Jordan Form of a Square Matrix

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1 Jordan Form of a Square Matrix Josh Engwer Texas Tech University June 3 KEY CONCEPTS & DEFINITIONS: R Set of all real numbers C Set of all complex numbers = {a + bi : a b R and i = } Field F = R or C Row reduction of a matrix : Gaussian Elimination (Gauss-Jordan Elimination is superfluous) Pivots of a row-reduced matrix are boxed. Row-swapping may be helpful sometimes but not necessary. Kernel of matrix A : ker A {v F : Av = } Nullspace of matrix A Kernel of matrix A Rank of matrix A : rank A # of pivots in A Eigenvalue λ with eigenvector v of square matrix A Av = λv Determinant of square matrix A : det(a) easiest to find using minors & cofactors Characteristic Equation of square matrix A : det(a λi) = (used to find eigenvalues of A) The j th -order Eigenspace of eigenvalue λ i : E j (λ i ) ker(a λ i I) j Algebraic Multiplicity of eigenvalue λ: AM[λ] multiplicity of λ in the characteristic equation Geometric Multiplicity of eigenvalue λ: GM[λ] # of independent eigenvectors for λ dim[e (λ)] i th Eigenchain corresponding to eigenvalue λ j : K (i) j THEOREMS: Rank-Nullity Theorem: If A is an mxn matrix then rank A + dim(ker A) = # of columns of A = n Diagonalization Theorem: If A is an nxn matrix with n distinct eigenvalues then A is diagonalizable. Jordan Form Theorem: Every nxn matrix can be written in Jordan Canonical Form: A = P JP Copyright Josh Engwer

2 EXAMPLE : Find the Jordan Form of A = STEP : Find all eigenvalues Characteristic equation : det(a λi) = λ 3 λ λ + = (λ + )(λ ) = Hence the eigenvalues of A are λ = { } {λ λ } AM[λ ] = and AM[λ ] =. STEP : Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): ker(a + I) = ker 4 6 = ker 3 = span span{r } Since = dim[e (λ )] = AM[λ ] = r is the one and only eigenvector for λ. STEP 3: Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): ker(a I) = ker 4 4 = ker = span span{v A} 5 Since = dim[e (λ )] < AM[λ ] = investigate E (λ ): STEP 4: Find nd -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I) : 4 8 ker(a I) = ker = ker 3 3 = span 36 4 Since = dim[e (λ )] = AM[λ ] = longest eigenchain length = Order[E (λ )] =. Since AM[λ ] = and longest eigenchain length = there must be eigenchain: K () Build eigenchain K () = {v w } from highest order eigenvector to lowest: w v STEP 5: Determine and pick one w {w A w B } s.t. w / span{v A }: v A w A w B = We need vector independence we must pick w = w A STEP 6: Determine v : v = (A λ I)w = = STEP 7: Build Matrix P consistent with eigenchain K () = {v w } and find Matrix P : P = [r K () ] = r v w = P = STEP 8: Create Jordan Matrix J consistent with P such that A = P JP : λ J = λ = Notice there are Jordan blocks. λ A = P JP : = span{w A w B }

3 EXAMPLE : Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues Characteristic equation : det(a λi) = λ 4 5λ 3 + 9λ 5λ = (λ ) 3 (λ ) = Hence the eigenvalues of A are λ = { } {λ λ } AM[λ ] = 3 and AM[λ ] =. STEP : Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): ker(a I) = ker = ker = span Since = dim[e (λ )] = AM[λ ] = r is the one and only eigenvector for λ. STEP 3: Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): ker(a I) = ker = ker = span Since = dim[e (λ )] < AM[λ ] = 3 investigate E (λ ): STEP 4: Find nd -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I) : 3 3 ker(a I) = ker = ker = span 5 3 Since = dim[e (λ )] < AM[λ ] = 3 investigate E 3 (λ ): span{r } span{v A } span{w A w B } STEP 5: Find the 3 rd -order eigenspace to eigenvalue λ E 3 (λ ) := ker(a λ I) 3 : ker = ker = span span{x A x B x C } 3 3 Since 3 = dim[e 3 (λ )] = AM[λ ] = 3 longest eigenchain length = Order[E 3 (λ )] = 3. Since AM[λ ] = 3 and longest eigenchain length = 3 there must be eigenchain: K () Eigenchain K () has length 3. Copyright Josh Engwer 3

