Difference between revisions of "The Inverse of a Linear Transformation"

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The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right.  Row reduce this matrix and the right half will be the inverse.  If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.
 
The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right.  Row reduce this matrix and the right half will be the inverse.  If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.
  
=== Example ===
+
=== Example 1===
 
Let <math> \mathrm{A} = \begin{bmatrix}
 
Let <math> \mathrm{A} = \begin{bmatrix}
 
   1 & 4 & 4\\
 
   1 & 4 & 4\\
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or right side only.
 
or right side only.
  
Example: Where  
+
===Definitions===
 +
Where  
 
<math> \pi:\mathbb{R}^3\to \mathbb{R}^2 </math> is the projection map  
 
<math> \pi:\mathbb{R}^3\to \mathbb{R}^2 </math> is the projection map  
  
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(An example of a function with no inverse on either side
 
(An example of a function with no inverse on either side
 
is the zero transformation on <math>\mathbb{R}^2</math>.)
 
is the zero transformation on <math>\mathbb{R}^2</math>.)
{{anchor|def:two sided inverse}}Some functions have a
 
'''two-sided inverse map''', another function
 
that is the inverse of the first, both from the left and from the right.
 
For instance, the map given by
 
<math>\vec{v}\mapsto 2\cdot \vec{v}</math> has the two-sided inverse
 
<math>\vec{v}\mapsto (1/2)\cdot\vec{v}</math>. 
 
In this subsection we will focus on two-sided inverses.
 
The appendix shows that a function
 
has a two-sided inverse if and only if it is both one-to-one and onto.
 
{{anchor|def:inverse function}}The appendix also shows that if a function <math>f</math> has a two-sided inverse then
 
it is unique, and so it is called
 
"the" inverse, and is denoted <math>f^{-1}</math>.
 
So our purpose in this subsection is, where a linear map <math>h</math> has an inverse,
 
to find the relationship between <math>{\rm Rep}_{B,D}(h)</math> and <math>{\rm Rep}_{D,B}(h^{-1})</math>
 
(recall that we have shown, in [[Linear Algebra/Rangespace and Nullspace#th:OOHomoEquivalence|Theorem II.2.21]]<!--\ref{th:OOHomoEquivalence}-->
 
of Section II of this chapter, that if a linear map has an inverse
 
then the inverse is a linear map also).
 
  
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+
Some functions have a '''two-sided inverse map''', another function that is the inverse of the first, both from the left and from the right. For instance, the map given by <math>\vec{v}\mapsto 2\cdot \vec{v}</math> has the two-sided inverse <math>\vec{v}\mapsto (1/2)\cdot\vec{v}</math>. 
;Definition 4.2{{anchor|def:inverse matrix}}:
+
In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. The appendix also shows that if a function <math>f</math> has a two-sided inverse then it is unique, and so it is called "the" inverse, and is denoted <math>f^{-1}</math>. So our purpose in this subsection is, where a linear map <math>h</math> has an inverse, to find the relationship between <math>{\rm Rep}_{B,D}(h)</math> and <math>{\rm Rep}_{D,B}(h^{-1})</math>.
 +
 
 
A matrix <math> G </math> is a '''left inverse matrix''' of the matrix <math> H </math> if <math> GH </math> is the identity matrix. It is a '''right inverse matrix''' if <math> HG </math> is the identity. A matrix <math>H</math> with a two-sided inverse is an '''invertible matrix'''. That two-sided inverse is called '''the inverse matrix''' and is denoted <math> H^{-1} </math>.
 
A matrix <math> G </math> is a '''left inverse matrix''' of the matrix <math> H </math> if <math> GH </math> is the identity matrix. It is a '''right inverse matrix''' if <math> HG </math> is the identity. A matrix <math>H</math> with a two-sided inverse is an '''invertible matrix'''. That two-sided inverse is called '''the inverse matrix''' and is denoted <math> H^{-1} </math>.
}}
 
  
Because of the correspondence between linear maps and matrices,
+
Because of the correspondence between linear maps and matrices, statements about map inverses translate into statements about matrix inverses.
statements about map inverses translate into statements about matrix inverses.
 
  
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+
===Lemmas and Theorems===
;Lemma 4.3{{anchor|le:LeftAndRightInvEqual}}:<!--\label{le:LeftAndRightInvEqual}-->
+
* If a matrix has both a left inverse and a right inverse then the two are equal.
If a matrix has both a left inverse and a right inverse then the two are equal.
 
