三角形式

It is now quite easy to prove the easiest one of the so-called canonical form theorems. Our assumption about the scalar field (namely, that it is algebraically closed) is still in force.

Theorem 1. If $A$ is any linear transformation on an $n$ -dimensional vector space $𝒱$ , then there exist $n + 1$ subspaces $ℳ_{0}, ℳ_{1}, \dots, ℳ_{n - 1}, ℳ_{n}$ with the following properties:

each $ℳ_{j}$ ( $j = 0, 1, \dots, n - 1, n$ ) is invariant under $A$ ,
the dimension of $ℳ_{j}$ is $j$ ,
( $𝒪 =$ ) $ℳ_{0} \subset ℳ_{1} \subset \dots \subset ℳ_{n - 1} \subset ℳ_{n}$ ( $= 𝒱$ ).

Proof. If $n = 0$ or $n = 1$ , the result is trivial; we proceed by induction, assuming that the statement is correct for $n - 1$ . Consider the dual transformation on ; since it has at least one proper vector, say , there exists a one-dimensional subspace $ℳ$ invariant under it, namely, the set of all multiples of . Let us denote by $ℳ_{n - 1}$ the annihilator (in ) of $ℳ$ , $ℳ_{n - 1} = ℳ^{0}$ ; then $ℳ_{n - 1}$ is an $(n - 1)$ -dimensional subspace of $𝒱$ , and $ℳ_{n - 1}$ is invariant under $A$ . Consequently we may consider $A$ as a linear transformation on $ℳ_{n - 1}$ alone, and we may find $ℳ_{0}, ℳ_{1}, \dots, ℳ_{n - 2}$ , $ℳ_{n - 1}$ , satisfying the conditions (i), (ii), (iii). We write $ℳ_{n} = 𝒱$ , and we are done. ◻

The chief interest of this theorem comes from its matricial interpretation. Since $ℳ_{1}$ is one-dimensional, we may find in it a vector $x_{1} \neq 0$ . Since $ℳ_{1} \subset ℳ_{2}$ , it follows that $x_{1}$ is also in $ℳ_{2}$ , and since $𝒩_{2}$ is two-dimensional, we may find in it a vector $x_{2}$ such that $x_{1}$ and $x_{2}$ span $ℳ_{2}$ . We proceed in this way by induction, choosing vectors $x_{j}$ so that $x_{1}, \dots, x_{j}$ lie in $ℳ_{j}$ and span $ℳ_{j}$ for $j = 1, \dots, n$ . We obtain finally a basis $𝒳 = {x_{1}, \dots, x_{n}}$ in $𝒱$ ; let us compute the matrix of $A$ in this coordinate system. Since $x_{j}$ is in $ℳ_{j}$ and since $ℳ_{j}$ is invariant under $A$ , it follows that $A x_{j}$ must be a linear combination of $x_{1}, \dots, x_{j}$ . Hence in the expression $A x_{j} = \sum_{i} α_{i j} x_{i}$ the coefficient of $x_{i}$ must vanish whenever $i > j$ ; in other words, $i > j$ implies $α_{i j} = 0$ . Hence the matrix of $A$ has the triangular form It is clear from this representation that $\det (A - α_{i i}) = 0$ for $i = 1, \dots, n$ , so that the $α_{i i}$ are the proper values of $A$ , appearing on the main diagonal of $[A]$ with the proper multiplicities. We sum up as follows.

Theorem 2. If $A$ is a linear transformation on an $n$ -dimensional vector space $𝒱$ , then there exists a basis $𝒳$ in $𝒱$ such that the matrix $[A; X]$ is triangular; or, equivalently, if $[A]$ is any matrix, there exists a non-singular matrix $[B]$ such that $[B]^{- 1} [A] [B]$ is triangular.

The triangular form is useful for proving many results about linear transformations. It follows from it, for example, that for any polynomial $p$ , the proper values of $p (A)$ , including their algebraic multiplicities, are precisely the numbers $p (λ)$ , where $λ$ runs through the proper values of $A$ .

A large part of the theory of linear transformations is devoted to improving the triangularization result just obtained. The best thing a matrix can be is not triangular but diagonal (that is, $α_{i j} = 0$ unless $i = j$ ); if a linear transformation is such that its matrix with respect to a suitable coordinate system is diagonal we shall call the transformation diagonable .

