Diagonalization of Matrices - Revision history

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2021-11-02T22:37:36Z

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Lila at 20:07, 8 October 2021

2021-10-08T20:07:52Z

Lila at 20:05, 8 October 2021

2021-10-08T20:05:26Z

Lila: Created page with "The prior subsection defines the relation of similarity and shows that, although similar matrices are necessarily matrix equivalent, the converse does not hold. Some matrix-eq..."

2021-10-08T20:02:06Z

Created page with "The prior subsection defines the relation of similarity and shows that, although similar matrices are necessarily matrix equivalent, the converse does not hold. Some matrix-eq..."

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The prior subsection defines the relation of similarity and shows that,
although similar matrices are necessarily matrix equivalent, the converse
does not hold.
Some matrix-equivalence classes break into two or more similarity
classes (the nonsingular <math>n \! \times \! n</math> matrices, for instance).
This means that the canonical form for matrix equivalence,
a block partial-identity, cannot be used as a canonical form
for matrix similarity because
the partial-identities cannot be in more than one
similarity class, so there are similarity classes without one.
This picture illustrates.
As earlier in this book, class representatives are shown with stars.
<center>
[[Image:Linalg_matrix_similarity_equiv_classes_2.png|x200px]]
</center>
We are developing a canonical form for representatives of
the similarity classes.
We naturally try to build on our previous work, meaning
first that the partial identity matrices should represent the similarity
classes into which they fall,
and beyond that, that the representatives should be as simple as possible.
The simplest extension of the partial-identity form is a diagonal form.

{{TextBox|1=
;Definition 2.1{{anchor|diagonalizable}}:
A transformation is '''diagonalizable''' if it has a diagonal representation with respect to the same basis for the codomain as for the domain. A '''diagonalizable matrix''' is one that is similar to a diagonal matrix: <math> T </math> is diagonalizable if there is a nonsingular <math> P </math> such that <math> PTP^{-1} </math> is diagonal.
}}

{{TextBox|1=
;Example 2.2{{anchor|ex:DiagTwoByTwo}}: 
The matrix

:<math>
\begin{pmatrix}
4 &-2 \\
1 &1
\end{pmatrix}
</math>

is diagonalizable.

:<math>
\begin{pmatrix}
2 &0 \\
0 &3
\end{pmatrix}
=
\begin{pmatrix}
-1 &2 \\
1 &-1
\end{pmatrix}
\begin{pmatrix}
4 &-2 \\
1 &1
\end{pmatrix}
\begin{pmatrix}
-1 &2 \\
1 &-1
\end{pmatrix}^{-1}
</math>
}}

{{TextBox|1=
;Example 2.3:
Not every matrix is diagonalizable.
The square of

:<math>
N=\begin{pmatrix}
0 &0 \\
1 &0
\end{pmatrix}
</math>

is the zero matrix. Thus, for any map <math>n</math> that <math> N </math> represents (with respect to the same basis for the domain as for the codomain), the composition <math> n\circ n </math> is the zero map. This implies that no such map <math> n </math> can be diagonally represented (with respect to any <math>B,B</math>) because no power of a nonzero diagonal matrix is zero. That is, there is no diagonal matrix in <math>N</math>'s similarity class.
}}

That example shows that a diagonal form will not do for a
canonical form— we cannot
find a diagonal matrix in each matrix similarity class.
However, the canonical form that we are developing has the property that if
a matrix can be diagonalized then the diagonal matrix is the canonical
representative of the similarity class.
The next result characterizes which maps can be diagonalized.

{{TextBox|1=
;Corollary 2.4{{anchor|cor:DiagIffBasisOfEigens}}:
A transformation <math> t </math> is diagonalizable if and only if there is a basis <math> B=\langle \vec{\beta}_1,\ldots,\vec{\beta}_n \rangle </math> and scalars <math> \lambda_1,\ldots,\lambda_n </math> such that <math> t(\vec{\beta}_i)=\lambda_i\vec{\beta}_i </math> for each <math> i </math>.
}}

{{TextBox|1=
;Proof:
This follows from the definition by
considering a diagonal representation matrix.