4 PART II: BUILD EIGENCHAINS Build eigenchain K () = {v w x } from highest order eigenvector to lowest: x w v STEP 6: Determine and pick one x {x A x B x C } s.t. x / span{w A w B }: w A w B x A x B x C We need vector independence we can pick either x B or x C. Here we ll pick x = x B = STEP 7: Determine w : w = (A λ I)x = STEP 8: Determine v : v = (A λ I)w = PART III: BUILD MATRICES = = STEP 9: Build Matrix P consistent with eigenchain K () = {v w x } and find Matrix P : P = [r K () ] = r v w x = P = STEP : Build Jordan Matrix J consistent with P such that A = P JP : λ J = λ λ = Notice there are Jordan blocks. λ A = P JP : = Copyright Josh Engwer 4

5 EXAMPLE 3: Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues Characteristic equation : det(a λi) = λ 4 λ λ 8λ + 8 = (λ 3) 4 = Hence the eigenvalues of A are λ = {3} {λ 3 } AM[λ 3 ] = 4. STEP : Find st -order eigenspace to eigenvalue λ 3 E (λ 3 ) := ker(a λ 3 I): ker(a 3I) = ker = ker = span span{v A } Since = dim[e (λ 3 )] < AM[λ 3 ] = 4 investigate E (λ 3 ): STEP 3: Find nd -order eigenspace to eigenvalue λ 3 E (λ 3 ) := ker(a λ 3 I) : ker(a 3I) = ker = ker = span span{w A w B } Since = dim[e (λ 3 )] < AM[λ 3 ] = 4 investigate E 3 (λ 3 ): STEP 4: Find the 3 rd -order eigenspace to eigenvalue λ 3 E 3 (λ 3 ) := ker(a λ 3 I) 3 : ker = ker = span span{x A x B x C } Since 3 = dim[e 3 (λ 3 )] < AM[λ 3 ] = 4 investigate E 4 (λ 3 ): STEP 5: Find the 4 th -order eigenspace to eigenvalue λ 3 E 4 (λ 3 ) := ker(a λ 3 I) 4 : ker(a 3I) 4 = ker = span span{y A y B y C y D } Since 4 = dim[e 4 (λ 3 )] = AM[λ 3 ] = 4 longest eigenchain length = Order[E 4 (λ 3 )] = 4. Since AM[λ 3 ] = 4 and longest eigenchain length = 4 there must be eigenchain: K () 3 Eigenchain K () 3 has length 4. Copyright Josh Engwer 5

6 PART II: BUILD EIGENCHAINS Build eigenchain K () 3 = {v 3 w 3 x 3 y 3 } from highest order eigenvector to lowest: y 3 x 3 w 3 v 3 STEP 6: Determine and pick one y 3 {y A y B y C y D } s.t. y 3 / span{x A x B x C }: x A x B x C y A y B y C y D = We need vector independence we must pick y 3 = y D = STEP 7: Determine x 3 : x 3 = (A λ 3 I)y 3 = STEP 8: Determine w 3 : w 3 = (A λ 3 I)x 3 = STEP 9: Determine v 3 : v 3 = (A λ 3 I)w 3 = PART III: BUILD MATRICES = = = STEP : Build Matrix P consistent with eigenchain K () 3 = {v 3 w 3 x 3 y 3 } and find Matrix P : P = [K () 3 ] = v 3 w 3 x 3 y 3 = P =. STEP : Build Jordan Matrix J consistent with P such that A = P JP : λ 3 3 J = λ 3 λ 3 = 3 3 Notice there is only Jordan block. λ 3 3 A = P JP : = Copyright Josh Engwer 6

7 EXAMPLE 4: Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues Characteristic equation : det(a λi) = (λ ) 4 = Hence the eigenvalues of A are λ = {} {λ } AM[λ ] = 4. STEP : Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): 4 ker(a I) = ker 4 = span 4 span{v A v B } Since = dim[e (λ )] < AM[λ ] = 4 investigate E (λ ): STEP 3: Find nd -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I) : ker(a I) = ker = span span{w A w B w C w D } Since 4 = dim[e (λ )] = AM[λ ] = 4 longest eigenchain length = Order[E (λ )] =. Since AM[λ ] = 4 and longest eigenchain length = there must be eigenchains: K () K (). Eigenchain K () has length. Eigenchain K () has length. Copyright Josh Engwer 7

8 PART II: BUILD EIGENCHAINS Build st eigenchain K () = {v () w () } from highest order eigenvector to lowest: w () v () STEP 4: Determine and pick one w () {w A w B w C w D } s.t. w () / span{v A v B }: v A v B w A w B w C w D We need vector independence we can pick w () from {w B w C }. Let s pick w () = w B = STEP 5: Determine v () : v () = (A λ I)w () = = 7 K() = Build nd eigenchain K () = {v () w () } from highest order eigenvector to lowest: w () v () STEP 6: Determine and pick w () {w B w C } s.t. w () w () : So we must pick w () = w C = STEP 7: Determine v () : v () = (A λ I)w () = PART III: BUILD MATRICES = 4 K() = STEP 8: Build Matrix P consistent with eigenchains K () K () and find Matrix P : P = [K () K () ] = v() w () v () w () = 7 4 P = STEP 9: Build Jordan Matrix J consistent with P such that A = P JP : λ J = λ λ = Notice there are Jordan blocks. λ A = P JP : =

