Proving existence of a linear transformation with given properties
$begingroup$
The question is as follows:
Suppose that $V$ is a vector space over $mathbb{C}$ of dimension $3$. Fix a non-zero vector $vin V$ and define
$$U:={Tinmathcal{L}(V):vmbox{ is and eigenvector of }T}.$$
Now fix an arbitrary basis $B$ of $V$. Show that there is a nonzero linear transformation $Sin U$ such that
$$mathcal{M}(S,B)=begin{bmatrix}a&a&c\b&a&c\b&b&cend{bmatrix}.$$
Note that the notation $mathcal{L}(V)$ is used to denote the linear operators on $V$ and $mathcal{M}(S,B)$ is the matrix of $S$ with respect to $B$. I'm not sure if this notation is conventional or not, and it's probably clear from context...
I am able to show that $U$ is a subspace of $mathcal{L}(V)$ of dimension $7$, but beyond that I'm not sure how to approach this. I know a nonconstructive solution exists, but I keep finding myself actually trying to compute values of $a,b,c$ (which is a poor approach even without knowledge of a nonconstructive solution, since the basis is arbitrarily chosen). This is a homework question, so I am requesting hints, not a full solution.
linear-algebra matrices linear-transformations
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Suppose that $V$ is a vector space over $mathbb{C}$ of dimension $3$. Fix a non-zero vector $vin V$ and define
$$U:={Tinmathcal{L}(V):vmbox{ is and eigenvector of }T}.$$
Now fix an arbitrary basis $B$ of $V$. Show that there is a nonzero linear transformation $Sin U$ such that
$$mathcal{M}(S,B)=begin{bmatrix}a&a&c\b&a&c\b&b&cend{bmatrix}.$$
Note that the notation $mathcal{L}(V)$ is used to denote the linear operators on $V$ and $mathcal{M}(S,B)$ is the matrix of $S$ with respect to $B$. I'm not sure if this notation is conventional or not, and it's probably clear from context...
I am able to show that $U$ is a subspace of $mathcal{L}(V)$ of dimension $7$, but beyond that I'm not sure how to approach this. I know a nonconstructive solution exists, but I keep finding myself actually trying to compute values of $a,b,c$ (which is a poor approach even without knowledge of a nonconstructive solution, since the basis is arbitrarily chosen). This is a homework question, so I am requesting hints, not a full solution.
linear-algebra matrices linear-transformations
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
$endgroup$
2
$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23
add a comment |
$begingroup$
The question is as follows:
Suppose that $V$ is a vector space over $mathbb{C}$ of dimension $3$. Fix a non-zero vector $vin V$ and define
$$U:={Tinmathcal{L}(V):vmbox{ is and eigenvector of }T}.$$
Now fix an arbitrary basis $B$ of $V$. Show that there is a nonzero linear transformation $Sin U$ such that
$$mathcal{M}(S,B)=begin{bmatrix}a&a&c\b&a&c\b&b&cend{bmatrix}.$$
Note that the notation $mathcal{L}(V)$ is used to denote the linear operators on $V$ and $mathcal{M}(S,B)$ is the matrix of $S$ with respect to $B$. I'm not sure if this notation is conventional or not, and it's probably clear from context...
I am able to show that $U$ is a subspace of $mathcal{L}(V)$ of dimension $7$, but beyond that I'm not sure how to approach this. I know a nonconstructive solution exists, but I keep finding myself actually trying to compute values of $a,b,c$ (which is a poor approach even without knowledge of a nonconstructive solution, since the basis is arbitrarily chosen). This is a homework question, so I am requesting hints, not a full solution.
linear-algebra matrices linear-transformations
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
$endgroup$
The question is as follows:
Suppose that $V$ is a vector space over $mathbb{C}$ of dimension $3$. Fix a non-zero vector $vin V$ and define
$$U:={Tinmathcal{L}(V):vmbox{ is and eigenvector of }T}.$$
Now fix an arbitrary basis $B$ of $V$. Show that there is a nonzero linear transformation $Sin U$ such that
$$mathcal{M}(S,B)=begin{bmatrix}a&a&c\b&a&c\b&b&cend{bmatrix}.$$
Note that the notation $mathcal{L}(V)$ is used to denote the linear operators on $V$ and $mathcal{M}(S,B)$ is the matrix of $S$ with respect to $B$. I'm not sure if this notation is conventional or not, and it's probably clear from context...
