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Geometric interpretation of why some matrices don't have eigenvalues
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I don't understand how to geometrically interpret the formula $Av = lambda v$
where $A$ is a matrix and $v, lambda$ are the corresponding eigenvectors and eigenvalues.
For instance, why does the matrix beginbmatrix
0 & 1\
-1 & 0
endbmatrix
not have any eigenvalues? How can I explain this, geometrically without just saying it's not the case because $lambda^2+1=0$ doesn't have any real solutions?
linear-algebra geometry eigenvalues-eigenvectors linear-transformations
edited 3 hours ago
Scientifica
5,07121331
asked 4 hours ago
novo
39318
 |Â
show 3 more comments
up vote
4
down vote
favorite
I don't understand how to geometrically interpret the formula $Av = lambda v$
where $A$ is a matrix and $v, lambda$ are the corresponding eigenvectors and eigenvalues.
For instance, why does the matrix beginbmatrix
0 & 1\
-1 & 0
endbmatrix
not have any eigenvalues? How can I explain this, geometrically without just saying it's not the case because $lambda^2+1=0$ doesn't have any real solutions?
linear-algebra geometry eigenvalues-eigenvectors linear-transformations
edited 3 hours ago
Scientifica
5,07121331
asked 4 hours ago
novo
39318
$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
2
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
1
Ok. Thank you for the help!
– novo
4 hours ago
1
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago
 |Â
show 3 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I don't understand how to geometrically interpret the formula $Av = lambda v$
where $A$ is a matrix and $v, lambda$ are the corresponding eigenvectors and eigenvalues.
For instance, why does the matrix beginbmatrix
0 & 1\
-1 & 0
endbmatrix
not have any eigenvalues? How can I explain this, geometrically without just saying it's not the case because $lambda^2+1=0$ doesn't have any real solutions?
linear-algebra geometry eigenvalues-eigenvectors linear-transformations
edited 3 hours ago
Scientifica
5,07121331
asked 4 hours ago
novo
39318
I don't understand how to geometrically interpret the formula $Av = lambda v$
where $A$ is a matrix and $v, lambda$ are the corresponding eigenvectors and eigenvalues.
For instance, why does the matrix beginbmatrix
0 & 1\
-1 & 0
endbmatrix
not have any eigenvalues? How can I explain this, geometrically without just saying it's not the case because $lambda^2+1=0$ doesn't have any real solutions?
linear-algebra geometry eigenvalues-eigenvectors linear-transformations
linear-algebra geometry eigenvalues-eigenvectors linear-transformations
edited 3 hours ago
Scientifica
5,07121331
asked 4 hours ago
novo
39318
edited 3 hours ago
Scientifica
5,07121331
asked 4 hours ago
novo
39318
edited 3 hours ago
Scientifica
5,07121331
edited 3 hours ago
Scientifica
5,07121331
edited 3 hours ago
Scientifica
5,07121331
5,07121331
asked 4 hours ago
novo
39318
asked 4 hours ago
novo
39318
asked 4 hours ago
novo
39318
39318
$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
2
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
1
Ok. Thank you for the help!
– novo
4 hours ago
1
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago
 |Â
show 3 more comments
$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
2
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
1
Ok. Thank you for the help!
– novo
4 hours ago
1
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago
$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
2
2
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
1
1
Ok. Thank you for the help!
– novo
4 hours ago
Ok. Thank you for the help!
– novo
4 hours ago
1
1
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
You can see an $n$-dimensional vector space as $mathbbR^n$. $Av=lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $thetainmathbb R$, the matrix
beginpmatrix
costheta & -sintheta\
sintheta & costheta
endpmatrix
represents the rotation of angle $theta$. You get your matrix for $theta=-dfracpi2$. You can see geometrically that such a rotation preserves no line.
answered 4 hours ago
Scientifica
5,07121331
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You can see an $n$-dimensional vector space as $mathbbR^n$. $Av=lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $thetainmathbb R$, the matrix
beginpmatrix
costheta & -sintheta\
sintheta & costheta
endpmatrix
represents the rotation of angle $theta$. You get your matrix for $theta=-dfracpi2$. You can see geometrically that such a rotation preserves no line.
answered 4 hours ago
Scientifica
5,07121331
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
You can see an $n$-dimensional vector space as $mathbbR^n$. $Av=lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $thetainmathbb R$, the matrix
beginpmatrix
costheta & -sintheta\
sintheta & costheta
endpmatrix
represents the rotation of angle $theta$. You get your matrix for $theta=-dfracpi2$. You can see geometrically that such a rotation preserves no line.
answered 4 hours ago
Scientifica
5,07121331
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You can see an $n$-dimensional vector space as $mathbbR^n$. $Av=lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $thetainmathbb R$, the matrix
beginpmatrix
costheta & -sintheta\
sintheta & costheta
endpmatrix
represents the rotation of angle $theta$. You get your matrix for $theta=-dfracpi2$. You can see geometrically that such a rotation preserves no line.
answered 4 hours ago
Scientifica
5,07121331
You can see an $n$-dimensional vector space as $mathbbR^n$. $Av=lambda v$ means that the image of $v$ is in the same direction as $v$. You can also interpret it as "$A$ fixes the line spanned by $v$". You can see $lambda$ as a factor that expresses how $v$ changes.
One can understand geometrically why the matrix you gave has no eigenvalues. In fact, you can clearly see why it doesn't fixe any line. Indeed, here's a thing: given $thetainmathbb R$, the matrix
beginpmatrix
costheta & -sintheta\
sintheta & costheta
endpmatrix
represents the rotation of angle $theta$. You get your matrix for $theta=-dfracpi2$. You can see geometrically that such a rotation preserves no line.
answered 4 hours ago
Scientifica
5,07121331
answered 4 hours ago
Scientifica
5,07121331
answered 4 hours ago
Scientifica
5,07121331
answered 4 hours ago
Scientifica
5,07121331
5,07121331
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
add a comment |Â
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That makes sense. What happens if the angle $theta$ was different? I'm thinking if $theta = 0$, then we have the trivial case of no rotation. But what if $theta = pi$. It would rotate the vector, but they would still lie on the same line and the equality would hold? Does this mean the matrix $A$ has no eigenvalues if it rotates the vector anything but $theta=pi$ (aswell as the trivial $theta = 0, 2pi$, etc...?
– novo
4 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
That's right! With this geometric thinking, the matrix has an eigenvalue (and actually ALL the vectors are eigenvectors) if and only if $theta=kpi$ for some $kinmathbb Z$. You can also show this using the characteristic polynomial I guess.
– Scientifica
3 hours ago
add a comment |Â
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$Av$ is simply not a scalar multiple of $v$, that is, $Av$ is not parallel to $v$. Try graphing $Bv$ and $v$ for $B$'s that do have eigenvalues.
– Jack Bauer
4 hours ago
2
Ok, does this occur because the matrix $A$ rotates any vector $v$ you multiply it with, therefore no eigenvalues can exist for such matrices, because if $A$ rotates $v$ you can never have $Av$ equal to a scalar multiple of $v$. Is that correct?
– novo
4 hours ago
novo, you explained better than I ever could. I just said not a scalar multiple. You made it is stronger to rotation. If you're right. I'm quite sleepy right now.
– Jack Bauer
4 hours ago
1
Ok. Thank you for the help!
– novo
4 hours ago
1
@Jack yes. Most comments did not show properly for me, so I was referring to the comment about rotations.
– Niki Di Giano
4 hours ago