}}
 
  
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+
* A matrix is invertible if and only if it is nonsingular.
;Theorem 4.4{{anchor|thm:invertable iff nonsingular}}:
+
** Proof: ''(For both results.)'' Given a matrix <math>H</math>, fix spaces of appropriate dimension for the domain and codomain. Fix bases for these spaces. With respect to these bases, <math>H</math> represents a map <math>h</math>. The statements are true about the map and therefore they are true about the matrix.
A matrix is invertible if and only if it is nonsingular.
 
}}
 
  
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* A product of invertible matrices is invertible&mdash; if <math> G </math> and <math> H </math> are invertible and if <math> GH </math> is defined then <math> GH </math> is invertible and <math> (GH)^{-1}=H^{-1}G^{-1} </math>.
;Proof:
+
** Proof: ''(This is just like the prior proof except that it requires two maps.)'' Fix appropriate spaces and bases and consider the represented maps <math> h </math> and <math> g </math>. Note that <math> h^{-1}g^{-1} </math> is a two-sided map inverse of <math> gh </math> since <math> (h^{-1}g^{-1})(gh) = h^{-1}(\mbox{id})h = h^{-1}h =\mbox{id} </math> and <math> (gh)(h^{-1}g^{-1}) = g(\mbox{id})g^{-1} = gg^{-1} = \mbox{id} </math>. This equality is reflected in the matrices representing the maps, as required.
''(For both results.)'' Given a matrix <math>H</math>, fix spaces of appropriate dimension for the domain and codomain. Fix bases for these spaces. With respect to these bases, <math>H</math> represents a map <math>h</math>. The statements are true about the map and therefore they are true about the matrix.
 
}}
 
  
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;Lemma 4.5{{anchor|lem:ProdInvIsInv}}: <!--\label{lem:ProdInvIsInv}-->
 
A product of invertible matrices is invertible&mdash; if <math> G </math> and <math> H </math> are invertible and if <math> GH </math> is defined then <math> GH </math> is invertible and <math> (GH)^{-1}=H^{-1}G^{-1} </math>.
 
}}
 
 
{{TextBox|1=
 
;Proof:
 
''(This is just like the prior proof except that it requires two maps.)''
 
Fix appropriate spaces and bases and consider the represented maps <math> h </math> and
 
<math> g </math>.
 
Note that <math> h^{-1}g^{-1} </math> is a two-sided map inverse of <math> gh </math> since
 
<math>
 
(h^{-1}g^{-1})(gh)
 
=
 
h^{-1}(\mbox{id})h
 
=h^{-1}h
 
=\mbox{id}
 
</math>
 
and
 
<math>
 
(gh)(h^{-1}g^{-1})
 
=
 
g(\mbox{id})g^{-1}
 
=gg^{-1}
 
=\mbox{id}
 
</math>.
 
This equality is reflected in the matrices representing the maps, as required.
 
}}
 
  
{{anchor|arrow diagram}}Here is the arrow diagram giving the relationship
+
Here is the arrow diagram giving the relationship
 
between map inverses and matrix inverses.  
 
between map inverses and matrix inverses.  
 
It is a special case
 
It is a special case
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\qquad\qquad (*)</math>
 
\qquad\qquad (*)</math>
  
By fixing spaces and bases (e.g., <math>\mathbb{R}^2,\mathbb{R}^2</math> and <math>\mathcal{E}_2,\mathcal{E}_2</math>),
+
By fixing spaces and bases (e.g., <math>\mathbb{R}^2, \mathbb{R}^2 </math> and <math>\mathcal{E}_2,\mathcal{E}_2</math>),
 
we take the matrix <math>H</math> to represent some map <math>h</math>.
 
we take the matrix <math>H</math> to represent some map <math>h</math>.
 
Then solving the system is the same as  
 
Then solving the system is the same as  
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to get <math>{\rm Rep}_{B}(\vec{x})</math>.
 
to get <math>{\rm Rep}_{B}(\vec{x})</math>.
  
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+
===Example 2===
;Example 4.6{{anchor|ex:InverseByLinSys}}:  <!--\label{ex:InverseByLinSys}-->
 
 
We can find a left inverse for the matrix just given
 
We can find a left inverse for the matrix just given
  
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=\begin{pmatrix} 5/3 \\ 4/3 \end{pmatrix}
 
=\begin{pmatrix} 5/3 \\ 4/3 \end{pmatrix}
 
</math>
 
</math>
}}
 
  
{{TextBox|1=
 
;Remark 4.7:
 
Why solve systems this way, when
 
Gauss' method takes less arithmetic
 
(this assertion can be made precise by counting the
 
number of arithmetic operations,
 
as computer algorithm designers do)?
 
