[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-657826-105":3,"detail-sidebar-cat-0-en-105":81,"doc-detail-657826-en":130},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":74,"head_meta":76,"extra_data":78,"updated_unix":80},105,"en","arrow-diagrams-and-generalized-eigen-vectors-arrow-diagrams","Arrow Diagrams and Generalized Eigen Vectors - Arrow Diagrams","","Arrow diagrams visualize the structure of eigenvectors and generalized eigenvectors for a linear system x' = Ax by organizing arrows into rows according to the kernels of (A − λI), (A − λI)^2, (A − λI)^3, and continuing until multiplicity m_λ is reached. The number of arrows in the bottom rows matches dim E_λ = ker(A − λI), dim ker((A − λI)^2), etc., producing constraints on possible diagram shapes and enabling Jordan-cycle identification. The document works through two concrete matrix examples, computing standard bases and constructing generalized vectors to build solutions and corresponding arrow diagrams.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/research-report/","Research & Report",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/arrow-diagrams-and-generalized-eigen-vectors-arrow-diagrams/657826/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/arrow-diagrams-and-generalized-eigen-vectors-arrow-diagrams/657826.png","ImageObject",300,407,{"name":42,"@type":43},"Dewi_Sukmawati","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-30","2026-10-26",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",7,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How do arrow diagrams determine the dimensions of eigenvector spaces?","Question",{"text":63,"@type":64},"For each eigenvalue λ_i, the bottom row count equals dim E_{λ_i} = ker(A − λ_i I). The bottom two rows match dim ker((A − λ_i I)^2), and the pattern continues for higher powers.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"What key structural feature do arrow diagrams always satisfy?",{"text":68,"@type":64},"For any eigenvalue λ, the number of arrows in a row cannot exceed the number of arrows in the lower row.",{"name":70,"@type":61,"acceptedAnswer":71},"How are generalized eigenvectors used to construct solutions and the arrow diagram?",{"text":72,"@type":64},"Generalized eigenvectors J are chosen so that (A − λI)J lies in E_λ. The document then solves for J, uses it to form expressions like x(t) = e^{λt}(J + t(A − λI)J) in the presented examples, and matches the resulting structure to the corresponding arrow diagram.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},657826,1793136847,{"code":4,"msg":82,"data":83},"success",[84,88,92,96,101,106,110,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":85,"show_sort_weight":86,"slug":87},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":89,"show_sort_weight":90,"slug":91},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":93,"show_sort_weight":94,"slug":95},"Exam",70,"exam",{"id":97,"doc_module":4,"doc_module_name":25,"category_name":98,"show_sort_weight":99,"slug":100},5,"Comic",60,"comic",{"id":102,"doc_module":