Day 23

Trace and Determinant

Definition: If $A = (a_{i j})$ is a matrix in $M_{n} (F)$ , then the trace of $A$ is $Tr (A) = \sum_{i = 1}^{n} a_{i i} .$

Proposition: The trace is linear in the matrix $A$ . Also

Tr (A B C) = Tr (B C A) = Tr (C A B) .

In particular $Tr (A B) = Tr (B A)$ .

Proof:

Let $X = A B$ . Then

x_{i k} = \sum_{j = 1}^{n} a_{i j} b_{j k} .

Now if $Y = X C$ then

y_{r s} = \sum_{t = 1}^{n} x_{r t} c_{t s} = \sum_{t = 1}^{n} \sum_{j = 1}^{n} a_{r j} b_{j t} c_{t s}

and then $Tr (Y)$ is

Tr (A B C) = \sum_{r = 1}^{n} y_{r r} = \sum_{r = 1}^{n} \sum_{t = 1}^{n} \sum_{j = 1}^{n} a_{r j} b_{j t} c_{t r} .

On the other hand

Tr (B C A) = \sum_{j = 1}^{n} \sum_{r = 1}^{n} \sum_{t = 1}^{n} b_{j t} c_{t r} a_{r j}

which is just a rearrangement of the sum; and similarly for $Tr (C A B)$ .

Trace of a linear map:

This allows us to define the trace of a linear map as the trace of any matrix representing it – different matrices differ by conjugation by the change of basis matrix.

Notice also that trace is a conjugacy class invariant in ${GL}_{n} (F)$ . Two conjugate matrices have the same trace.

Multilinear Functions

Definition: A function $H : V_{1} \times V_{2} \times \dots \times V_{k} \to W$ is multilinear if it is linear as a function of each variable, with the other variables held fixed.

A function $H : \overset{n}{\overset{⏞}{V \times \dots \times V}} \to F$ is called an $n$ -multilinear form.

If $A = {a_{1}, \dots, a_{n}}$ is a basis for $V$ , then an n-multilinear form is determined by its values $H (x_{1}, \dots, x_{n})$ where each $x_{i}$ is chosen from $A$ . There are $n^{n}$ such values.

For example if $\dim V = 2$ with basis $a_{1}, a_{2}$ then

H (x_{11} a_{1} + x_{12} a_{2}, x_{21} a_{1} + x_{22} a_{2}) = x_{11} x_{21} f (a_{1}, a_{1}) + x_{11} x_{22} F (a_{1}, a_{2}) + x_{12} x_{21} F (a_{2}, a_{1}) + x_{12} x_{22} F (a_{2}, a_{2})

The “dot product” is a 2-linear form (a “bilinear” form) on $R^{n}$ or more generally on $F^{n}$ .

If we think of the trace as a function of the column vectors of a matrix, it is a multilinear form.

Symmetric and Alternating forms

A multilinear form $H : V^{n} \to F$ is called alternating if $H (v_{1}, \dots, v_{n}) = 0$ whenever two adjacent $v_{i}$ are equal to each other. It is called symmetric if $H (v_{1}, \dots, v_{n})$ stays the same under rearrangement of the $v_{i}$ .

The dot product is a symmetric bilinear form since $H (v, w) = H (w, v)$ .

Lemma: If $H$ is an alternating multilinear form, then $H (v_{1}, \dots, v_{n}) = 0$ whenever two of the $v_{i}$ coincide; and $H (v_{1}, \dots, v_{n})$ changes sign whenever two of the $v_{i}$ are interchanged. More generally,

H (v_{σ (1)}, v_{σ (2)}, \dots, v_{σ (n)}) = sgn (σ) H (v_{1}, \dots, v_{n})

where $σ$ is a permutation in $S_{n}$ and $sgn (σ)$ is the sign character.

Proof: Suppose $H_{i, i + 1} (x, y)$ is the function $H$ with fixed entries in all positions except $i$ and $i + 1$ . Then $H_{i, i + 1} (x + y, x + y) = 0$ by the alternating property; but $H_{i, i + 1} (x + y, x + y) = H_{i, i + 1} (x, x) + H_{i, i + 1} (x, y) + H_{i, i + 1} (y, x) + H_{i, i + 1} (y, y)$ by multilinearity. Since the outer terms are zero by the alternating property, we get $H_{i, i + 1} (x, y) = - H_{i, i + 1} (y, x) .$

Now if $H_{i, j} (x, y)$ is $H$ with all positions fixed except $i$ and $j$ , notice that we can progressively swap adjacent values until $y$ is in position $i + 1$ , changing signs each time. Therefore $H_{i, j} (x, y) = \pm H_{i, i + 1} (x, y)$ . In particular $H_{i, i + 1} (x, x) = 0 = H_{i, j} (x, x) .$

Therefore $H (v_{1}, \dots, v_{n}) = 0$ whenever any two of the $v_{i}$ coincide; and repeating the argument we used for adjacent entries we get that $H (v_{1}, \dots, v_{n})$ changes sign when we swap any two variables.

Since an arbitrary permutation is a product of transpositions, and the sign character is defined as $(- 1)^{k}$ where $k$ is the number of such transpositions, we get the formula for a general permutation.

Remark: Why not define alternating to mean $H (v_{1}, \dots, v_{n})$ changes sign if we swap adjacent entries? Look at characteristic two.

Corollary: If $H$ is alternating, and $w_{i} = v_{i}$ except that $w_{j} = v_{j} + a v_{k}$ for some $j$ , then $H (w_{1}, \dots, w_{n}) = H (v_{1}, \dots, v_{n})$ . Use linearity in the $j$ slot to see this.

An alternating multilinear form is defined by its values $H (a_{1}, \dots, a_{n})$ where the $a_{i}$ are chosen from a basis of $V$ , but these elements of $F$ have to satisfy the permutation property and must vanish if any basis elements are repeated.

If $V$ is $n$ -dimensional and $a_{1}, \dots, a_{n}$ is a basis, then an $n$ -multilinear form is determined by a single value $H (a_{1}, a_{2}, \dots, a_{n})$ and if $v_{i} = \sum x_{j i} a_{j}$ then $H (v_{1}, \dots, v_{n}) = H (a_{1}, \dots, a_{n}) \sum_{σ} sgn (σ) x_{σ (1) 1} x_{σ (2) 2} \dots x_{σ (n) n} .$

Definition: The determinant is the unique alternating multilinear map $det : M_{n} (F) \to F$ such that $det (I) = 1$ . Here $det$ is viewed as a function of the columns of a matrix $A$ .

Lemma: The determinant of $A$ and its transpose are the same.

Proposition: $det (A B) = det (A) det (B)$ .

Proof: Let $C = A B$ . Then the columns of $C$ are linear combinations of the columns of $A$ . In fact $C_{j} = \sum_{i = 1}^{n} b_{i j} A_{i}$ where $C_{j}$ and $A_{i}$ are the corresponding columns of $C$ and $A$ . So $det C = \sum (sgn (σ) b_{σ (1) 1} b_{σ (2) 2} \dots b_{σ (n) n}) det (A_{1}, \dots A_{n}) = det (B) det (A)$