:<math>
{\rm Rep}_{B,B}(t)=
\left(\begin{array}{c|c|c}
\vdots & &\vdots \\
{\rm Rep}_{B}(t(\vec{\beta}_1)) &\cdots &{\rm Rep}_{B}(t(\vec{\beta}_n)) \\
\vdots & &\vdots
\end{array}\right)
=
\left(\begin{array}{c|c|c}
\lambda_1 & &0 \\
\vdots &\ddots &\vdots \\
0 & &\lambda_n
\end{array}\right)
</math>

This representation is equivalent to the existence of a basis satisfying the stated conditions simply by the definition of matrix representation.
}}

{{TextBox|1=
;Example 2.5{{anchor|ex:DiagUpperTrian}}:
To diagonalize

:<math>
T=\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
</math>

we take it as the representation of a transformation with respect to the
standard basis <math>T={\rm Rep}_{\mathcal{E}_2,\mathcal{E}_2}(t)</math> and we look for a basis
<math> B=\langle \vec{\beta}_1,\vec{\beta}_2 \rangle </math> such that

:<math>
{\rm Rep}_{B,B}(t)
=
\begin{pmatrix}
\lambda_1 &0 \\
0 &\lambda_2
\end{pmatrix}
</math>

that is, such that
<math>t(\vec{\beta}_1)=\lambda_1\vec{\beta}_1</math>
and <math>t(\vec{\beta}_2)=\lambda_2\vec{\beta}_2</math>.

:<math>
\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
\vec{\beta}_1=\lambda_1\cdot\vec{\beta}_1
\qquad
\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
\vec{\beta}_2=\lambda_2\cdot\vec{\beta}_2
</math>

We are looking for scalars <math> x </math> such that this equation

:<math>
\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}=x\cdot\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}
</math>

has solutions <math>b_1</math> and <math>b_2</math>, which are not both zero.
Rewrite that as a linear system.

:<math>
\begin{array}{*{2}{rc}r}
(3-x)\cdot b_1 &+ &2\cdot b_2 &= &0 \\
& &(1-x)\cdot b_2 &= &0
\end{array}
\qquad (*)</math>

In the bottom equation
the two numbers multiply to give zero only if
at least one of them is zero so there are two possibilities,
<math>b_2=0</math> and <math>x=1</math>.
In the <math> b_2=0 </math> possibility,
the first equation gives that either <math>b_1=0</math> or <math> x=3 </math>.
Since the case of both <math>b_1=0</math> and <math>b_2=0</math> is disallowed,
we are left looking at the possibility of <math>x=3</math>.
With it, the first equation in (<math>*</math>) is <math>0\cdot b_1+2\cdot b_2=0</math>
and so associated with <math>3</math> are vectors
with a second component of zero and a first component that is free.

:<math>
\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
\begin{pmatrix} b_1 \\ 0 \end{pmatrix}=3\cdot\begin{pmatrix} b_1 \\ 0 \end{pmatrix}
</math>

That is, one solution to (<math>*</math>) is <math>\lambda_1=3</math>, and we have a
first basis vector.

:<math>
\vec{\beta}_1=\begin{pmatrix} 1 \\ 0 \end{pmatrix}
</math>

In the <math>x=1</math> possibility,
the first equation in (<math>*</math>) is <math>2\cdot b_1+2\cdot b_2=0</math>, and so
associated with <math>1</math> are vectors whose
second component is the negative of their first component.

:<math>
\begin{pmatrix}
3 &2 \\
0 &1
\end{pmatrix}
\begin{pmatrix} b_1 \\ -b_1 \end{pmatrix}=1\cdot\begin{pmatrix} b_1 \\ -b_1 \end{pmatrix}
</math>

Thus, another solution is <math>\lambda_2=1</math> and
a second basis vector is this.

:<math>
\vec{\beta}_2=\begin{pmatrix} 1 \\ -1 \end{pmatrix}
</math>

To finish, drawing the similarity diagram

:[[Image:Linalg_matrix_equivalent_cd_3.png|x150px]]

and noting that the matrix <math>{\rm Rep}_{B,\mathcal{E}_2}(\mbox{id})</math> is easy
leads to this diagonalization.

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

In the next subsection, we will expand on that example by considering
more closely the property of [[#cor:DiagIffBasisOfEigens|Corollary 2.4]].
This includes seeing another way,
the way that we will routinely use, to find the <math>\lambda</math>'s.

