The Geometric Interpretation of the Determinant - Revision history

Khanh at 20:27, 5 November 2021

2021-11-05T20:27:39Z

Lila at 02:50, 11 October 2021

2021-10-11T02:50:07Z

Lila: Created page with "This parallelogram picture x200px is familiar from the construction of the sum of the two vectors. One way to compute the..."

2021-10-08T20:11:12Z

Created page with "This parallelogram picture <center> x200px </center> is familiar from the construction of the sum of the two vectors. One way to compute the..."

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This parallelogram picture
<center>
[[Image:Linalg_parallelogram.png|x200px]]
</center>
is familiar from the construction of the sum of the two vectors.
One way to compute the area that it encloses is to draw this rectangle
and subtract the area of each subregion.
<center>
[[Image:Linalg_parallelogram_area.png|x200px]]        
<math>\begin{array}{l}
\text{area of parallelogram} \\
\quad
=\text{area of rectangle}
-\text{area of }A-\text{area of }B \\
\qquad
-\cdots-\text{area of }F \\
\quad
=(x_1+x_2)(y_1+y_2)-x_2y_1-x_1y_1/2 \\
\qquad
-x_2y_2/2-x_2y_2/2-x_1y_1/2-x_2y_1 \\
\quad
=x_1y_2-x_2y_1
\end{array}</math>
</center>
The fact that the area equals the value of the determinant

:<math>
\begin{vmatrix}
x_1 &x_2 \\
y_1 &y_2
\end{vmatrix}
=x_1y_2-x_2y_1
</math>

is no coincidence.
{{anchor|size}}The properties in the definition of determinants
make reasonable postulates for a function
that measures the size of the region
enclosed by the vectors in the matrix.

For instance, this shows
the effect of multiplying one of the box-defining vectors by
a scalar (the scalar used is <math>k=1.4</math>).
<center>
[[Image:Linalg_parallelogram_2.png|x100px]]
        
[[Image:Linalg_parallelogram_3.png|x100px]]
</center>
The region formed by <math>k\vec{v}</math> and <math>\vec{w}</math>
is bigger, by a factor of <math> k </math>,
than the shaded region enclosed by <math>\vec{v}</math> and <math>\vec{w}</math>.
That is,
<math> \text{size}\, (k\vec{v},\vec{w})=k\cdot\text{size}\, (\vec{v},\vec{w}) </math> and
in general we expect of the size measure that
<math>\text{size}\, (\dots,k\vec{v},\dots)=k\cdot\text{size}\, (\dots,\vec{v},\dots)</math>.
Of course, this postulate is already familiar as
one of the properties in the defintion of determinants.

Another property of determinants is that
they are unaffected by pivoting.
Here are before-pivoting and
after-pivoting boxes (the scalar used is <math>k=0.35</math>).
<center>
[[Image:Linalg_parallelogram_4.png|x100px]]
    
[[Image:Linalg_parallelogram_5.png|x100px]]
</center>
Although the region on the right, the box formed by <math>v</math> and
<math>k\vec{v}+\vec{w}</math>, is
more slanted than the shaded region, the two have the
same base and the same height and hence the same area.
This illustrates that
<math> \text{size}\, (\vec{v},k\vec{v}+\vec{w})=\text{size}\, (\vec{v},\vec{w}) </math>.
Generalized,
<math>\text{size}\, (\dots,\vec{v},\dots,\vec{w},\dots)
=\text{size}\, (\dots,\vec{v},\dots,k\vec{v}+\vec{w},\dots)</math>,
which is a restatement of the determinant postulate.

Of course, this picture
<center>
[[Image:Linalg_parallelogram_basis.png|x100px]]
</center>
shows that <math> \text{size}\, (\vec{e}_1,\vec{e}_2)=1 </math>,
and we naturally extend that to any number of dimensions
<math>\text{size}\,(\vec{e}_1,\dots,\vec{e}_n)=1</math>,
which is a restatement of the property that the determinant of
the identity matrix is one.

With that, because property (2) of determinants is
redundant (as remarked right after the definition),
we have that all of the properties of determinants are reasonable
to expect of a function that gives the size of boxes.
We can now cite the work done in the prior section to show
that the determinant exists and is unique to be
assured that these postulates are consistent and sufficient
(we do not need any more postulates).
That is, we've got an intuitive
justification to interpret <math> \det(\vec{v}_1,\dots,\vec{v}_n) </math> as the
size of the box formed by the vectors.
(''Comment.''
An even more basic approach, which also leads to the definition
below, is in {{harv|Weston|1959}}.

