find a recursive relation for the characteristic polynomial of the $k times k $ matrix?
find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$
and compute the polynomial for $kle 5$
My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have
$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$
after that im not able proceed further
Any hints/solution will be appreciated
linear-algebra
asked yesterday
jasmine
1,589416
|
show 4 more comments
find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$
and compute the polynomial for $kle 5$
My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have
$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$
after that im not able proceed further
Any hints/solution will be appreciated
linear-algebra
asked yesterday
jasmine
1,589416
2
do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
2
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
1
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday
|
show 4 more comments
find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$
and compute the polynomial for $kle 5$
My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have
$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$
after that im not able proceed further
Any hints/solution will be appreciated
linear-algebra
asked yesterday
jasmine
1,589416
find a recursive relation for the characteristic polynomial of the $k times k $ matrix $$begin{pmatrix} 0 & 1 \ 1 & 0 & 1 \ mbox{ } & 1 & . & . \ mbox{ } &mbox{ } & . & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & 0end{pmatrix}$$
and compute the polynomial for $kle 5$
My attempt : Let $M_k$ be the $ktimes k$ matrix and $P_k(x)=det(M_k-xI_k)$ be its characteristic polynomial. We have
$$P_k(x)=detbegin{pmatrix} -x & 1 \ 1 & -x & 1 \ mbox{ } & 1 & ddots & ddots \ mbox{ } &mbox{ } & ddots & . &. & mbox{ } \ mbox{ } &mbox{ } &mbox{ } & . & . & 1 \ mbox{ } & mbox{ } & mbox{ } & mbox{ } & 1 & -xend{pmatrix}$$
after that im not able proceed further
Any hints/solution will be appreciated
linear-algebra
linear-algebra
asked yesterday
jasmine
1,589416
asked yesterday
jasmine
1,589416
asked yesterday
jasmine
1,589416
asked yesterday
jasmine
1,589416
asked yesterday
jasmine
1,589416
1,589416
2
do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
2
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
1
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday
|
show 4 more comments
2
do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
2
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
1
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday
2
2
do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
2
2
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
1
1
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday
|
show 4 more comments
1 Answer
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Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.
Solution:
$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$
edited yesterday
J.G.
23.1k22137
answered yesterday
A. P
1085
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A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.
Solution:
$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$
edited yesterday
J.G.
23.1k22137
answered yesterday
A. P
1085
New contributor
A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.
Solution:
$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$
edited yesterday
J.G.
23.1k22137
answered yesterday
A. P
1085
New contributor
A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.
Solution:
$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$
edited yesterday
J.G.
23.1k22137
answered yesterday
A. P
1085
New contributor
A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hint: Look at the formula for the determinant of using the first row.
Then you get a recursive definition for the determinant.
Solution:
$P_k(x)=-xP_{k-1}(x)-P_{k-1}(-1)$
edited yesterday
J.G.
23.1k22137
answered yesterday
A. P
1085
New contributor
A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
J.G.
23.1k22137
edited yesterday
J.G.
23.1k22137
edited yesterday
J.G.
23.1k22137
23.1k22137
answered yesterday
A. P
1085
New contributor
A. P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
A. P
1085
answered yesterday
A. P
1085
1085
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New contributor
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do it first for $k=2$ and $k=3. ;$ I found a quote "There is nothing impossible to him who will try."
– Will Jagy
yesterday
ya ,,,im trying @WillJagy.....haa
– jasmine
yesterday
what do you get for $k=2?$
– Will Jagy
yesterday
2
now do $k=3.$ By the way, there is a pretty good recipe for writing the char poly in dimension 3, where $sigma_1$ is the trace and $sigma_3$ is the determinant, we also need $sigma_2$ which is the sum of the little 2 by 2 prinipal minor determintns, Then char poly is $x^3 - sigma_1 x^2 + sigma_2 x - sigma_3.$ I think there is also a recipe for $sigma_2$ involving traces of the matrix squared, I will see if I can find that. Worth getting right and memorizing for dimension 2 and 3
– Will Jagy
yesterday
1
Alright, worked it out, for matrix $M,$ $$sigma_2 = frac{1}{2} left( sigma_1^2 - operatorname{trace} M^2 right)$$ where $sigma_1 = operatorname{trace} M$ as i indicated
– Will Jagy
yesterday