7—Operators and Matrices 159and the last term is zero. Translate this into components. Use the common notation for a determinant,
a square array with vertical bars, butforget that you know how to compute this symbol! I’m going
to use it simply as a notation by keep track of vector manipulations. The numerical value will come
out at the end as the computed value of a volume. ~vi=f
(
~ei
)
=
∑
jfji~ej, thenΛ
(
~v 1 ,~v 2 ,~v 3
)
=
Λ
(
~v 1 ,~v 2 +α~v 1 ,~v 3
)
=
∣∣
∣∣
∣
∣
f 11 f 12 +αf 11 f 13
f 21 f 22 +αf 21 f 23
f 31 f 32 +αf 31 f 33
∣∣
∣∣
∣
∣
=
∣∣
∣∣
∣
∣
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
∣∣
∣∣
∣
∣
+α
∣∣
∣∣
∣
∣
f 11 f 11 f 13
f 21 f 21 f 23
f 31 f 31 f 33
∣∣
∣∣
∣
∣
=
∣∣
∣∣
∣
∣
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
∣∣
∣∣
∣
∣
To evaluate this object, simply chooseα to make the elementf 12 +αf 11 = 0. Then repeat the
operation, adding a multiple of the first column to the third, making the elementf 13 +βf 11 = 0. This
operation doesn’t change the original value ofΛ
(
~v 1 ,~v 2 ,~v 3
)
.
Λ
(
~v 1 ,~v 2 +α~v 1 ,~v 3 +β~v 1
)
=
∣∣
∣
∣∣
∣
f 11 0 0
f 21 f 22 +αf 21 f 23 +βf 21
f 31 f 32 +αf 31 f 33 +βf 31
∣∣
∣
∣∣
∣
=
∣∣
∣
∣∣
∣
f 11 0 0
f 21 f 22 ′ f 23 ′
f 31 f 32 ′ f 33 ′
∣∣
∣
∣∣
∣
Repeat the process to eliminatef 23 ′ , addingγ~v 2 ′to the third argument, whereγ=−f 23 ′ /f 22 ′.
=
∣
∣
∣∣
∣∣
f 11 0 0
f 21 f 22 ′ f 23 ′
f 31 f 32 ′ f 33 ′
∣
∣
∣∣
∣∣=
∣
∣
∣∣
∣∣
f 11 0 0
f 21 f 22 ′ f 23 ′ +γf 22 ′
f 31 f 32 ′ f 33 ′ +γf 32 ′
∣
∣
∣∣
∣∣=
∣
∣
∣∣
∣∣
f 11 0 0
f 21 f 22 ′ 0
f 31 f 32 ′ f 33 ′′
∣
∣
∣∣
∣∣ (7.44)
Written in the last form, as a triangular array, the final result for the determinantdoes not depend
on the elementsf 21 ,f 31 ,f 32 ′. They may as well be zero. Why? Just do the same sort of column
operations, but working toward the left. Eliminatef 31 andf 32 ′ by adding a constant times the third
column to the first and second columns. Then eliminatef 21 by using the second column. You don’t
actually have to do this, you just have to recognize that it can be done so that you can ignore the lower
triangular part of the array.
Translate this back to the original vectors andΛis unchanged:
Λ(
~v 1 ,~v 2 ,~v 3
)
= Λ
(
f 11 ~e 1 ,f 22 ′ ~e 2 ,f 33 ′′~e 3
)
=f 11 f 22 ′ f 33 ′′Λ
(
~e 1 ,~e 2 ,~e 3
)
The volume of the original box isΛ(
~e 1 ,~e 2 ,~e 3
)
, so the quotient of the new volume to the old
one is
det =f 11 f 22 ′ f 33 ′′ (7.45)
The fact thatΛis unique up to a constant factor doesn’t matter. Do you want to measure volume
in cubic feet, cubic centimeters, or cubic light-years? This algorithm is called Gauss elimination.
It’s development started with the geometry and used vector manipulations to recover what you may
recognize from elsewhere as the traditional computed value of the determinant.
Did I leave anything out in this computation of the determinant? Yes, one point. What if
in Eq. (7.44) the numberf 22 ′ = 0? You can’t divide by it then. You can however interchange any
two arguments ofΛ, causing simply a sign change. If this contingency occurs then you need only
interchange the two columns to get a component of zero where you want it. Just keep count of such
switches whenever they occur.
Trace
There’s a property closely related to the determinant of an operator. It’s called the trace. If you have