{{TextBox|1=
;Example 1.1{{anchor|ex:VolParPiped}}: 
The volume of this parallelepiped, which can be found by the usual
formula from high school geometry, is <math>12</math>.
<center>
[[Image:Linalg_parallelepiped.png|x300px]]
        
<math>\begin{vmatrix}
2 &0 &-1\\
0 &3 &0 \\
2 &1 &1
\end{vmatrix}=12</math>
</center>
}}

{{TextBox|1=
;Remark 1.2:
Although property (2) of the definition of
determinants is redundant, it raises an important point.
Consider these two.
<center>
<TABLE border=0px cellpadding=15px>
<TR>
<TD>[[Image:Linalg_parallelogram_orientation_1.png|x200px]]
<TD>[[Image:Linalg_parallelogram_orientation_2.png|x200px]]
<TR>
<TD><math>\begin{vmatrix}
4 &1 \\
2 &3
\end{vmatrix}=10</math>
<TD><math>\begin{vmatrix}
1 &4 \\
3 &2
\end{vmatrix}=-10</math>
</TABLE>
</center>
The only difference between them is in the order in which the vectors
are taken.
If we take <math>\vec{u}</math> first and then go to <math>\vec{v}</math>,
follow the counterclockwise arc shown,
then the sign is positive.
Following a clockwise arc gives a negative sign.
{{anchor|orientation}}The sign returned by the size function reflects the
"orientation" or "sense" of the box.
(We see the same thing if we picture the effect of scalar multiplication
by a negative scalar.)

Although it is both interesting and important, the idea of orientation turns out to be tricky. It is not needed for the development below, and so we will pass it by. (See [[#exer:BasisOrient|Problem 20]].)
}}

{{TextBox|1=
;Definition 1.3{{anchor|box}}:
The '''box''' (or '''parallelepiped''') formed by <math> \langle
\vec{v}_1,\dots,\vec{v}_n \rangle </math> (where each vector is from
<math>\mathbb{R}^n</math>) includes all of the set <math>
\{t_1\vec{v}_1+\dots+t_n\vec{v}_n \,\big|\, t_1,\ldots,t_n\in [0..1]\}
</math>. The '''volume''' of a box is the absolute value of the
determinant of the matrix with those vectors as columns.}}

{{TextBox|1=
;Example 1.4:
Volume, because it is an absolute value,
does not depend on the order in which the vectors are given.
The volume of the parallelepiped in [[#ex:VolParPiped|Example 1.1]],
can also be computed as the absolute value of this determinant.

:<math>
\begin{vmatrix}
0 &2 &-1 \\
3 &0 &0 \\
1 &2 &1
\end{vmatrix}=-12
</math>
}}

The definition of volume gives a geometric interpretation to
something in the space, boxes made from vectors.
The next result relates the geometry to the functions that operate on
spaces.

{{TextBox|1=
;Theorem 1.5{{anchor|th:MatChVolByDetMat}}:
A transformation <math> t:\mathbb{R}^n\to \mathbb{R}^n </math> changes
the size of all boxes by the same factor, namely the size of the image
of a box <math>\left|t(S)\right|</math> is <math>\left|T\right|</math>
times the size of the box <math>\left|S\right|</math>, where
<math>T</math> is the matrix representing <math>t</math> with respect
to the standard basis. That is, for all <math>n \! \times \! n</math>
matrices, the determinant of a product is the product of the
determinants
<math>\left|TS\right|=\left|T\right|\cdot\left|S\right|</math>.
}}

The two sentences state the same idea, first in map terms and then
in matrix terms.
Although we tend to prefer a map point of view,
the second sentence, the matrix version, is more convienent for the proof
and is also the way that we shall use this result later.
(Alternate proofs are given as
[[#exer:DetProdEqProdDetsFcn|Problem 16]] and
[[#exer:DetProdEqProdDetsPerms|Problem 21]]].)

{{TextBox|1=
;Proof:
The two statements are equivalent because
<math>\left|t(S)\right|=\left|TS\right|</math>, as both give the size of the box that is the
image of the unit box <math>\mathcal{E}_n</math> under the composition <math>t\circ s</math>
(where <math>s</math> is the map represented by <math>S</math> with respect to the standard basis).