9 EXAMPLE 5: Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues Characteristic equation : det(a λi) = (λ 5) 4 = Hence the eigenvalues of A are λ = {5} {λ 5 } AM[λ 5 ] = 4. STEP : Find st -order eigenspace to eigenvalue λ 5 E (λ 5 ) := ker(a λ 5 I): ker(a 5I) = ker = span span{v A v B } Since = dim[e (λ 5 )] < AM[λ 5 ] = 4 investigate E (λ 5 ): STEP 3: Find nd -order eigenspace to eigenvalue λ 5 E (λ 5 ) := ker(a λ 5 I) : ker(a 5I) = ker = span span{w A w B w C } Since 3 = dim[e (λ 5 )] < AM[λ 5 ] = 4 investigate E 3 (λ 5 ): STEP 4: Find 3 rd -order eigenspace to eigenvalue λ 5 E 3 (λ 5 ) := ker(a λ 5 I) 3 : ker(a 5I) 3 = ker = span span{x A x B x C x D } Since 4 = dim[e 3 (λ 5 )] < AM[λ 5 ] = 4 longest eigenchain length = Order[E 3 (λ 5 )] = 3. Since AM[λ 5 ] = 4 and longest eigenchain length = 3 there must be eigenchains: K () 5 K () 5. K () K () 5 has length 3. 5 has length. Copyright Josh Engwer 9

10 PART II: BUILD EIGENCHAINS Build st eigenchain K () 5 = {v () 5 w () 5 x () 5 } from highest order eigenvector to lowest: x () 5 w () 5 v () 5 STEP 5: Determine and pick one x () 5 {x A x B x C x D } s.t. x () 5 / span{w A w B w C }: w A w B w C x A x B x C x D We need vector independence we must pick x () 5 = x A = STEP 6: Determine w () 5 : w () 5 = (A λ 5 I)x () 5 = STEP 7: Determine v () 5 : v () 5 = (A λ 5 I)w () 5 = Build nd eigenchain K () 5 = {v () 5 } = =. K() 5 = STEP 8: Determine and pick v () 5 {v A v B }: Since eigenchain K () 5 has length simply pick v () 5 {v A v B }. Let s pick v () 5 = v B = PART III: BUILD MATRICES K() 5 = STEP : Build Matrix P consistent with eigenchains K () 5 K () 5 and find Matrix P : P = [K () 5 K () 5 ] = v() 5 w () 5 x () 5 v () 5 = P = STEP : Build Jordan Matrix J consistent with P such that A = P JP : λ 5 5 J = λ 5 λ 5 = 5 5 Notice there are Jordan blocks. λ 5 5 A = P JP : = / / / /

11 EXAMPLE 6: Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues Characteristic equation : det(a λi) = (λ 4) 5 = Hence the eigenvalues of A are λ = {4} {λ 4 } AM[λ 4 ] = 5. STEP : Find st -order eigenspace to eigenvalue λ 4 E (λ 4 ) := ker(a λ 4 I): 6 ker(a 4I) = ker = span span{v A v B } 3 Since = dim[e (λ 4 )] < AM[λ 4 ] = 5 investigate E (λ 4 ): STEP 3: Find nd -order eigenspace to eigenvalue λ 4 E (λ 4 ) := ker(a λ 4 I) : 3 6 ker(a 4I) = ker = span span{w A w B w C w D } 3 Since 4 = dim[e (λ 4 )] < AM[λ 4 ] = 5 investigate E 3 (λ 4 ): STEP 4: Find 3 rd -order eigenspace to eigenvalue λ 4 E 3 (λ 4 ) := ker(a λ 4 I) 3 : ker = span span{x A x B x C x D x E } Since 5 = dim[e 3 (λ 4 )] = AM[λ 4 ] = 5 longest eigenchain length = Order[E 3 (λ 4 )] = 3. Since AM[λ 4 ] = 5 and longest eigenchain length = 3 there must be eigenchains: K () 4 K () 4. Eigenchain K () 4 has length 3. Eigenchain K () 4 has length. Copyright Josh Engwer

