Eigen Value of differential equation [closed]
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How can I find eigenvalue and eigenvector of the following:
$$Sp(x)= p"(x)-2xp'(x)$$
In class we only covered matrices for eigenvector. I have no clue how to approach differential equation. Someone please help.
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Sep 13 at 10:39
amWhy
191k27223437
asked Sep 11 at 2:40
Misha
126
closed as off-topic by Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy Sep 14 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy
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show 7 more comments
up vote
0
down vote
favorite
How can I find eigenvalue and eigenvector of the following:
$$Sp(x)= p"(x)-2xp'(x)$$
In class we only covered matrices for eigenvector. I have no clue how to approach differential equation. Someone please help.
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Sep 13 at 10:39
amWhy
191k27223437
asked Sep 11 at 2:40
Misha
126
closed as off-topic by Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy Sep 14 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy
1
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
1
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
2
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
1
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48
|
show 7 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can I find eigenvalue and eigenvector of the following:
$$Sp(x)= p"(x)-2xp'(x)$$
In class we only covered matrices for eigenvector. I have no clue how to approach differential equation. Someone please help.
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Sep 13 at 10:39
amWhy
191k27223437
asked Sep 11 at 2:40
Misha
126
How can I find eigenvalue and eigenvector of the following:
$$Sp(x)= p"(x)-2xp'(x)$$
In class we only covered matrices for eigenvector. I have no clue how to approach differential equation. Someone please help.
linear-algebra eigenvalues-eigenvectors linear-transformations
linear-algebra eigenvalues-eigenvectors linear-transformations
edited Sep 13 at 10:39
amWhy
191k27223437
asked Sep 11 at 2:40
Misha
126
edited Sep 13 at 10:39
amWhy
191k27223437
asked Sep 11 at 2:40
Misha
126
edited Sep 13 at 10:39
amWhy
191k27223437
edited Sep 13 at 10:39
amWhy
191k27223437
edited Sep 13 at 10:39
amWhy
191k27223437
191k27223437
asked Sep 11 at 2:40
Misha
126
asked Sep 11 at 2:40
Misha
126
asked Sep 11 at 2:40
Misha
126
126
closed as off-topic by Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy Sep 14 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy
closed as off-topic by Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy Sep 14 at 0:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, José Carlos Santos, Adrian Keister, Theoretical Economist, amWhy
1
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
1
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
2
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
1
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48
|
show 7 more comments
1
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
1
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
2
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
1
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48
1
1
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
1
1
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
2
2
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
1
1
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48
|
show 7 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$$
S 1 = 0\
S x = -2 x\
S x^2 = 2-4x^2\
= 2*1 + 0*x + (-4)*x^2
$$
If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $lambda=0$ for the eigenvector $p(x)=1$.
Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $lambda=-2$ for the eigenvector $p(x)=x$.
Put the matrix together in this basis $ 1,x,x^2$
$$
S = beginpmatrix
0 & 0 & 2\
0 & -2 & 0\
0 & 0 & -4\
endpmatrix
$$
This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.
answered Sep 11 at 4:12
AHusain
2,7502816
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$
S 1 = 0\
S x = -2 x\
S x^2 = 2-4x^2\
= 2*1 + 0*x + (-4)*x^2
$$
If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $lambda=0$ for the eigenvector $p(x)=1$.
Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $lambda=-2$ for the eigenvector $p(x)=x$.
Put the matrix together in this basis $ 1,x,x^2$
$$
S = beginpmatrix
0 & 0 & 2\
0 & -2 & 0\
0 & 0 & -4\
endpmatrix
$$
This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.
answered Sep 11 at 4:12
AHusain
2,7502816
add a comment |
up vote
2
down vote
accepted
$$
S 1 = 0\
S x = -2 x\
S x^2 = 2-4x^2\
= 2*1 + 0*x + (-4)*x^2
$$
If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $lambda=0$ for the eigenvector $p(x)=1$.
Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $lambda=-2$ for the eigenvector $p(x)=x$.
Put the matrix together in this basis $ 1,x,x^2$
$$
S = beginpmatrix
0 & 0 & 2\
0 & -2 & 0\
0 & 0 & -4\
endpmatrix
$$
This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.
answered Sep 11 at 4:12
AHusain
2,7502816
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$
S 1 = 0\
S x = -2 x\
S x^2 = 2-4x^2\
= 2*1 + 0*x + (-4)*x^2
$$
If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $lambda=0$ for the eigenvector $p(x)=1$.
Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $lambda=-2$ for the eigenvector $p(x)=x$.
Put the matrix together in this basis $ 1,x,x^2$
$$
S = beginpmatrix
0 & 0 & 2\
0 & -2 & 0\
0 & 0 & -4\
endpmatrix
$$
This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.
answered Sep 11 at 4:12
AHusain
2,7502816
$$
S 1 = 0\
S x = -2 x\
S x^2 = 2-4x^2\
= 2*1 + 0*x + (-4)*x^2
$$
If you let $p(x)=1$, you get $Sp(x) = 0 = 0*p(x)$ so $lambda=0$ for the eigenvector $p(x)=1$.
Another one is also apparent already. Let $p(x)=x$, you get $Sp(x)=-2x$, so $lambda=-2$ for the eigenvector $p(x)=x$.
Put the matrix together in this basis $ 1,x,x^2$
$$
S = beginpmatrix
0 & 0 & 2\
0 & -2 & 0\
0 & 0 & -4\
endpmatrix
$$
This is the matrix you seek to find the eigenvalues and eigenvectors for. There are the two that we've already found. The only one left will turn out to be $lambda=-4$ with $p(x)=-1+0*x+2*x^2$. In that case $Sp(x)=4-8x^2=(-4)*p(x)$.
answered Sep 11 at 4:12
AHusain
2,7502816
answered Sep 11 at 4:12
AHusain
2,7502816
answered Sep 11 at 4:12
AHusain
2,7502816
answered Sep 11 at 4:12
AHusain
2,7502816
2,7502816
add a comment |
add a comment |
1
The definition of eigenvector and eigenvalue is similar in this context: you need to find a non-zero function p(x) (often called an eigenfunction) satisfying $Sp(x) = lambda p(x)$ for some $lambda$.
– Matt
Sep 11 at 2:52
hey..i have tried with p(x)=1+x+x^2. And got Sp(x)= -4x^2-2x+2.. is the eigen vector (2,-2,-4). thank you!
– Misha
Sep 11 at 3:05
1
That $S(p(x))=-4x^2-2x+2$ is not proportional with a $lambda$ to $p(x)=1+x+x^2$. So that is not an eigenvector.
– AHusain
Sep 11 at 3:13
2
It is fairly likely that the problem is intended for the vector space of polynomials with a fixed degree bound (usually 2 or 3), and real coefficients.
– Will Jagy
Sep 11 at 3:17
1
In that case, you might as well take a fixed basis of "vectors" $1,x,x^2.$ Calculate what $S$ does to each of these. For each vector, the column of the matrix are the coefficients of what $S$ does to it... Then you want, finally, the eigenvalues of that matrix, and the eigenvectors (which are of the form $a + b x + c x^2 $)
– Will Jagy
Sep 11 at 3:48