Geometric understanding of subtracting lambda from diagonals
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Given the definition of eigenvalues/eigenvectors:
$Av = lambda v $
you could rearrange the terms to be:
$(A - lambda I)v = 0$
Geometrically, the first equation says that multiplying by $A$ is the same as scaling the vector $v$ by $lambda$. However, in the second equation, how do you visualize the effect of subtracting the matrix $lambda I$ from matrix $A$ and how does that induce a linearly dependent set of basis vectors?
TL;DR: I understand that the new matrix $(A-lambda I)$ collapses the span of $v$ into a lower dimension but I don't understand how $A$ relates to $(A-lambda I)$ geometrically.
linear-algebra geometry eigenvalues-eigenvectors visualization
asked Nov 24 at 16:44
hlinee
634
add a comment |
up vote
6
down vote
favorite
Given the definition of eigenvalues/eigenvectors:
$Av = lambda v $
you could rearrange the terms to be:
$(A - lambda I)v = 0$
Geometrically, the first equation says that multiplying by $A$ is the same as scaling the vector $v$ by $lambda$. However, in the second equation, how do you visualize the effect of subtracting the matrix $lambda I$ from matrix $A$ and how does that induce a linearly dependent set of basis vectors?
TL;DR: I understand that the new matrix $(A-lambda I)$ collapses the span of $v$ into a lower dimension but I don't understand how $A$ relates to $(A-lambda I)$ geometrically.
linear-algebra geometry eigenvalues-eigenvectors visualization
asked Nov 24 at 16:44
hlinee
634
3
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given the definition of eigenvalues/eigenvectors:
$Av = lambda v $
you could rearrange the terms to be:
$(A - lambda I)v = 0$
Geometrically, the first equation says that multiplying by $A$ is the same as scaling the vector $v$ by $lambda$. However, in the second equation, how do you visualize the effect of subtracting the matrix $lambda I$ from matrix $A$ and how does that induce a linearly dependent set of basis vectors?
TL;DR: I understand that the new matrix $(A-lambda I)$ collapses the span of $v$ into a lower dimension but I don't understand how $A$ relates to $(A-lambda I)$ geometrically.
linear-algebra geometry eigenvalues-eigenvectors visualization
asked Nov 24 at 16:44
hlinee
634
Given the definition of eigenvalues/eigenvectors:
$Av = lambda v $
you could rearrange the terms to be:
$(A - lambda I)v = 0$
Geometrically, the first equation says that multiplying by $A$ is the same as scaling the vector $v$ by $lambda$. However, in the second equation, how do you visualize the effect of subtracting the matrix $lambda I$ from matrix $A$ and how does that induce a linearly dependent set of basis vectors?
TL;DR: I understand that the new matrix $(A-lambda I)$ collapses the span of $v$ into a lower dimension but I don't understand how $A$ relates to $(A-lambda I)$ geometrically.
linear-algebra geometry eigenvalues-eigenvectors visualization
linear-algebra geometry eigenvalues-eigenvectors visualization
asked Nov 24 at 16:44
hlinee
634
asked Nov 24 at 16:44
hlinee
634
asked Nov 24 at 16:44
hlinee
634
asked Nov 24 at 16:44
hlinee
634
asked Nov 24 at 16:44
hlinee
634
634
3
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13
add a comment |
3
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13
3
3
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13
add a comment |
1 Answer
1
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2
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May I ask what do you understand by geometric relation or visualization? What you said
$(A-lambda I)$ collapses the span of $v$ into a lower dimension
is exactly how I visualize it. To be more precisely, it completely removes the image of ${rm span}(v)$. Thus only the co-space of ${rm span}(v)$ is possibly nontrivially acted upon.
The action on the remaining co-space is also modified, namely by subtracting $lambda$ times the input vector from the image, $Aw - lambda w$ (for $w perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.
answered Nov 24 at 19:51
rehctawrats
21018
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
May I ask what do you understand by geometric relation or visualization? What you said
$(A-lambda I)$ collapses the span of $v$ into a lower dimension
is exactly how I visualize it. To be more precisely, it completely removes the image of ${rm span}(v)$. Thus only the co-space of ${rm span}(v)$ is possibly nontrivially acted upon.
The action on the remaining co-space is also modified, namely by subtracting $lambda$ times the input vector from the image, $Aw - lambda w$ (for $w perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.
answered Nov 24 at 19:51
rehctawrats
21018
add a comment |
up vote
2
down vote
May I ask what do you understand by geometric relation or visualization? What you said
$(A-lambda I)$ collapses the span of $v$ into a lower dimension
is exactly how I visualize it. To be more precisely, it completely removes the image of ${rm span}(v)$. Thus only the co-space of ${rm span}(v)$ is possibly nontrivially acted upon.
The action on the remaining co-space is also modified, namely by subtracting $lambda$ times the input vector from the image, $Aw - lambda w$ (for $w perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.
answered Nov 24 at 19:51
rehctawrats
21018
add a comment |
up vote
2
down vote
up vote
2
down vote
May I ask what do you understand by geometric relation or visualization? What you said
$(A-lambda I)$ collapses the span of $v$ into a lower dimension
is exactly how I visualize it. To be more precisely, it completely removes the image of ${rm span}(v)$. Thus only the co-space of ${rm span}(v)$ is possibly nontrivially acted upon.
The action on the remaining co-space is also modified, namely by subtracting $lambda$ times the input vector from the image, $Aw - lambda w$ (for $w perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.
answered Nov 24 at 19:51
rehctawrats
21018
May I ask what do you understand by geometric relation or visualization? What you said
$(A-lambda I)$ collapses the span of $v$ into a lower dimension
is exactly how I visualize it. To be more precisely, it completely removes the image of ${rm span}(v)$. Thus only the co-space of ${rm span}(v)$ is possibly nontrivially acted upon.
The action on the remaining co-space is also modified, namely by subtracting $lambda$ times the input vector from the image, $Aw - lambda w$ (for $w perp v$). In the case that $v$ has geometric multiplicity one, then this just happens to not be collapsed to zero.
answered Nov 24 at 19:51
rehctawrats
21018
answered Nov 24 at 19:51
rehctawrats
21018
answered Nov 24 at 19:51
rehctawrats
21018
answered Nov 24 at 19:51
rehctawrats
21018
21018
add a comment |
add a comment |
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3
Interesting question. One possible idea. If $A$ happened to be diagonal then you can visualize the parallelepiped formed by stretching the basis vectors using the diagonal elements as scale factors. Subtracting a constant from the diagonal changes that parallelepiped (not simply by scaling it). When you subtract an eigenvalue the pppd has zero volume.
– Ethan Bolker
Nov 24 at 16:53
First, how do you visualize $A$ geometrically?
– Rahul
Nov 24 at 20:13