AB diagonalizable then BA also diagonalizable












1












$begingroup$


If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?



I read that if AB=BA then both AB and BA are diagonalizable.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
    $endgroup$
    – Lord Shark the Unknown
    2 hours ago










  • $begingroup$
    @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
    $endgroup$
    – kccu
    1 hour ago
















1












$begingroup$


If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?



I read that if AB=BA then both AB and BA are diagonalizable.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
    $endgroup$
    – Lord Shark the Unknown
    2 hours ago










  • $begingroup$
    @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
    $endgroup$
    – kccu
    1 hour ago














1












1








1





$begingroup$


If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?



I read that if AB=BA then both AB and BA are diagonalizable.






linear-algebra







share|cite|improve this question









$endgroup$




If A and B don't commute are there counterexamples that AB is diagonalizable but BA not?



I read that if AB=BA then both AB and BA are diagonalizable.






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









craftcraft

132




132












  • $begingroup$
    If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
    $endgroup$
    – Lord Shark the Unknown
    2 hours ago










  • $begingroup$
    @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
    $endgroup$
    – kccu
    1 hour ago


















  • $begingroup$
    If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
    $endgroup$
    – Lord Shark the Unknown
    2 hours ago










  • $begingroup$
    @LordSharktheUnknown They are asking what can happen if $AB neq BA$.
    $endgroup$
    – kccu
    1 hour ago
















$begingroup$
If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
$endgroup$
– Lord Shark the Unknown
2 hours ago




$begingroup$
If $AB=BA$ then $AB$, $BA$ are either both diagonalisable or neither is diagonalisable. I hope that's obvious.
$endgroup$
– Lord Shark the Unknown
2 hours ago












$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
1 hour ago




$begingroup$
@LordSharktheUnknown They are asking what can happen if $AB neq BA$.
$endgroup$
– kccu
1 hour ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Try $A = begin{bmatrix}0&0\1&0end{bmatrix}$ and $B = begin{bmatrix}0&0\0&1end{bmatrix}$






share|cite|improve this answer









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    1 Answer
    1






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    active

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    active

    oldest

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    4












    $begingroup$

    Try $A = begin{bmatrix}0&0\1&0end{bmatrix}$ and $B = begin{bmatrix}0&0\0&1end{bmatrix}$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Try $A = begin{bmatrix}0&0\1&0end{bmatrix}$ and $B = begin{bmatrix}0&0\0&1end{bmatrix}$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Try $A = begin{bmatrix}0&0\1&0end{bmatrix}$ and $B = begin{bmatrix}0&0\0&1end{bmatrix}$






        share|cite|improve this answer









        $endgroup$



        Try $A = begin{bmatrix}0&0\1&0end{bmatrix}$ and $B = begin{bmatrix}0&0\0&1end{bmatrix}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Michael BiroMichael Biro

        11.5k21831




        11.5k21831






























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