Similar Matrices And Diagonalization
Similar Matrices and DiagonalizationLinear AlgebraMATH 2076Section 5.3Similarity & Diagonalization24 March 20171 / 10
Similar Matrices and Diagonalizable MatricesSection 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is,Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A PDP 1 .Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A PDP 1 .An n n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is,Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A PDP 1 .An n n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is, there is a basis { v1 , v2 , . . . , vn } for Rn such thateach vector vi is an eigenvector for A.Section 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A PDP 1 .An n n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is, there is a basis { v1 , v2 , . . . , vn } for Rn such thateach vector vi is an eigenvector for A. When this holds, say withA vi λi vi , we haveSection 5.3Similarity & Diagonalization24 March 20172 / 10
Similar Matrices and Diagonalizable MatricesTwo n n matrices A and B are similar if and only if there is an invertiblematrix P such that A PBP 1 (and then we also haveB P 1 AP QAQ 1 where Q P 1 ).An n n matrix A is diagonalizable if and only if it is similar to a diagonalmatrix; that is, there are a diagonal matrix D and an invertible matrix Psuch that A PDP 1 .An n n matrix A is diagonalizable if and only if there is an eigenbasisassoc’d with A; that is, there is a basis { v1 , v2 , . . . , vn } for Rn such thateach vector vi is an eigenvector for A. When this holds, say with A vi λi vi , we haveλ1 0 . . . 0 0 λ2 . . . 0 hi A PDP 1 where P v1 v2 . . . vnand D . . .0 0Section 5.3Similarity & Diagonalization0. . . λn24 March 20172 / 10
A 2 2 Example 1 2The matrix A has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 .2 1Section 5.3Similarity & Diagonalization24 March 20173 / 10
A 2 2 Example 1 2The matrix A has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 .2 1Therefore,A PDPSection 5.3 1 where 1 1P 2 1Similarity & Diagonalization 5 0and D .0 124 March 20173 / 10
A 2 2 Example 1 2The matrix A has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 .2 1Therefore,A PDP 1 where 1 1P 2 1 5 0and D .0 1But what does this mean?Section 5.3Similarity & Diagonalization24 March 20173 / 10
A 3 3 Example 4 1 0The matrix A 1 5 1 has simple eigenvalues 3, 4, 6 with0 1 4associated eigenvectorsSection 5.3Similarity & Diagonalization24 March 20174 / 10
A 3 3 Example 4 1 0The matrix A 1 5 1 has simple eigenvalues 3, 4, 6 with0 1 4 1 11associated eigenvectors v1 1 , v2 0 , v3 2 .111Section 5.3Similarity & Diagonalization24 March 20174 / 10
A 3 3 Example 4 1 0The matrix A 1 5 1 has simple eigenvalues 3, 4, 6 with0 1 4 1 11associated eigenvectors v1 1 , v2 0 , v3 2 .111Therefore,A PDP 1Section 5.3where 1 1 1P 1 0 2 1 11Similarity & Diagonalization 3 0 0and D 0 4 0 .0 0 624 March 20174 / 10
A 3 3 Example 4 1 0The matrix A 1 5 1 has simple eigenvalues 3, 4, 6 with0 1 4 1 11associated eigenvectors v1 1 , v2 0 , v3 2 .111Therefore,A PDP 1where 1 1 1P 1 0 2 1 11 3 0 0and D 0 4 0 .0 0 6But what does this mean?Section 5.3Similarity & Diagonalization24 March 20174 / 10
3 3 Matrices with Simple and Double EigenvaluesSection 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3with associated eigenvectorsSection 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1Section 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , soSection 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , so A2 is diagonalizable.Section 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , so A2 is diagonalizable. 5 6 0A3 1 2 0 has one simple eigenvalue 4 and one double4 6 1eigenvalue 1 with associated eigenvectorsSection 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , so A2 is diagonalizable. 5 6 0A3 1 2 0 has one simple eigenvalue 4 and one double4 6 1 60eigenvalue 1 with associated eigenvectors v1 1 , v2 0 .61Section 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , so A2 is diagonalizable. 5 6 0A3 1 2 0 has one simple eigenvalue 4 and one double4 6 1 60eigenvalue 1 with associated eigenvectors v1 1 , v2 0 .61There is no eigenbasis assoc’d with A3 , soSection 5.3Similarity & Diagonalization24 March 20175 / 10
3 3 Matrices with Simple and Double Eigenvalues 1 4 4A2 0 3 2 has one simple eigenvalue 5 and one double eigenvalue 20 2 3 210with associated eigenvectors v1 1 , v2 0 , v3 1 .10 1{ v1 , v2 , v3 } is an eigenbasis assoc’d with A2 , so A2 is diagonalizable. 5 6 0A3 1 2 0 has one simple eigenvalue 4 and one double4 6 1 60eigenvalue 1 with associated eigenvectors v1 1 , v2 0 .61There is no eigenbasis assoc’d with A3 , so A3 is not diagonalizable.Section 5.3Similarity & Diagonalization24 March 20175 / 10
Recall that A 1 2has simple eigenvalues λ1 5 and λ2 1 with4 3assoc’d eigenvectors
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 .2 1Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }.Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }.Recall that A Suppose x c1 v1 c2 v2 ; so
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look atA x Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5c1A x c2B
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5 0c15c1A x c20 1 c2B
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5 05 0 c15c1 x B .A x c20 1 c20 1B
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5c1A x c2BThus using B-coordinates, thediagonal matrix 5 05 0 c1 x B . 0 1 c20 1action of A is just multiplication by the
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5 05 0 c15c1 x B .A x c20 1 c20 1BThus using B-coordinates, the action of A is just multiplication by thediagonal matrix 5 0D 0 1
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5 05 0 c15c1 x B .A x c20 1 c20 1BThus using B-coordinates, the action of A is just multiplication by thediagonal matrix 5 0λ1 0D .0 10 λ2
1 2has simple eigenvalues λ1 5 and λ2 1 with4 3 11assoc’d eigenvectors v1 and v2 . Since A’s two eigenvectors2 1are LI, they form an eigenbasis B { v1 , v2 }. cSuppose x c1 v1 c2 v2 ; so x B 1 .c2Look at A x A c1 v1 c2 v2 c1 A v1 c2 A v2 5c1 v1 c2 v2Recall that A which says that h i5 05 0 c15c1 x B .A x c20 1 c20 1BThus using B-coordinates, the action of A is just multiplication by thediagonal matrix 5 0λ1 0D .0 10 λ2But, how do we get A x ?
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too.
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. x plane c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. c x B x P c x plane c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 c x B x P c v2 x plane c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1B { v1 , v2 } c x B x P c v2 x plane c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane cHere c x B 1 meansc2 c plane
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane c plane hi cHere c x B 1 means x c1 v1 c2 v2 v1 v2 c P x B ; soc2
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane c plane hi cHere c x B 1 means x c1 v1 c2 v2 v1 v2 c P x B ; so thec2B to S change of coordinates matrix P is given by
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1B { v1 , v2 } x P c c x B c x B x P c v2 x plane c plane hi cHere c x B 1 means x c1 v1 c2 v2 v1 v2 c P x B ; so thec2B to S change of coordinates matrix P is given byP PSB
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane c plane hi cHere c x B 1 means x c1 v1 c2 v2 v1 v2 c P x B ; so thec2B to S change of coordinates matrix P is given byhiP PSB v1 v2
Let’s draw a picture for the coordinate mapping x 7 x B , and also forthe inverse map too. v1 x P cB { v1 , v2 } c x B c x B x P c v2 x plane c plane hi cHere c x B 1 means x c1 v1 c2 v2 v1 v2 c P x B ; so thec2B to S change of coordinates matrix P is given by hi1 1.P PSB v1 v2 2 1
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B andw
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B and w B P 1 w w
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B and w B P 1 w wwhere the B to S change of coordinates matrix P is given by
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B and w B P 1 w wwhere the B to S change of coordinates matrix P is given byP PSB
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B and w B P 1 w wwhere the B to S change of coordinates matrix P is given byhiP PSB v1 v2
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v1 , v2 .2 1 in R2 we haveRecall that for any vector w P w B and w B P 1 w wwhere the B to S change of coordinates matrix P is given by hi1 1P PSB v1 v2 .2 1
From anslide: WTF A x and we know h earlieri 5 01 2A x D x B where D and A 0 14 3Band we have an eigenbasis assoc’d with A given by 11B { v1 , v2 } where v
Section 5.3 Similarity & Diagonalization 24 March 2017 2 / 10. Similar Matrices and Diagonalizable Matrices Two n n matrices A and B are similar if and only if there is an invertible matrix P such that A PBP 1
Diagonalization Math 240 Change of Basis Diagonalization Uses for diagonalization The change of basis matrix De nition Suppose V is a vector space with two bases B fv 1; 2;:::; n gand C w 1 2 n: The change of basis matrix from Bto Cis the matrix S [s ij], where v j s 1jw 1 s 2jw 2 s
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Joint diagonalization of square matrices is an important problem of numeric computation. Many applications make use of joint diagonalization techniques as their main algorithmic tool, for example c 2004 Andreas Zie
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Applications of Diagonalization Hsiu-Hau Lin hsiuhau@phys.nthu.edu.tw (Apr 12, 2010) The notes cover applications of matrix diagonalization (Boas 3.12). Quadratic curves Consider the quadratic curve, 5x2 4xy 2y2 30: (1) It can be casted into the matrix form and then brought into diagonal
From this point of view, it is natural to look for a diagonalization basis, namely, a basis of eigenvectors (eigen modes) for FN. In this regard, the main conceptual difficulty comes from the fact that the diagonalization problem is
Diagonalization Diagonalization problem: For a square matrix A, does there exist an invertible matrix P such that P-1AP is diagonal? Diagonalizable matrix: A square matrix A is called diagonalizable if there exists an invertible matrix P such that P-1AP is a diagonal matrix. Notes: (P diagonalizes
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