First consider the case that <math>\left|T\right|=0</math>.
A matrix has a zero determinant if and only if it is not invertible.
Observe that if <math> TS </math> is invertible, so that there
is an <math>M</math> such that <math> (TS)M=I </math>, then
the associative property of matrix multiplication
<math> T(SM)=I </math> shows that <math> T </math> is also invertible (with inverse <math>SM</math>).
Therefore, if <math> T </math> is not invertible then neither is <math> TS </math> —
if <math>\left|T\right|=0</math> then <math>\left|TS\right|=0</math>,
and the result holds in this case.

Now consider the case that <math>\left|T\right|\neq 0</math>, that <math>T</math> is nonsingular.
Recall that any nonsingular matrix can be factored into a product
of elementary matrices, so that <math>TS=E_1E_2\cdots E_rS</math>.
In the rest of this argument,
we will verify that if <math>E</math> is an elementary matrix then
<math> \left|ES\right|=\left|E\right|\cdot\left|S\right| </math>.
The result will follow because then
<math>\left|TS\right|=\left|E_1\cdots E_rS\right|=\left|E_1\right|\cdots\left|E_r\right|\cdot\left|S\right|
=\left|E_1\cdots E_r\right|\cdot\left|S\right|=\left|T\right|\cdot\left|S\right|</math>.

If the elementary matrix <math>E</math> is <math>M_i(k)</math> then <math>M_i(k)S</math> equals <math>S</math> except that row <math>i</math> has been multiplied by <math>k</math>. The third property of determinant functions then gives that <math>\left|M_i(k)S\right|=k\cdot\left|S\right|</math>. But <math>\left|M_i(k)\right|=k</math>, again by the third property because <math>M_i(k)</math> is derived from the identity by multiplication of row <math>i</math> by <math>k</math>, and so <math> \left|ES\right|=\left|E\right|\cdot\left|S\right| </math> holds for <math>E=M_i(k)</math>. The <math>E=P_{i,j}=-1</math> and <math>E=C_{i,j}(k)</math> checks are similar.
}}

{{TextBox|1=
;Example 1.6:
Application of the map <math>t</math> represented with respect to the standard
bases by

:<math>
\begin{pmatrix}
1 &1 \\
-2 &0
\end{pmatrix}
</math>

will double sizes of boxes, e.g., from this
<center>
[[Image:Linalg_parallelogram_doubled_1.png|x200px]]
        
<math>\begin{vmatrix}
2 &1 \\
1 &2
\end{vmatrix}=3</math>
</center>
to this
<center>
[[Image:Linalg_parallelogram_doubled_2.png|x300px]]
        
<math>\begin{vmatrix}
3 &3 \\
-4 &-2
\end{vmatrix}=6</math>
</center>
}}

{{TextBox|1=
;Corollary 1.7:
If a matrix is invertible then the determinant of its inverse is the inverse of its determinant <math>\left|T^{-1}\right|=1/\left|T\right|</math>.
}}

{{TextBox|1=
;Proof:
<math>1=\left|I\right|=\left|TT^{-1}\right|=\left|T\right|\cdot\left|T^{-1}\right|</math>
}}

Recall that determinants are not additive homomorphisms,
<math>\det(A+B)</math> need not equal <math>\det(A)+\det(B)</math>.
The above theorem says, in contrast, that determinants are
multiplicative homomorphisms:
<math>\det(AB)</math> does equal <math>\det(A)\cdot \det(B)</math>.

==Resources==
* [https://en.wikibooks.org/wiki/Linear_Algebra/Determinants_as_Size_Functions Determinants as Size Functions], Wikibooks: Linear Algebra

@@ Line 276: / Line 276: @@
 <math>\det(AB)</math> does equal <math>\det(A)\cdot \det(B)</math>.
-==Resources==
+== Licensing ==
-* [https://en.wikibooks.org/wiki/Linear_Algebra/Determinants_as_Size_Functions Determinants as Size Functions], Wikibooks: Linear Algebra
+Content obtained and/or adapted from:

@@ Line 34: / Line 34: @@
 is no coincidence.
-{{anchor|size}}The properties in the definition of determinants
 make reasonable postulates for a function
 that measures the size of the region
@@ Line 98: / Line 99: @@
 justification to interpret <math> \det(\vec{v}_1,\dots,\vec{v}_n) </math> as the
 size of the box formed by the vectors.
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Example 1.1{{anchor|ex:VolParPiped}}: <!--\label{ex:VolParPiped}-->
+'''Example 1.1:'''
 The volume of this parallelepiped, which can be found by the usual
 formula from high school geometry, is <math>12</math>.
@@ Line 115: / Line 113: @@
 \end{vmatrix}=12</math>
 </center>
-}}
+</blockquote>
-{{TextBox|1=
-;Remark 1.2:
+'''Remark 1.2:'''
 Although property (2) of the definition of
 determinants is redundant, it raises an important point.
@@ Line 144: / Line 142: @@
 then the sign is positive.
 Following a clockwise arc gives a negative sign.
-{{anchor|orientation}}The sign returned by the size function reflects the
 "orientation" or "sense" of the box.
 (We see the same thing if we picture the effect of scalar multiplication
 by a negative scalar.)
-Although it is both interesting and important, the idea of orientation turns out to be tricky. It is not needed for the development below, and so we will pass it by. (See [[#exer:BasisOrient|Problem 20]]<!--\ref{exer:BasisOrient}-->.)
+Although it is both interesting and important, the idea of orientation turns out to be tricky. It is not needed for the development below, and so we will pass it by.
-}}
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
+'''Definition 1.3:'''
 The '''box''' (or '''parallelepiped''') formed by <math> \langle
 \vec{v}_1,\dots,\vec{v}_n \rangle  </math> (where each vector is from
@@ Line 159: / Line 156: @@
 \{t_1\vec{v}_1+\dots+t_n\vec{v}_n \,\big|\, t_1,\ldots,t_n\in [0..1]\}
 </math>. The '''volume''' of a box is the absolute value of the
-determinant of the matrix with those vectors as columns.}}
+determinant of the matrix with those vectors as columns.
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Example 1.4:
+Example 1.4:
 Volume, because it is an absolute value,
 does not depend on the order in which the vectors are given.
-The volume of the parallelepiped in [[#ex:VolParPiped|Example 1.1]]<!--\ref{ex:VolParPiped}-->,
+The volume of the parallelepiped in can also be computed as the absolute value of this determinant.
 :<math>
@@ Line 175: / Line 172: @@
 \end{vmatrix}=-12
 </math>
-}}
+</blockquote>
 The definition of volume gives a geometric interpretation to
@@ Line 182: / Line 179: @@
 spaces.
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Theorem 1.5{{anchor|th:MatChVolByDetMat}}:<!--\label{th:MatChVolByDetMat}-->
 A transformation <math> t:\mathbb{R}^n\to \mathbb{R}^n </math> changes
 the size of all boxes by the same factor, namely the size of the image
@@ Line 193: / Line 191: @@
 determinants
 <math>\left|TS\right|=\left|T\right|\cdot\left|S\right|</math>.
-}}
+</blockquote>
 The two sentences state the same idea, first in map terms and then
@@ Line 200: / Line 198: @@
 the second sentence, the matrix version, is more convienent for the proof
 and is also the way that we shall use this result later.
-{{TextBox|1=
+'''Proof:'''
 The two statements are equivalent because
 <math>\left|t(S)\right|=\left|TS\right|</math>, as both give the size of the box that is the
@@ Line 232: / Line 226: @@
 If the elementary matrix <math>E</math> is <math>M_i(k)</math> then <math>M_i(k)S</math> equals <math>S</math> except that row <math>i</math> has been multiplied by <math>k</math>. The third property of determinant functions then gives that <math>\left|M_i(k)S\right|=k\cdot\left|S\right|</math>. But <math>\left|M_i(k)\right|=k</math>, again by the third property because <math>M_i(k)</math> is derived from the identity by multiplication of row <math>i</math> by <math>k</math>, and so <math> \left|ES\right|=\left|E\right|\cdot\left|S\right| </math> holds for <math>E=M_i(k)</math>. The <math>E=P_{i,j}=-1</math> and <math>E=C_{i,j}(k)</math> checks are similar.
-{{TextBox|1=
-;Example 1.6:
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 Application of the map <math>t</math> represented with respect to the standard
 bases by
@@ Line 264: / Line 258: @@
 \end{vmatrix}=6</math>
 </center>
-}}
+</blockquote>
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Corollary 1.7:
+'''Corollary 1.7:'''
 If a matrix is invertible then the determinant of its inverse is the inverse of its determinant <math>\left|T^{-1}\right|=1/\left|T\right|</math>.
-}}
+</blockquote>
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Proof:
+Proof:
 <math>1=\left|I\right|=\left|TT^{-1}\right|=\left|T\right|\cdot\left|T^{-1}\right|</math>
-}}
+</blockquote>
 Recall that determinants are not additive homomorphisms,