Beyond its conceptual appeal of fitting into our program of
 
discovering how to represent the various map operations,
 
solving linear systems by using the matrix inverse has
 
at least two advantages.
 
  
First, once the work of finding an inverse has been done,  
+
====Remark====
solving a system with the
+
Why solve systems this way, when Gauss' method takes less arithmetic (this assertion can be made precise by counting the number of arithmetic operations, as computer algorithm designers do)? Beyond its conceptual appeal of fitting into our program of discovering how to represent the various map operations, solving linear systems by using the matrix inverse has at least two advantages.
same coefficients but different constants is easy and fast: if
+
 
we change the entries on the right of the system (<math>*</math>)  
+
First, once the work of finding an inverse has been done, solving a system with the same coefficients but different constants is easy and fast: if we change the entries on the right of the system (<math>*</math>) then we get a related problem
then we get a related problem
 
  
 
:<math>
 
:<math>
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usually used to find the inverse matrix.
 
usually used to find the inverse matrix.
  
{{TextBox|1=
+
* A matrix is invertible if and only if it can be written as the product of elementary reduction matrices. The inverse can be computed by applying to the identity matrix the same row steps, in the same order, as are used to Gauss-Jordan reduce the invertible matrix.  
;Lemma 4.8{{anchor|lem:ComputeInvMat}}: <!--\label{lem:ComputeInvMat}-->
+
** Proof: A matrix <math>H</math> is invertible if and only if it is nonsingular and thus Gauss-Jordan reduces to the identity. This reduction can be done with elementary matrices
A matrix is invertible if and only if it can be written as the product of elementary reduction matrices. The inverse can be computed by applying to the identity matrix the same row steps, in the same order, as are used to Gauss-Jordan reduce the invertible matrix.  
+
:::<math> R_{r}\cdot R_{r-1}\cdot\dots R_{1}\cdot H=I </math>.
}}
+
:::This equation gives the two halves of the result.
  
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+
:::First, elementary matrices are invertible and their inverses are also elementary. Applying <math>R_r^{-1}</math> to the left of both sides of that equation, then <math>R_{r-1}^{-1}</math>, etc., gives <math>H</math> as the product of elementary matrices <math>H=R_1^{-1}\cdots R_r^{-1}\cdot I</math> (the <math>I</math> is here to cover the trivial <math>r=0</math> case).
;Proof:
 
A matrix <math>H</math> is invertible if and only if it is nonsingular and thus
 
Gauss-Jordan reduces to the identity.
 
By [[Linear Algebra/Mechanics of Matrix Multiplication#cor:ReducViaMatrices|Corollary 3.22]]<!--\ref{cor:ReducViaMatrices}--> this reduction can
 
be done with elementary matrices  
 
<math> R_r\cdot R_{r-1}\dots R_1\cdot H=I </math>.
 
This equation gives the two halves of the result.
 
  
First, elementary matrices are invertible and their inverses are also
+
:::Second, matrix inverses are unique and so comparison of the above equation with <math>H^{-1}H=I</math> shows that <math>H^{-1}=R_r\cdot R_{r-1}\dots R_1\cdot I</math>. Therefore, applying <math>R_1</math> to the identity, followed by <math>R_2</math>, etc., yields the inverse of <math>H</math>.
elementary.
 
Applying <math>R_r^{-1}</math> to the left of both sides of that equation, then
 
<math>R_{r-1}^{-1}</math>, etc., gives <math>H</math> as the product of
 
elementary matrices <math>H=R_1^{-1}\cdots R_r^{-1}\cdot I</math>
 
(the <math>I</math> is here to cover the trivial <math>r=0</math> case).
 
