{"id":19,"date":"2020-12-01T21:39:06","date_gmt":"2020-12-01T21:39:06","guid":{"rendered":"http:\/\/blog.metu.edu.tr\/e235505\/?p=19"},"modified":"2020-12-01T23:08:10","modified_gmt":"2020-12-01T23:08:10","slug":"if-abi-then-bai","status":"publish","type":"post","link":"https:\/\/blog.metu.edu.tr\/e235505\/2020\/12\/01\/if-abi-then-bai\/","title":{"rendered":"If AB=I, then BA=I"},"content":{"rendered":"\n

Proposition: Let A and B be nxn matrices. Also, let<\/p>\n\n\n\n

AB=I<\/p>\n\n\n\n

then<\/p>\n\n\n\n

BA=I<\/p>\n\n\n\n

Proof: Every matrix is row equivalent to a single row-reduced echelon matrix<\/a>. That is, for every matrix, there is a finite sequence of elementary row operations such that if we apply this sequence of row operations on this matrix, we get a row-reduced echelon matrix. If we denote R to be this matrix,<\/p>\n\n\n\n

er<\/sub>(er-1<\/sub>(………e1<\/sub>(R)…..))=A<\/p>\n\n\n\n

for some elementary row operations, er<\/sub>,er-1<\/sub>,…….,e1<\/sub>. Every elementary row operation on any matrix corresponds to multiplication of an elementary matrix with that matrix. This elementary matrix is obtained by doing the row operation on Imxm<\/sub> where m is equal to row number of the matrix that we apply the elementary row operation. Denote Ei<\/sub> to be correspondent elementary matrix of ei<\/sub>. Then,<\/p>\n\n\n\n

Er<\/sub>Er-1<\/sub>…….E1<\/sub>R=A<\/p>\n\n\n\n

(Er<\/sub>Er-1<\/sub>…….E1<\/sub>R)B=I<\/p>\n\n\n\n

multiply both sides Er<\/sub>‘<\/sup>, inverse of Er<\/sub><\/p>\n\n\n\n

Er<\/sub>‘<\/sup>(Er<\/sub>Er-1<\/sub>…….E1<\/sub>R)B=Er<\/sub>‘<\/sup>I<\/p>\n\n\n\n

keep multiplying,<\/p>\n\n\n\n

Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>(Er<\/sub>Er-1<\/sub>…….E1<\/sub>R)B=Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I<\/p>\n\n\n\n

E1<\/sub>‘<\/sup>……..Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>(Er<\/sub>Er-1<\/sub>…….E1<\/sub>R)B=E1<\/sub>‘<\/sup>……….Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I<\/p>\n\n\n\n

RB=E1<\/sub>‘<\/sup>……….Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I<\/p>\n\n\n\n

Can R contain a zero row? No, because if it were, then RB would contain a zero row and any matrix that is row equivalent to I can’t contain a zero row. Why? Because if there were such matrix, we would obtain a contradiction. Let’s assume that RB contains some zero rows and find (RB)’ (equal to RB without zero rows) matrix’s row-reduced echelon matrix, and then add zero rows at the end of this row-reduced echelon matrix. Then, this final matrix, say M, is row reduced echelon matrix of RB, and clearly it is not equal to I. This is a contradiction, since we have two distinct row reduced echelon matrix of RB; one is I and the other is M. This is not possible.<\/p>\n\n\n\n

So, R must not have zero row, then it must be I (easy to verify). Then,<\/p>\n\n\n\n

RB=IB=B=E1<\/sub>‘<\/sup>……….Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I<\/p>\n\n\n\n

Is BA=I?<\/p>\n\n\n\n

BA=(E1<\/sub>‘<\/sup>……….Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I)(Er<\/sub>Er-1<\/sub>…….E1<\/sub>R)<\/p>\n\n\n\n

=(E1<\/sub>‘<\/sup>……….Er-1<\/sub>‘<\/sup>Er<\/sub>‘<\/sup>I)(Er<\/sub>Er-1<\/sub>…….E1<\/sub>I)=I<\/p>\n\n\n\n

Yes, it is<\/p>\n","protected":false},"excerpt":{"rendered":"

Proposition: Let A and B be nxn matrices. Also, let AB=I then BA=I Proof: Every matrix is row equivalent to a single row-reduced echelon matrix. That is, for every matrix, there is a finite sequence of elementary row operations such that if we apply this sequence of row operations on this matrix, we get a … <\/p>\n

Continue reading “If AB=I, then BA=I”<\/span><\/a><\/p>\n","protected":false},"author":6755,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_links_to":"","_links_to_target":""},"categories":[1],"tags":[],"class_list":["post-19","post","type-post","status-publish","format-standard","hentry","category-uncategorized","entry"],"_links":{"self":[{"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/posts\/19","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/users\/6755"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/comments?post=19"}],"version-history":[{"count":0,"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/posts\/19\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/media?parent=19"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/categories?post=19"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.metu.edu.tr\/e235505\/wp-json\/wp\/v2\/tags?post=19"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}