Mixing
About a some basis for linear mapping on a plane
$begingroup$
Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
$p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.
Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form
$
left [ begin {array}{lr}
alpha & -beta\
beta & alpha
end{array} right ].
$
Is there a basis in which matrix of $f$ is of the form
$
left [ begin {array}{lr}
0 & -c \
1& -b
end{array} right ]?
$
Thanks.
linear-algebra
asked Jan 31 at 20:06
AlexAlex
699519
$endgroup$
add a comment |
$begingroup$
Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
$p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.
Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form
$
left [ begin {array}{lr}
alpha & -beta\
beta & alpha
end{array} right ].
$
Is there a basis in which matrix of $f$ is of the form
$
left [ begin {array}{lr}
0 & -c \
1& -b
end{array} right ]?
$
Thanks.
linear-algebra
asked Jan 31 at 20:06
AlexAlex
699519
$endgroup$
add a comment |
$begingroup$
Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
$p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.
Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form
$
left [ begin {array}{lr}
alpha & -beta\
beta & alpha
end{array} right ].
$
Is there a basis in which matrix of $f$ is of the form
$
left [ begin {array}{lr}
0 & -c \
1& -b
end{array} right ]?
$
Thanks.
linear-algebra
asked Jan 31 at 20:06
AlexAlex
699519
$endgroup$
Let $f: Vrightarrow V$ be a linear mapping on a real 2-dimensional vector space $V$ with characteristic polynomial
$p(x)=x^2+bx+c$ and with negative discriminant $Delta=b^2-4ac<0$.
Let $t_1=alpha+ibeta$, $t_2=alpha-ibeta$ be its roots.
Then I know that there exists a basis $e_1, e_2$ in $V$ in which thematrix of $f$ is of the form
$
left [ begin {array}{lr}
alpha & -beta\
beta & alpha
end{array} right ].
$
Is there a basis in which matrix of $f$ is of the form
$
left [ begin {array}{lr}
0 & -c \
1& -b
end{array} right ]?
$
Thanks.
linear-algebra
linear-algebra
asked Jan 31 at 20:06
AlexAlex
699519
asked Jan 31 at 20:06
AlexAlex
699519
asked Jan 31 at 20:06
AlexAlex
699519
asked Jan 31 at 20:06
AlexAlex
699519
asked Jan 31 at 20:06
AlexAlex
699519
699519
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.
What is the matrix of $f$ in the basis $(a,f(a))$?
We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!
The answer to your question is positive.
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.
What is the matrix of $f$ in the basis $(a,f(a))$?
We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!
The answer to your question is positive.
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.
What is the matrix of $f$ in the basis $(a,f(a))$?
We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!
The answer to your question is positive.
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.
What is the matrix of $f$ in the basis $(a,f(a))$?
We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!
The answer to your question is positive.
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
If for all $x in V$, $x$ and $f(x)$ are collinear, $f$ would be an homothetic transformation. As this is not the case, it exists $a in V$ such that ${a, f(a)}$ is linearly independent.
What is the matrix of $f$ in the basis $(a,f(a))$?
We have $f(a)= 0.a + 1.f(a)$ and $f(f(a))= f^2(a) = (-c).a +(-b).f(a)$ using the equation of the characteristic polynomial. This is exactly the matrix we were looking for!
The answer to your question is positive.
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 31 at 20:39
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
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