  
Second, matrix inverses are unique and so comparison of the above equation with <math>H^{-1}H=I</math> shows that <math>H^{-1}=R_r\cdot R_{r-1}\dots R_1\cdot I</math>. Therefore, applying <math>R_1</math> to the identity, followed by <math>R_2</math>, etc., yields the inverse of <math>H</math>.
+
===Example 3===
}}
 
 
 
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;Example 4.9:
 
 
To find the inverse of
 
To find the inverse of
  
 
:<math>
 
:<math>
 
\begin{pmatrix}
 
\begin{pmatrix}
1  &1  \\
+
1  & 1  \\
2  &-1
+
2  & -1
 
\end{pmatrix}
 
\end{pmatrix}
 
</math>
 
</math>
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:<math>
 
:<math>
 
\begin{pmatrix}
 
\begin{pmatrix}
1  &1  \\
+
1  & 1  \\
2  &-1
+
2  & -1
 
\end{pmatrix}^{-1}
 
\end{pmatrix}^{-1}
 
=
 
=
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\end{pmatrix}
 
\end{pmatrix}
 
</math>
 
</math>
}}
 
  
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+
===Example 4===
;Example 4.10{{anchor|exam:ThreeByThreeMatInv}}: <!--\label{exam:ThreeByThreeMatInv}-->
 
 
This one happens to start with a row swap.
 
This one happens to start with a row swap.
  
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\end{array}
 
\end{array}
 
</math>
 
</math>
}}
 
  
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===Example 5===
;Example 4.11:
 
 
A non-invertible matrix is detected by the fact that the left half won't
 
A non-invertible matrix is detected by the fact that the left half won't
 
reduce to the identity.
 
reduce to the identity.
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\end{array}\right)
 
\end{array}\right)
 
</math>
 
</math>
}}
 
  
 
This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.
 
This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.
 
The <math>2 \! \times \! 2</math> case is handy.
 
The <math>2 \! \times \! 2</math> case is handy.
  
{{TextBox|1=
+
==Licensing==
;Corollary 4.12{{anchor|cor:TwoByTwoInv}}:<!--\label{cor:TwoByTwoInv}-->
+
Content obtained and/or adapted from:
The inverse for a <math> 2 \! \times \! 2 </math> matrix exists and equals
+
* [https://en.wikibooks.org/wiki/Linear_Algebra/Inverses Inverses, Wikibooks: Linear Algebra] under a CC BY-SA license
 
+
* [https://en.wikibooks.org/wiki/Linear_Algebra/The_Inverse_of_a_Matrix The Inverse of a Matrix, Wikibooks: Linear Algebra] under a CC BY-SA license
:<math>
 
\begin{pmatrix}
 
a  &b  \\
 
c  &d
 
\end{pmatrix}^{-1}
 
=
 
\frac{1}{ad-bc}
 
\begin{pmatrix}
 
d  &-b \\
 
-c  &a
 
\end{pmatrix}
 
</math>
 
 
 
if and only if <math> ad-bc\neq 0 </math>.
 
}}
 
{{TextBox|1=
 
;Proof:
 
This computation is [[#exer:TwoByTwoInv|Problem 10]]<!--\ref{exer:TwoByTwoInv}-->.
 
}}
 
 
 
We have seen here, as in the Mechanics of Matrix Multiplication subsection,
 
that we can exploit the correspondence between
 
linear maps and matrices.
 
So we can fruitfully study both maps and matrices, translating back and forth
 
to whichever helps us the most.
 
 
 
Over the entire four subsections of
 
this section we have developed an algebra system for matrices.
 
We can compare it with the familiar algebra system for the real numbers.
 
Here we are working not with numbers but with matrices.
 
We have matrix addition and subtraction operations,  
 
and they work in much the same
 
way as the real number operations, except that they only combine same-sized
 
matrices.
 
We also have a matrix multiplication operation
 
and an operation inverse to multiplication.
 
These are somewhat like the familiar real number operations
 
(associativity, and distributivity over addition, for example), but
 
there are differences (failure of commutativity, for example).
 
And, we have scalar multiplication, which is in some ways another extension
 
of real number multiplication.
 
This matrix system provides an example that algebra
 
systems other than the
 
elementary one can be interesting and useful.
 

Latest revision as of 07:28, 3 November 2021

Inverse of an n-by-n matrix

An n-by-n matrix A is the inverse of n-by-n matrix B (and B the inverse of A) if BA = AB = I, where I is an identity matrix.

The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right. Row reduce this matrix and the right half will be the inverse. If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.

Example 1

Let

We begin by expanding and partitioning A to include the identity matrix, and then proceed to row reduce A until we reach the identity matrix on the left-hand side.




The matrix is then the inverse of the original matrix A.


Inverse of a Linear Transformation

We now consider how to represent the inverse of a linear map. We start by recalling some facts about function inverses. Some functions have no inverse, or have an inverse on the left side or right side only.

Definitions

Where is the projection map

and is the embedding

the composition is the identity map on .

