HowtoFindJordanCanonicalForms Here is a metho d to nd a Jordan canonical form of matrices and some examples showing the metho d at work MethodOutline i For a transformation  use a basis to get a matr
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HowtoFindJordanCanonicalForms Here is a metho d to nd a Jordan canonical form of matrices and some examples showing the metho d at work MethodOutline i For a transformation use a basis to get a matr

If you are just given a matrix use that matrix ii Compute det xI the characteristic p olynomial of the transformation and factor is as 1 where for and for all If the characteristic p olynomial do es not split we will not b e able to put the tran

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HowtoFindJordanCanonicalForms Here is a metho d to nd a Jordan canonical form of matrices and some examples showing the metho d at work MethodOutline i For a transformation use a basis to get a matr




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HowtoFindJordanCanonicalForms Here is a metho d to nd a Jordan canonical form of matrices and some examples showing the metho d at work: MethodOutline (i) For a transformation , use a basis to get a matrix = [ for the transformation. If you are just given a matrix, use that matrix. (ii) Compute ) = det( xI , the characteristic p olynomial of the transformation and factor is as =1 where for and for all . If the characteristic p olynomial do es not split, we will not b e able to put the transformation into Jordan canonical form. (iii) For each , compute the dimensions of (( for

= 1 ,... until dim (( ) = say dim (( λI ) = . Letting = dim (( , this will give us a sequence 0 = < d (iv) The sequence ,...,d determines how many Jordan blo cks corresp onding to we have and their resp ective sizes. Here is one way to interpret the sequence: tells us how many blo cks of size at least one there are. Then, tells how many of those blo cks are actually of size at least two. Then, tells us how many of those blo cks are actually of size at least three. Rep eating this, we can see exactly how many blo cks of each size we have. Let's lo ok at some examples. Example1. Let b e

a transformation with matrix 2 2 3 1 3 3 (i) This is already in matrix form. (ii) We see det( xI ) = det 1 2 + 2 = ( 2)(( 3)( + 2) + 6) + 2( + 2) + 3) 3( 3)) = ( 2)( ) + 2( + 1) 3( + 1) which is + 2 + 2 + 3 3 = + 3 1 = ( 1) (iii) For = 1 , we have 1 2 3 1 2 3 which has rank 1. Thus, = dim (( )) = 2 Now, = 0 , so = 3 (note that the sequence strictly increases until it reaches , so b ecause = 3 for = 1 and dim (( )) = 2 , we could actually immediately conclude dim (( ) = 3 without any computation). (iv) The = 2 tells us the Jordan canonical form of our matrix has 2 blo cks with eigenvalue . The

= 1 condition tells us one of these blo cks has size at least 2, and so the other has size 1. Thus, a Jordan canonical form for is 1 0 0 0 1 1 0 0 1 Example2. Let b e a transformation with matrix 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 (i) This is already in matrix form.
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(ii) det( xI ) = det 1 0 0 0 0 + 1 0 0 1 1 1 0 1 0 0 0 1 0 0 = ( 1) det + 1 0 0 1 0 1 0 0 1 0 0 = ( 1)( 1) 1) 1)( ) = ( 1) 1) + 1) This is 1) (iii) For = 1 , we already have = 1 . For = 0 , we see 0 = 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 which has rank 3 and hence = dim ) = 2 . Now, 1 0 0 0 0 1 0 0 0 0 1 0

0 0 0 1 0 0 1 0 0 0 0 which has rank 2, so = dim ) = 3 . We see 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 and so = dim ) = 4 (note this computation was unnecessary as we must have >d as = 4 ). (iv) From the ab ove, we have 1 Jordan blo ck with eigenvalue 1 and 2 Jordan blo cks with eigenvalue 0. The sequence tells us there are 2 blo cks, with 2 = 1 having size at least 2 and 3 = 1 having size at least 3. That is, there is 1 blo ck of size 1 and 1 of size 3. Hence, a Jordan canonical form for is 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 Example3. Let b e a transformation with

matrix 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 1 0 0 0 1 0 0 2 0 0 0 0 0 0 1 (i) This is already in matrix form. (ii) det( xI ) = det 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0 0 2 0 0 0 0 0 0 = ( 1) det 0 0 1 1 1 0 0 1 0 1 0 0 1) det 0 1 1 1 1 0 This is 1) 1)( 2)) + ( 1) 1) = ( 1) 1) = ( 1) + 3 1) = ( 1)
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(iii) For = 1 , we have 0 0 0 0 0 0 1 0 0 1 1 1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 which has rank 3. Thus, = dim (( )) = 3 . Now, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 , which has rank 1. Thus, = dim (( ) = 5 . As ab ove, from here we can

conclude = 6 (iv) As = 3 , we have 3 Jordan blo cks corresp onding to = 1 . As = 5 3 = 2 , 2 of these blo cks have size at least 2. As = 6 5 = 1 , 1 of these blo cks has size at least 3. Thus, we have 1 blo ck of size 1, 1 blo ck of size 2, and 1 blo ck of size 3. Hence, a Jordan canonical form for is 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1