4,"doc_module_name":25,"category_name":103,"show_sort_weight":104,"slug":105},6,"Technology",50,"technology",{"id":55,"doc_module":4,"doc_module_name":25,"category_name":107,"show_sort_weight":108,"slug":109},"Healthcare",40,"healthcare",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":112,"slug":113},8,30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":97,"slug":129},19,"General","general",{"code":4,"msg":82,"data":131},{"doc_id":79,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":111,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":123,"language":139,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":12,"update_tm":143,"read_time":144},962094243415,"https://ap-avatar.wpscdn.com/davatar_3d24733baf745e90a7e4bdd5f77d97b2","Arrow Diagrams and Generalized Eigen Vectors  \nArrow Diagrams  \nAn arrow diagram is a table of arrows that represents the distribution of eigenvectors and generalized eigenvectors of a matrix A 2 Mn (R) .  \nLet 􀀕 1 ; 􀀕 2 ; : : : ; 􀀕 k be the eigenvalues of A. For each 􀀕i you will get a rectangular grid of arrows, for example  \n8  \n􀀕i : \u003C  \n:  \nwith the following property: the number of arrows on the bottom row is the dimension of E􀀕 = ker(A􀀀􀀕I), the number of arrows in the bottom two rows is the dimension of ker((A 􀀀 􀀕I)2 ), the number of arrows in the bottom three rows is the dimension of ker((A 􀀀 􀀕I)3 ), and so on. Continue until the total number of arrows equals the multiplicity of 􀀕i , m􀀕 i . For example, if you \\have all your eigenvectors\" for an eigenvalue 􀀕 of multiplicity m􀀕 , that is to say dim(E􀀕 ) = m􀀕 , then the arrow diagram for that 􀀕 will be a single row of m􀀕 arrows:  \n􀀕 : 􀀈 􀀁 􀀁 􀀁  \nThe diagram always has one important feature: for any eigenvalue 􀀕, the number of arrows in a row can not exceed the number of arrows in a lower row.  \nThus, if you have an eigenvalue 􀀕 of multiplicity m􀀕 and you only have one eigenvector, that is to say dim(E􀀕 ) = 1, then the arrow diagram for that 􀀕 will be a single column of m􀀕 arrows:  \n8  \n>>  \n>  \n􀀕 : \u003C ...  \n>>  \n>  \n:  \nand this is exactly the situation in which we get a Jordan cycle.  \nIf you have a 4 􀀂 4 with eigenvalue 􀀀2 of multiplicity m􀀀2 = 1 and eigenvalue 􀀀1 of multiplicity m􀀀1 = 3, then the possible arrow diagrams are  \n(a)  \n􀀀2 : 􀀈 􀀀 1 : 􀀈  \n(b)  \n􀀀2 : 􀀈 􀀀 1 : 􀀚  \n(c)  \n8  \n􀀀2 : 􀀈 􀀀 1 : \u003C  \n:  \nA 4 􀀂 4 with a single eigenvalue, 􀀕, has the following possible arrow diagrams:  \n8  \n8 >  \n(a) 􀀕 : 􀀈 (b) 􀀕 : 􀀚 (c) 􀀕 : 􀀚 (d) 􀀕 : \u003C: (e) 􀀕 : :  \nExample 1  \nConsider  \nddt~x = A~x where A =  \n2 4  \n0 􀀀3 3 1 4 􀀀1 􀀀2 􀀀2 5  \n3 5  \nhas eigenvalue 􀀕 = 3; m3 = 3 :  \n1. What is the standard basis of E3 ?  \n2. Give the solution(s) generated by the standard vector(s) J 2 ker((A 􀀀 􀀕I)2 ) n ker(A 􀀀 􀀕I) . These are the vectors represented by arrows in the second row (from the bottom) of the arrow diagram.  \n3. What is the arrow diagram for the system?  \nSolution:  \n1. The standard basis of E3 :  \n(A 􀀀 3I) =  \n2 4  \n􀀀3 􀀀3 3 1 1 􀀀1 􀀀2 􀀀2 2  \n3 5  \n! RREF(A 􀀀 3I) = 2 10 10 1 3 4 0 0 0 5  \nso the standard basis of E3 is  \n8 \u003C ~v31 =:  \n2 4  \n􀀀1  \n1 0  \n3 2 1 3 5 ; ~v23 = 4 01 5  \n9 =  \n;  \nSince there are two eigenvectors, the bottom row of our arrow diagram has two arrows, and now we know the arrow diagram:  \n3 : 􀀚  \n2. Solution(s) generated by generalized eigenvector(s) J in the second row: We are looking for vectors J 2 R3 such that (A 􀀀 3I)J is an eigenvector, that is  \n8  \n(A 􀀀 3I)J 2 E3 = Span \u003C:  \n2 4  \n􀀀1  \n1 0  \n3 2 5 ; 4  \n1 0  \n1  \n3 5  \n9 =  \n;  \n2 4  \n􀀀3 􀀀3 3 1 1 􀀀1 􀀀2 􀀀2 2  \n3 5  \nJ = 􀀋 􀀁  \n2 4  \n􀀀1  \n1 0  \n3 5  \n+ 􀀌 􀀁 2 10 3 = 4 1 5  \n2 4  \n􀀀􀀋 + 􀀌  \n􀀋  \n􀀌  \n3 5  \nSo we row reduce  \n2 4  \n􀀀3 􀀀3 3 1 1 􀀀1 􀀀2 􀀀2 2  \n􀀀􀀋 + 􀀌  \n􀀋  \n􀀌  \n3 5  \nR2 + 3R1 􀀀!  \nR3 + 2R1  \nR 1 $ R2 􀀀!  \n2 4  \n1 0  \n0  \n1 0  \n0  \n􀀀1  \n0 0  \n2 4  \n1  \n􀀀3  \n􀀀2  \n􀀋  \n2􀀋 +  \n2􀀋 +  \n1 􀀀1  \n􀀀3 3  \n􀀀2 2  \n􀀌 3  \n􀀌 5  \n􀀋  \n􀀀􀀋 + 􀀌  \n􀀌  \n3 5  \nThis system is consistent only when the entries in the augmented column of rows with all zeroes are also zero. This gives us a homogeneous system  \n2􀀋 + 􀀌 = 0  \n2􀀋 + 􀀌 = 0  \nWe solve this using parametric form and get  \n􀀔 22 11 􀀕 􀀀! 􀀔 1 1=0 02 􀀕 ) 􀀋 = 􀀀 12􀀌 ) 􀀔 􀀋􀀌 􀀕 = 􀀌 􀀔 􀀀1~~1~~2 􀀕  \nThis gives us one linearly independent solution, 􀀋 = 􀀀1=2; 􀀌 = 1, so we will get one vector J. We can 􀀌nd J by solving  \n2  \n4 and  \n1 0  \n0  \n1 0  \n0  \n􀀀1  \n0 0  \n(A 􀀀 3I)J =  \n2 4  \n􀀀􀀋 + 􀀌  \n􀀋  \n􀀌  \n􀀀 1  \n2  \n0 0  \n3 5  \n) J =  \n2 4  \n􀀀 1  \n2  \n0 0  \n3 5  \n3 5  \n=  \n2 4  \n3 2  \n􀀀 1  \n2  \n1  \n3 5  \nThis gives us a solution  \n~x = e3t (J + t(A 􀀀 3I)J) = e3t 02 @4  \n3. Arrow diagram  \n3 : 􀀚  \n􀀀 1  \n2  \n0 0  \n3 5  \n+ t  \n2 4  \n3 2  \n􀀀 1  \n2  \n1  \n3 5  \n1 = e3t A  \n2 4  \n􀀀 ~~1~~2 + ~~3~~2t  \n􀀀 ~~1~~2t  \nt  \n3 5  \nExample 2  \nConsider  \nddt~x = A~x where A = 26  4  \n􀀀1  \n􀀀2  \n􀀀1  \n􀀀1  \n0  \n􀀀3  \n􀀀1  \n􀀀1","cbCaij7Lpzsd2Nla","https://ap.wps.com/l/cbCaij7Lpzsd2Nla","pdf",119986,"English","# Arrow Diagrams\n## Eigenvalues and kernel dimensions\n## Constraints on diagram shape\n## Example 1\n## Example 2\n## Solution construction and arrow diagrams","[{\"question\":\"How do arrow diagrams determine the dimensions of eigenvector spaces?\",\"answer\":\"For each eigenvalue λ_i, the bottom row count equals dim E_{λ_i} = ker(A − λ_i I). The bottom two rows match dim ker((A − λ_i I)^2), and the pattern continues for higher powers.\"},{\"question\":\"What key structural feature do arrow diagrams always satisfy?\",\"answer\":\"For any eigenvalue λ, the number of arrows in a row cannot exceed the number of arrows in the lower row.\"},{\"question\":\"How are generalized eigenvectors used to construct solutions and the arrow diagram?\",\"answer\":\"Generalized eigenvectors J are chosen so that (A − λI)J lies in E_λ. The document then solves for J, uses it to form expressions like x(t) = e^{λt}(J + t(A − λI)J) in the presented examples, and matches the resulting structure to the corresponding arrow diagram.\"}]","Arrow Diagrams and Generalized Eigen Vectors - Arrow Diagrams | PDF",1792991558,25]