12 PART II: BUILD EIGENCHAINS Build st eigenchain K () 4 = {v () 4 w () 4 x () 4 } from highest order eigenvector to lowest: x () 4 w () 4 v () 4 STEP 5: Determine and pick one x () 4 {x A x B x C x D x E } s.t. x () 4 / span{w A w B w C w D }: w A w B w C w D x A x B x C x D x E We need vector independence we must pick x () 4 = x A. STEP 6: Determine w () 4 then v () 4 : w () 4 = (A λ 4 I)x () 4 = 6 3 v() 4 = (A λ 4 I)w () 4 = 6 3 K() 4 = 6 3 Build nd eigenchain K () 4 = {v () 4 w () 4 } from highest order eigenvector to lowest: w () 4 v () 4 STEP 7: Determine and pick one w () 4 {w A w B w C w D } s.t. w () 4 / span{v A v B }: v A v B w A w B w C w D We need vector independence we can pick from {w A w C }. Let s pick w () 4 = w A. STEP 8: Determine v () 4 : v () 4 = (A λ 4 I)w () 4 = 3 K() 4 = Copyright Josh Engwer

13 PART III: BUILD MATRICES STEP 9: Build Matrix P consistent with eigenchains K () 4 K () 4 and find Matrix P : P = v () 4 w () 4 x () 4 v () 4 w () 4 = P = STEP : Build Jordan Matrix J consistent with P such that A = P JP : λ 4 4 λ 4 J = λ 4 λ 4 = 4 4 Notice there are Jordan blocks. 4 λ 4 4 A = = / 3 / 3 / 3 / 3 / 3 / 3 / 3 / 3 Copyright Josh Engwer 3

14 EXAMPLE 7: Find the Jordan Form of A = PART I: FIND ALL RELEVANT EIGENSPACES STEP : Find all eigenvalues. Characteristic equation : det(a λi) = (λ ) 5 = Hence the eigenvalues of A are λ = {} {λ } AM[λ ] = 5. STEP : Find st -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I): ker(a I) = ker = span span{v A v B v C } Since 3 = dim[e (λ )] < AM[λ ] = 5 investigate E (λ ): STEP 3: Find nd -order eigenspace to eigenvalue λ E (λ ) := ker(a λ I) : ker(a I) = ker = span span{w A w B w C w D } Since 4 = dim[e (λ )] < AM[λ ] = 5 investigate E 3 (λ ): STEP 4: Find 3 rd -order eigenspace to eigenvalue λ E 3 (λ ) := ker(a λ I) 3 : ker = span span{x A x B x C x D x E } Since 5 = dim[e 3 (λ )] = AM[λ ] = 5 longest eigenchain length = Order[E 3 (λ )] = 3. Since AM[λ ] = 5 and longest eigenchain length = 3 there must be 3 eigenchains: K () K () K (3). Eigenchain K () has length 3. Eigenchain K () has length. Eigenchain K (3) has length. Copyright Josh Engwer 4

15 PART II: BUILD EIGENCHAINS Build st eigenchain K () = {v () w () x () } from highest order eigenvector to lowest: x () w () v () STEP 5: Determine and pick one x () {x A x B x C x D x E } s.t. x () / span{w A w B w C w D }: w A w B w C w D x A x B x C x D x E We need vector independence we must pick x () = x C. STEP 6: Determine w () then v () : w () = (A λ I)x () = 4 38 Build nd eigenchain K () = {v () } v() = (A λ I)w () = 6 4 K() = STEP 7: Pick one v () span{v A v B v C }: 5 Let s pick v () = v A = K() = 5 Build 3 rd eigenchain K (3) = {v (3) } STEP 8: Pick one v (3) span{v A v B v C } s.t. v (3) v () : Let s pick v (3) = v C = K(3) = Copyright Josh Engwer 5

16 PART III: BUILD MATRICES STEP 9: Build Matrix P consistent with eigenchains K () K () K (3) and find Matrix P : P = [K () K () K (3) ] = P = 5 / 6 5 / 6 / / 5 / / / / 3 / 3 / v () w () x () v () v (3) = STEP : Build Jordan Matrix J consistent with P such that A = P JP : λ λ J = λ λ = Notice there are 3 Jordan blocks. λ A = = / 6 5 / 6 / / 5 / / / / 3 / 3 /

17 References [] S. Axler Linear Algebra Done Right. Springer New York NY nd Edition 997. [] S. H. Friedberg A. J. Insel L. E. Spence Linear Algebra. Pearson Upper Saddle River NJ 4th Edition 3. [3] D. C. Lay Linear Algebra and its Applications. Addison-Wesley Reading MA nd Edition. [4] G. Strang Introduction to Linear Algebra. Wellesley-Cambridge Press Wellesley MA 3rd Edition 3. [5] S. H. Weintraub Jordan Canonical Form: Theory and Practice. Morgan & Claypool Lexington KY st Edition 9. Copyright Josh Engwer 7

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