How to show that eigenvalues of AB and BA are same?
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Let $A$ be an $mtimes n$ matrix and $B$ be an $ntimes m$ matrix. How to use Jordan canonical form to prove the matrix shown below are similar.
Given matrices are $$begin{bmatrix} AB&0\ B&0end{bmatrix}$$ and $$begin{bmatrix} 0&0\ B&BAend{bmatrix}$$
How to find their Jordan forms? I have no idea.
matrices eigenvalues-eigenvectors jordan-normal-form
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Let $A$ be an $mtimes n$ matrix and $B$ be an $ntimes m$ matrix. How to use Jordan canonical form to prove the matrix shown below are similar.
Given matrices are $$begin{bmatrix} AB&0\ B&0end{bmatrix}$$ and $$begin{bmatrix} 0&0\ B&BAend{bmatrix}$$
How to find their Jordan forms? I have no idea.
matrices eigenvalues-eigenvectors jordan-normal-form
No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21
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down vote
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up vote
0
down vote
favorite
Let $A$ be an $mtimes n$ matrix and $B$ be an $ntimes m$ matrix. How to use Jordan canonical form to prove the matrix shown below are similar.
Given matrices are $$begin{bmatrix} AB&0\ B&0end{bmatrix}$$ and $$begin{bmatrix} 0&0\ B&BAend{bmatrix}$$
How to find their Jordan forms? I have no idea.
matrices eigenvalues-eigenvectors jordan-normal-form
Let $A$ be an $mtimes n$ matrix and $B$ be an $ntimes m$ matrix. How to use Jordan canonical form to prove the matrix shown below are similar.
Given matrices are $$begin{bmatrix} AB&0\ B&0end{bmatrix}$$ and $$begin{bmatrix} 0&0\ B&BAend{bmatrix}$$
How to find their Jordan forms? I have no idea.
matrices eigenvalues-eigenvectors jordan-normal-form
matrices eigenvalues-eigenvectors jordan-normal-form
edited Nov 23 at 12:13
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 23 at 11:50
Mittal G
1,192515
1,192515
No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21
add a comment |
No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21
No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21
No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21
add a comment |
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No need to use any Jordan form. See, e.g., here.
– Algebraic Pavel
Nov 23 at 15:21