We say is a left inverse map of or, what is the same thing, that is a right inverse map of . However, composition in the other order doesn't give the identity map— here is a vector that is not sent to itself under .

In fact, the projection has no left inverse at all. For, if were to be a left inverse of then we would have

for all of the infinitely many 's. But no function can send a single argument to more than one value.

(An example of a function with no inverse on either side is the zero transformation on .)

Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right. For instance, the map given by has the two-sided inverse . In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. The appendix also shows that if a function has a two-sided inverse then it is unique, and so it is called "the" inverse, and is denoted . So our purpose in this subsection is, where a linear map has an inverse, to find the relationship between and .

A matrix is a left inverse matrix of the matrix if is the identity matrix. It is a right inverse matrix if is the identity. A matrix with a two-sided inverse is an invertible matrix. That two-sided inverse is called the inverse matrix and is denoted .

Because of the correspondence between linear maps and matrices, statements about map inverses translate into statements about matrix inverses.

Lemmas and Theorems

  • If a matrix has both a left inverse and a right inverse then the two are equal.
  • A matrix is invertible if and only if it is nonsingular.
    • Proof: (For both results.) Given a matrix , fix spaces of appropriate dimension for the domain and codomain. Fix bases for these spaces. With respect to these bases, represents a map . The statements are true about the map and therefore they are true about the matrix.
  • A product of invertible matrices is invertible— if and are invertible and if is defined then is invertible and .
    • Proof: (This is just like the prior proof except that it requires two maps.) Fix appropriate spaces and bases and consider the represented maps and . Note that is a two-sided map inverse of since and . This equality is reflected in the matrices representing the maps, as required.


Here is the arrow diagram giving the relationship between map inverses and matrix inverses. It is a special case of the diagram for function composition and matrix multiplication.

Linalg matinv arrow.png

Beyond its place in our general program of seeing how to represent map operations, another reason for our interest in inverses comes from solving linear systems. A linear system is equivalent to a matrix equation, as here.

By fixing spaces and bases (e.g., and ), we take the matrix to represent some map . Then solving the system is the same as asking: what domain vector is mapped by to the result ? If we could invert then we could solve the system by multiplying to get .

Example 2

We can find a left inverse for the matrix just given

by using Gauss' method to solve the resulting linear system.

Answer: , , , and . This matrix is actually the two-sided inverse of , as can easily be checked. With it we can solve the system () above by applying the inverse.


Remark

Why solve systems this way, when Gauss' method takes less arithmetic (this assertion can be made precise by counting the number of arithmetic operations, as computer algorithm designers do)? Beyond its conceptual appeal of fitting into our program of discovering how to represent the various map operations, solving linear systems by using the matrix inverse has at least two advantages.

First, once the work of finding an inverse has been done, solving a system with the same coefficients but different constants is easy and fast: if we change the entries on the right of the system () then we get a related problem

with a related solution method.

In applications, solving many systems having the same matrix of coefficients is common.

Another advantage of inverses is that we can explore a system's sensitivity to changes in the constants. For example, tweaking the on the right of the system () to

can be solved with the inverse.

to show that changes by of the tweak while moves by of that tweak. This sort of analysis is used, for example, to decide how accurately data must be specified in a linear model to ensure that the solution has a desired accuracy. }}

We finish by describing the computational procedure usually used to find the inverse matrix.

  • A matrix is invertible if and only if it can be written as the product of elementary reduction matrices. The inverse can be computed by applying to the identity matrix the same row steps, in the same order, as are used to Gauss-Jordan reduce the invertible matrix.
    • Proof: A matrix is invertible if and only if it is nonsingular and thus Gauss-Jordan reduces to the identity. This reduction can be done with elementary matrices
.
This equation gives the two halves of the result.
First, elementary matrices are invertible and their inverses are also elementary. Applying to the left of both sides of that equation, then , etc., gives as the product of elementary matrices (the is here to cover the trivial case).
Second, matrix inverses are unique and so comparison of the above equation with shows that . Therefore, applying to the identity, followed by , etc., yields the inverse of .

Example 3

To find the inverse of

we do Gauss-Jordan reduction, meanwhile performing the same operations on the identity. For clerical convenience we write the matrix and the identity side-by-side, and do the reduction steps together.

This calculation has found the inverse.

Example 4

This one happens to start with a row swap.

Example 5

A non-invertible matrix is detected by the fact that the left half won't reduce to the identity.

This procedure will find the inverse of a general matrix. The case is handy.

Licensing

Content obtained and/or adapted from: