_{~t,&pi.(~s)} prod{~a_{~i,&pi.(~i)},~i ≠ ~s, _ }

|B| _ = _ |A| _ _ _ + _ _ _ λsum{&zeta.(&pi.) ~a_{~t,&pi.(~s)} prod{~a_{~i,&pi.(~i)},~i ≠ ~s, _ }, &pi. &in. P_{~n}, _ }

The sum in this expression can be rewritten as:

sum{&zeta.(&pi.) ~a_{~t,&pi.(~s)} ~a_{~t,&pi.(~t)} prod{~a_{~i,&pi.(~i)},~i &nin. \{{~s, ~t}\}, _ }, &pi. &in. P_{~n}, _ }

_ = _ sum{&zeta.(&pi.) ~a_{~t,&pi.(~s)} ~a_{~t,&pi.(~t)} prod{~a_{~i,&pi.(~i)},~i &nin. \{{~s, ~t}\}, _ }, &pi. &in. A_{~n}, _ }

_ _ _ _ + _ _ _ sum{&zeta.(&pi.) ~a_{~t,&pi.(~s)} ~a_{~t,&pi.(~t)} prod{~a_{~i,&pi.(~i)},~i &nin. \{{~s, ~t}\}, _ }, &pi. &in. S_{~n}, _ }

where A_{~n} is the set of even permutations (sign = +1), and S_{~n} is the set of odd permutations (sign = &minus.1).
Consider _ &sigma. = &pi.&comp.&tau._{~s,~t} _ where &tau._{~s,~t} is the transformation on the elements ~s and ~t. _
I.e. &tau._{~s,~t}(~i) = ~i , _ ~i &nin. \{~s, ~t\} ; _ &tau._{~s,~t}(~s) = ~t ; _ and _ &tau._{~s,~t}(~t) = ~s .
So _ &sigma.(~i) = &pi.(~i) , _ ~i &nin. \{~s, ~t\} ; _ &sigma.(~s) = &pi.(~t) ; _ and _ &sigma.(~t) = &pi.(~s). _ Also &zeta.(&sigma.) = &minus.&zeta.(&pi.), so &pi. &in. S_{~n} &imply. &sigma. &in. A_{~n} _ Substituting this in the second sum above, the expression becomes:

sum{&zeta.(&pi.) ~a_{~t,&pi.(~s)} ~a_{~t,&pi.(~t)} prod{~a_{~i,&pi.(~i)},~i &nin. \{{~s, ~t}\}, _ }, &pi. &in. A_{~n}, _ }

_ _ _ - _ _ _ sum{&zeta.(&sigma.) ~a_{~t,&sigma.(~t)} ~a_{~t,&sigma.(~s)} prod{~a_{~i,&sigma.(~i)},~i &nin. \{{~s, ~t}\}, _ }, &sigma. &in. A_{~n}, _ }

These two sums are the same, so their difference is zero. So |B| = |A| + (λ # 0) _ i.e. _ |B| = |A| .

## Full Rank Matrices

An ~n # ~n matrix A is said to be of #{~{full rank}} if rank A = &rho. = ~n.

If an ~n # ~n matrix A is not of full rank, i.e. rank A = &rho. &le. ~n, then |A| = 0.

Proof:
Suppose _ #{~r}_1^T ... #{~r}_{~n}^T _ are the rows of A, they are linearly dependent, _ i.e. &exist. λ_{~i} not all zero such that &sum._{~i} λ_{~i} #{~r}_{~i}^T = #0. _ Suppose λ_{~s} ≠ 0, _ then

sum{ fract{λ_{~i},λ_{~s}} #{~r}_{~i}^T ,~i = 1,~n} _ = _ #{~r}_{~s}^T + sum{ fract{λ_{~i},λ_{~s}} #{~r}_{~i}^T ,~i ≠ ~s, _ } _ = _ #0

Performing the corresponding row operations on A we obtain a matrix with a row of zeros. _ &therefore. its determinant is zero.

## Specific Determinants

### Diagonal Matrices

An ~n # ~n matrix A is #{~{diagonal}} if _ ~a_{~i ~j} = 0 _ when _ ~i ≠ ~j.

If A is diagonal then _ |A| = &prod._{~i} ~a_{~i ~i} .

Proof:

|A| _ = _ sum{&zeta.(&pi.)prod{~a_{~i,&pi.(~i)},~i = 1,~n} , &pi. &in. P_{~n}, _ }

But only the ~a_{~i ~i} are non-zero, so we only need consider the identity permutation &epsilon., where &epsilon.(~i) = ~i _ &forall. ~i. Note that &zeta.(&epsilon.) = 1, hence the result.

### Identity Matrix

As I_{~n} is a diagonal matrix we have _ |I_{~n}| = 1.

### Elementary Row Matrices

For the elementary row matrices we have:

1. |R_{λ~s}| _ = _ λ
2. |R_{~s,~t}| _ = _ &minus.1
3. |R_{~s+λ~t}| _ = _ 1

since R = R I_{~n} in all cases, and the result follows from the previous proposition.

So when R is an elementary row matrix _ |R A| = |R| |A|

### Scalar Product

|λA| = λ^{~n} |A|

λA is obtained from A by multiplying each row by λ, _ i.e. _ λA = R_{λ1} ... R_{λ~n} A _ so _ |λA| = |R_{λ1}| ... |R_{λ~n}| |A| = λ^{~n} |A|

## Product of Matrices

If A and B are both ~n # ~n matrices then _ |A B| _ = _ |A| |B| _ ( _ = _ |B A| )

Proof:

If rank(A) < ~n then |A| = 0, and _ rank(A B) _ &le. _ min\{rank(A), rank(B)\} _ = 0 _ &imply. _ |A B| = 0

If rank(A) = ~n then A is regular and A = R_1R_2 ... R_{~k} (elementary row matrices).
So |A B| = |R_1 ... R_{~k} B| = |R_1| ... |R_{~k}| |B| = |R_1 ... R_{~k}| |B| = |A| |B|

### Inverse Matrices

|A^{&minus.1}| = |A| ^{&minus.1}

|A^{&minus.1}| |A| = |A^{&minus.1}A| = |I_{~n}| = |A| = 1 _ &imply. _ |A^{&minus.1}| = |A| ^{&minus.1} .