EXERCISES

Exercise 1. Interpret the following matrices as linear transformations on $ℂ^{2}$ and, in each case, find a basis of $ℂ^{2}$ such that the matrix of the transformation with respect to that basis is triangular.

$[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ .
$[\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}]$ .
$[\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}]$ .
$[\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]$ .
$[\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ .
$[\begin{matrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ .

Exercise 2. Two commutative linear transformations on a finite-dimensional vector space $𝒱$ over an algebraically closed field can be simultaneously triangularized. In other words, if $A B = B A$ , then there exists a basis $𝒳$ such that both $[A; 𝒳]$ and $[B; 𝒳]$ are triangular. (Hint: to imitate the proof in Section: Triangular form , it is desirable to find a subspace $ℳ$ of $𝒱$ invariant under both $A$ and $B$ . With this in mind, consider any proper value $λ$ of $A$ and examine the set of all solutions of $A x = λ x$ for the role of $ℳ$ .)

Exercise 3. Formulate and prove the analogues of the results of Section: Triangular form for triangular matrices below the diagonal (instead of above it).

Exercise 4. Suppose that $A$ is a linear transformation over an $n$ -dimensional vector space. For every alternating $n$ -linear form $w$ , write $\overset{―}{A} w$ for the function defined by

Since $\overset{―}{A} w$ is an alternating $n$ -linear form, and, in fact, $\overset{―}{A}$ is a linear transformation on the (one-dimensional) space of such forms, it follows that $\overset{―}{A} w = τ (A) \cdot w$ , where $τ (A)$ is a scalar.

$τ (0) = 0$ .
$τ (1) = n$ .
$τ (A + B) = τ (A) + τ (B)$ .
$τ (α A) = α τ (A)$ .
If the scalar field has characteristic zero and if $A$ is a projection, then $τ (A) = ρ (A)$ .
If $(α_{i j})$ is the matrix of $A$ in some coordinate system, then $τ (A) = \sum_{i} α_{i i}$ .
.
$τ (A B) = τ (B A)$ .
For which permutations $π$ of the integers $1, \dots, k$ is it true that $τ (A_{1} \dots A_{k}) = τ (A_{π (1)} \dots A_{π (k)})$ for all $k$ -tuples $(A_{1}, \dots, A_{k})$ of linear transformations?
If the field of scalars is algebraically closed, then $τ (A) = tr A$ . (For this reason trace is usually defined to be $τ$ ; the most popular procedure is to use (f) as the definition.)

Exercise 5.

Suppose that the scalar field has characteristic zero. Prove that if $E_{1}, \dots, E_{k}$ and $E_{1} + \dots + E_{k}$ are projections, then $E_{i} E_{j} = 0$ whenever $i \neq j$ . (Hint: from the fact that $tr (E_{1} + \dots + E_{k}) = tr (E_{1}) + \dots + tr (E_{k})$ conclude that the range of $E_{1} + \dots + E_{k}$ is the direct sum of the ranges of $E_{1}, \dots, E_{k}$ .)
If $A_{1}, \dots, A_{k}$ are linear transformations on an $n$ -dimensional vector space, and if $A_{1} + \dots + A_{k} = 1$ and $ρ (A_{1}) + \dots + ρ (A_{k}) \leq n$ , then each $A_{i}$ is a projection and $A_{i} A_{j} = 0$ whenever $i \neq j$ . (Start with $k = 2$ and proceed by induction; use a direct sum argument as in (a).)

Exercise 6.

If $A$ is a linear transformation on a finite-dimensional vector space over a field of characteristic zero, and if $tr A = 0$ , then there exists a basis $𝒳$ such that if $[A; 𝒳] = (α_{i j})$ , then $α_{i i} = 0$ for all $i$ . (Hint: using the fact that $A$ is not a scalar, prove first that there exists a vector $x$ such that $x$ and $A x$ are linearly independent. This proves that $α_{11}$ can be made to vanish; proceed by induction.)
Show that if the characteristic is not zero, the conclusion of (a) is false. (Hint: if the characteristic is $2$ , compute $B C - C B$ , where $B = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ and $C = [\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}]$ .)