Mixing
$J={J_r}({lambda})$ is a Jordan Block matrix for $lambda$, $s{leq}r$ is an integer. Find formula for $J^s$.
$begingroup$
So The title is part (i) and part (ii) is "use the formula to show that if A is a square matrix with identity $A^l$ for some $l$, then A is diagonalizable.
I'm totally new to Jordan block matrices and Jordan normal form and this is one of our workshop questions. I've got lecture notes in front of me but they don't really seem to be helping, can anyone else help explain how to go about this question.
I do understand that A is diagonalizable if and only if each Jordan block of A has size 1 and wouldn't that then mean r=1. so I'm not sure how s can be less than r.
linear-algebra jordan-normal-form
asked Jan 10 at 14:20
L GL G
248
$endgroup$
add a comment |
$begingroup$
So The title is part (i) and part (ii) is "use the formula to show that if A is a square matrix with identity $A^l$ for some $l$, then A is diagonalizable.
I'm totally new to Jordan block matrices and Jordan normal form and this is one of our workshop questions. I've got lecture notes in front of me but they don't really seem to be helping, can anyone else help explain how to go about this question.
I do understand that A is diagonalizable if and only if each Jordan block of A has size 1 and wouldn't that then mean r=1. so I'm not sure how s can be less than r.
linear-algebra jordan-normal-form
asked Jan 10 at 14:20
L GL G
248
$endgroup$
$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33
add a comment |
$begingroup$
So The title is part (i) and part (ii) is "use the formula to show that if A is a square matrix with identity $A^l$ for some $l$, then A is diagonalizable.
I'm totally new to Jordan block matrices and Jordan normal form and this is one of our workshop questions. I've got lecture notes in front of me but they don't really seem to be helping, can anyone else help explain how to go about this question.
I do understand that A is diagonalizable if and only if each Jordan block of A has size 1 and wouldn't that then mean r=1. so I'm not sure how s can be less than r.
linear-algebra jordan-normal-form
asked Jan 10 at 14:20
L GL G
248
$endgroup$
So The title is part (i) and part (ii) is "use the formula to show that if A is a square matrix with identity $A^l$ for some $l$, then A is diagonalizable.
I'm totally new to Jordan block matrices and Jordan normal form and this is one of our workshop questions. I've got lecture notes in front of me but they don't really seem to be helping, can anyone else help explain how to go about this question.
I do understand that A is diagonalizable if and only if each Jordan block of A has size 1 and wouldn't that then mean r=1. so I'm not sure how s can be less than r.
linear-algebra jordan-normal-form
linear-algebra jordan-normal-form
asked Jan 10 at 14:20
L GL G
248
asked Jan 10 at 14:20
L GL G
248
edited Jan 10 at 14:27
L G
asked Jan 10 at 14:20
L GL G
248
asked Jan 10 at 14:20
L GL G
248
asked Jan 10 at 14:20
L GL G
248
248
$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33
add a comment |
$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33
$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33
add a comment |
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$begingroup$
If $A^l=I$ then what is the characteristic polynomial of $A$?
$endgroup$
– Levent
Jan 10 at 14:30
$begingroup$
Try to compute some powers of $J$ for the case that $r=3$, and try if you can see a pattern. For part (ii), you might want to use that if $J^l$ is the identity for some $l>0$, then $r=1$ (which should follow if you solved part (i)).
$endgroup$
– Uncountable
Jan 10 at 14:31
$begingroup$
$sle r$, so if you end up proving $s=r$, there is no contradiction
$endgroup$
– Calvin Khor
Jan 10 at 14:33