What does the following notation mean: $(L-lambda I)$
$begingroup$
In my tekstbook powervectors are discussed where
$(L-lambda I)^p textbf{v}=0$ was mentioned. Here is $L$ a linear operator on the vectorspace $V$, $lambda$ a scalar and $textbf{v} in V $.
My question: How is $L-lambda I$ definied? I'm confused because $L$ is here a function and not a matrix.
linear-algebra notation definition
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
$endgroup$
add a comment |
$begingroup$
In my tekstbook powervectors are discussed where
$(L-lambda I)^p textbf{v}=0$ was mentioned. Here is $L$ a linear operator on the vectorspace $V$, $lambda$ a scalar and $textbf{v} in V $.
My question: How is $L-lambda I$ definied? I'm confused because $L$ is here a function and not a matrix.
linear-algebra notation definition
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
$endgroup$
1
$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31
add a comment |
$begingroup$
In my tekstbook powervectors are discussed where
$(L-lambda I)^p textbf{v}=0$ was mentioned. Here is $L$ a linear operator on the vectorspace $V$, $lambda$ a scalar and $textbf{v} in V $.
My question: How is $L-lambda I$ definied? I'm confused because $L$ is here a function and not a matrix.
linear-algebra notation definition
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
$endgroup$
In my tekstbook powervectors are discussed where
$(L-lambda I)^p textbf{v}=0$ was mentioned. Here is $L$ a linear operator on the vectorspace $V$, $lambda$ a scalar and $textbf{v} in V $.
My question: How is $L-lambda I$ definied? I'm confused because $L$ is here a function and not a matrix.
linear-algebra notation definition
linear-algebra notation definition
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
edited Dec 10 '18 at 16:10
Keep_On_Cruising
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
asked Dec 10 '18 at 14:29
Keep_On_CruisingKeep_On_Cruising
1719
1719
1
$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31
add a comment |
1
$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31
1
1
$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31
$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$I$ is the identity operator on the vector space: $Ix = x$ for all $x in V$.
So $$(L - lambda I) x = L x - lambda I x = L x - lambda x$$
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
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add a comment |
$begingroup$
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$frac{mathrm{d}^2 u(t)}{mathrm{d}t^2}+2beta frac{mathrm{d}u(t)}{mathrm{d}t}+omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2beta D+omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=frac{mathrm{d}u(t)}{mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$left(frac{1}{c^2} partial_t^2-partial_x^2+mu Iright)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $partial_t to (-i omega)$ and $partial_x to (ik)$ in the case of a plane wave.
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
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add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
$I$ is the identity operator on the vector space: $Ix = x$ for all $x in V$.
So $$(L - lambda I) x = L x - lambda I x = L x - lambda x$$
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
$I$ is the identity operator on the vector space: $Ix = x$ for all $x in V$.
So $$(L - lambda I) x = L x - lambda I x = L x - lambda x$$
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
$I$ is the identity operator on the vector space: $Ix = x$ for all $x in V$.
So $$(L - lambda I) x = L x - lambda I x = L x - lambda x$$
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
$endgroup$
$I$ is the identity operator on the vector space: $Ix = x$ for all $x in V$.
So $$(L - lambda I) x = L x - lambda I x = L x - lambda x$$
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
answered Dec 10 '18 at 14:31
Robert IsraelRobert Israel
323k23213467
323k23213467
add a comment |
add a comment |
$begingroup$
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$frac{mathrm{d}^2 u(t)}{mathrm{d}t^2}+2beta frac{mathrm{d}u(t)}{mathrm{d}t}+omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2beta D+omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=frac{mathrm{d}u(t)}{mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$left(frac{1}{c^2} partial_t^2-partial_x^2+mu Iright)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $partial_t to (-i omega)$ and $partial_x to (ik)$ in the case of a plane wave.
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
$endgroup$
add a comment |
$begingroup$
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$frac{mathrm{d}^2 u(t)}{mathrm{d}t^2}+2beta frac{mathrm{d}u(t)}{mathrm{d}t}+omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2beta D+omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=frac{mathrm{d}u(t)}{mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$left(frac{1}{c^2} partial_t^2-partial_x^2+mu Iright)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $partial_t to (-i omega)$ and $partial_x to (ik)$ in the case of a plane wave.
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
$endgroup$
add a comment |
$begingroup$
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$frac{mathrm{d}^2 u(t)}{mathrm{d}t^2}+2beta frac{mathrm{d}u(t)}{mathrm{d}t}+omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2beta D+omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=frac{mathrm{d}u(t)}{mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$left(frac{1}{c^2} partial_t^2-partial_x^2+mu Iright)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $partial_t to (-i omega)$ and $partial_x to (ik)$ in the case of a plane wave.
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
$endgroup$
Robert did fully answer the question, but I'd like to add a different example: The differential equation of the harmonic oscillator is
$$frac{mathrm{d}^2 u(t)}{mathrm{d}t^2}+2beta frac{mathrm{d}u(t)}{mathrm{d}t}+omega^2 u(t)=0$$
And it can be rewritten as
$$(D^2+2beta D+omega^2 I)u(t)=0$$
Where $D$ is the differential operator: $Du(t)=frac{mathrm{d}u(t)}{mathrm{d}t}$.
Or the Klein-Gordon equation in $1+1$ dimensions:
$$left(frac{1}{c^2} partial_t^2-partial_x^2+mu Iright)u(t)=0$$
This kind of notation is very useful, because the differential operator will take it's eigenvalue on it's eigenfunction, so you can easily get the dispersion relation, with just substitutions of the eigenvalues: $partial_t to (-i omega)$ and $partial_x to (ik)$ in the case of a plane wave.
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
edited Dec 10 '18 at 15:35
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
answered Dec 10 '18 at 15:09
BotondBotond
5,8532832
5,8532832
add a comment |
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$begingroup$
You can think " I " as an identity function not a matrix
$endgroup$
– Euduardo
Dec 10 '18 at 14:31