Mixing
Semisimple matrix
$begingroup$
I have to find all possible values of a,b,c, and d for which the $ 2times 2 $ matrix [a b; c d] is semisimple. I know that a matrix S is semisimple if there is a nonsingular matrix P such that $ P^-1SP=A $ is diagonal. How can I use this to find all the values a, b, c and d?
matrices semi-simple-rings
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
$endgroup$
add a comment |
$begingroup$
I have to find all possible values of a,b,c, and d for which the $ 2times 2 $ matrix [a b; c d] is semisimple. I know that a matrix S is semisimple if there is a nonsingular matrix P such that $ P^-1SP=A $ is diagonal. How can I use this to find all the values a, b, c and d?
matrices semi-simple-rings
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
$endgroup$
add a comment |
$begingroup$
I have to find all possible values of a,b,c, and d for which the $ 2times 2 $ matrix [a b; c d] is semisimple. I know that a matrix S is semisimple if there is a nonsingular matrix P such that $ P^-1SP=A $ is diagonal. How can I use this to find all the values a, b, c and d?
matrices semi-simple-rings
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
$endgroup$
I have to find all possible values of a,b,c, and d for which the $ 2times 2 $ matrix [a b; c d] is semisimple. I know that a matrix S is semisimple if there is a nonsingular matrix P such that $ P^-1SP=A $ is diagonal. How can I use this to find all the values a, b, c and d?
matrices semi-simple-rings
matrices semi-simple-rings
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
edited Apr 24 '17 at 12:36
Martin Sleziak
44.9k10121275
44.9k10121275
asked Mar 20 '17 at 8:50
SiriSiri
163
asked Mar 20 '17 at 8:50
SiriSiri
163
asked Mar 20 '17 at 8:50
SiriSiri
163
163
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
$endgroup$
add a comment |
$begingroup$
Let $A=begin{pmatrix}a&b\c&dend{pmatrix}in M_n(mathbb{C})$ (we work over $mathbb{C}$).
If $A$ is not diagonal, then
$A$ is diagonalizable IFF $(a-d)^2+4bcnot=0$.
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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oldest
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$begingroup$
Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
$endgroup$
add a comment |
$begingroup$
Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
$endgroup$
add a comment |
$begingroup$
Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
$endgroup$
Since you already know semisimple is equivalent to diagonalizable (it is, in the case of complex matrices), then you can consider the possible Jordan forms for 2x2 matrices. You want to avoid the situation of a Jordan block. That means the two remaining situations are: you have a multiple of the identity matrix, or the characteristic polynomial has two roots. Both of these situations can be expressed in terms of a, b, c, and d.
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
answered Apr 4 '17 at 20:54
user25959user25959
1,573916
1,573916
add a comment |
add a comment |
$begingroup$
Let $A=begin{pmatrix}a&b\c&dend{pmatrix}in M_n(mathbb{C})$ (we work over $mathbb{C}$).
If $A$ is not diagonal, then
$A$ is diagonalizable IFF $(a-d)^2+4bcnot=0$.
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
$endgroup$
add a comment |
$begingroup$
Let $A=begin{pmatrix}a&b\c&dend{pmatrix}in M_n(mathbb{C})$ (we work over $mathbb{C}$).
If $A$ is not diagonal, then
$A$ is diagonalizable IFF $(a-d)^2+4bcnot=0$.
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
$endgroup$
add a comment |
$begingroup$
Let $A=begin{pmatrix}a&b\c&dend{pmatrix}in M_n(mathbb{C})$ (we work over $mathbb{C}$).
If $A$ is not diagonal, then
$A$ is diagonalizable IFF $(a-d)^2+4bcnot=0$.
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
$endgroup$
Let $A=begin{pmatrix}a&b\c&dend{pmatrix}in M_n(mathbb{C})$ (we work over $mathbb{C}$).
If $A$ is not diagonal, then
$A$ is diagonalizable IFF $(a-d)^2+4bcnot=0$.
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
answered Jan 26 at 14:55
loup blancloup blanc
24k21851
24k21851
add a comment |
add a comment |
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