==Resources==
* [https://en.wikibooks.org/wiki/Linear_Algebra/Diagonalizability Diagonalizability], Wikibooks: Linear Algebra

← Older revision		Revision as of 22:37, 2 November 2021
Line 254:		Line 254:
	==Resources==		==Resources==
	* [https://en.wikibooks.org/wiki/Linear_Algebra/Diagonalizability Diagonalizability], Wikibooks: Linear Algebra		* [https://en.wikibooks.org/wiki/Linear_Algebra/Diagonalizability Diagonalizability], Wikibooks: Linear Algebra
		+
		+	==Licensing==
		+	Content obtained and/or adapted from:
		+	* [https://en.wikibooks.org/wiki/Linear_Algebra/Diagonalizability Diagonalizability, Wikibooks: Linear Algebra] under a CC BY-SA license

@@ Line 83: / Line 83: @@
 The next result characterizes which maps can be diagonalized.
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Corollary 2.4{{anchor|cor:DiagIffBasisOfEigens}}:<!--\label{cor:DiagIffBasisOfEigens}-->
+'''Corollary 2.4''':
-A transformation <math> t </math> is diagonalizable if and only if there is a basis <math> B=\langle \vec{\beta}_1,\ldots,\vec{\beta}_n  \rangle  </math> and scalars <math> \lambda_1,\ldots,\lambda_n </math> such that <math> t(\vec{\beta}_i)=\lambda_i\vec{\beta}_i </math> for each <math> i </math>.
+:A transformation <math> t </math> is diagonalizable if and only if there is a basis <math> B=\langle \vec{\beta}_1,\ldots,\vec{\beta}_n  \rangle  </math> and scalars <math> \lambda_1,\ldots,\lambda_n </math> such that <math> t(\vec{\beta}_i)=\lambda_i\vec{\beta}_i </math> for each <math> i </math>.
-}}
+</blockquote>
-{{TextBox|1=
+Proof:
 This follows from the definition by
 considering a diagonal representation matrix.
@@ Line 109: / Line 108: @@
 This representation is equivalent to the existence of a basis satisfying the stated conditions simply by the definition of matrix representation.
-{{TextBox|1=
-;Example 2.5{{anchor|ex:DiagUpperTrian}}:<!--\label{ex:DiagUpperTrian}-->
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 To diagonalize
@@ Line 246: / Line 245: @@
 \end{pmatrix}
 </math>
-}}
+</blockquote>
 In the next subsection, we will expand on that example by considering

@@ Line 22: / Line 22: @@
 The simplest extension of the partial-identity form is a diagonal form.
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">'''Definition 2.1''':
-;Definition 2.1{{anchor|diagonalizable}}:
+: A transformation is '''diagonalizable''' if it has a diagonal representation with respect to the same basis for the codomain as for the domain. A '''diagonalizable matrix''' is one that is similar to a diagonal matrix: <math> T </math> is diagonalizable if there is a nonsingular <math> P </math> such that <math> PTP^{-1} </math> is diagonal.
-A transformation is '''diagonalizable''' if it has a diagonal representation with respect to the same basis for the codomain as for the domain. A '''diagonalizable matrix''' is one that is similar to a diagonal matrix: <math> T </math> is diagonalizable if there is a nonsingular <math> P </math> such that <math> PTP^{-1} </math> is diagonal.
+</blockquote>
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Example 2.2{{anchor|ex:DiagTwoByTwo}}: <!--\label{ex:DiagTwoByTwo}-->
+'''Example 2.2''':
 The matrix
@@ Line 59: / Line 58: @@
 \end{pmatrix}^{-1}
 </math>
-}}
+</blockquote>
-{{TextBox|1=
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
-;Example 2.3:
+'''Example 2.3''':
 Not every matrix is diagonalizable.
 The square of
@@ Line 74: / Line 73: @@
 is the zero matrix. Thus, for any map <math>n</math> that <math> N </math> represents (with respect to the same basis for the domain as for the codomain), the composition <math> n\circ n </math> is the zero map. This implies that no such map <math> n </math> can be diagonally represented (with respect to any <math>B,B</math>) because no power of a nonzero diagonal matrix is zero. That is, there is no diagonal matrix in <math>N</math>'s similarity class.
-}}
+</blockquote>
 That example shows that a diagonal form will not do for a