I am able to show that $U$ is a subspace of $mathcal{L}(V)$ of dimension $7$, but beyond that I'm not sure how to approach this. I know a nonconstructive solution exists, but I keep finding myself actually trying to compute values of $a,b,c$ (which is a poor approach even without knowledge of a nonconstructive solution, since the basis is arbitrarily chosen). This is a homework question, so I am requesting hints, not a full solution.
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
asked Dec 7 '18 at 4:56
AtsinaAtsina
791116
791116
2
$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23
add a comment |
2
$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23
2
2
$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $dim V=9$, $dim U=7$, and $dim W=3$, it follows that $dim(Ucap W)ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3times3$ matrices is $9$, and $mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
$endgroup$
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
|
show 5 more comments
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$begingroup$
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $dim V=9$, $dim U=7$, and $dim W=3$, it follows that $dim(Ucap W)ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3times3$ matrices is $9$, and $mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
$endgroup$
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
|
show 5 more comments
$begingroup$
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $dim V=9$, $dim U=7$, and $dim W=3$, it follows that $dim(Ucap W)ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3times3$ matrices is $9$, and $mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
$endgroup$
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
|
show 5 more comments
$begingroup$
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $dim V=9$, $dim U=7$, and $dim W=3$, it follows that $dim(Ucap W)ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3times3$ matrices is $9$, and $mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
$endgroup$
Edit: Here is the solution presented in the comments below. My original solution is after this.
Note that $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Let $W$ be the subspace of linear operators with matrices of the form of $M$. Then since $dim V=9$, $dim U=7$, and $dim W=3$, it follows that $dim(Ucap W)ge1$. The desired result follows.
Observe that the dimension of the space of linear operators with matrices of the form of $M$ (with respect to $B$) is $3$. Since the dimension of the full space of $3times3$ matrices is $9$, and $mbox{dim }U=7$, by the pigeonhole principle, the desired result follows.
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
edited Dec 8 '18 at 2:43
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
answered Dec 7 '18 at 5:30
AtsinaAtsina
791116
791116
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
|
show 5 more comments
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
It isn't pigeonhole principle, but a consequence of $dim Vgedim(U+W)=dim U+dim W-dim(Ucap W)$. Put $dim V=9, dim U=7$ and $dim W=3$, we get $dim(Ucap W)ge1$.
$endgroup$
– user1551
Dec 7 '18 at 11:43
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
@user1551 Your argument that the intersection of the spaces must be non-zero is effectively the same as saying you have 10 pigeons and only 9 pigeonholes, is it not?
$endgroup$
– Atsina
Dec 7 '18 at 17:01
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
How? Please define your "pigeons" and "pigeonholes" rigourously. You seem to be bending a simple dimension argument into one using the pigeonhole principle. This of course can be done technically, but I would say that's a very contrived way to disguise a technical detail as a hand wavy, "intuitive"-looking argument that sounds familiar to the laymen.
$endgroup$
– user1551
Dec 7 '18 at 17:22
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 I don't see how it's hand wavy. We have a vector space of dimension $9$, a subspace of dimension $7$, and another subspace of dimension $3$...these subspaces must intersect.
$endgroup$
– Atsina
Dec 7 '18 at 17:38
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
$begingroup$
@user1551 That's where the pigeonhole principle arises. If they didn't intersect, then we would somehow have $10$ dimensions hiding inside a $9$ dimensional space.
$endgroup$
– Atsina
Dec 7 '18 at 17:41
|
show 5 more comments
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$begingroup$
I think you are meant to use that 7+3>9. 3 being the dimension of space of matrices of the form M and 9 being the dimension of space of all matrices
$endgroup$
– user25959
Dec 7 '18 at 5:20
$begingroup$
@user25959 Ha! It's obvious when you see it. Pigeonhole principle. Thanks!
$endgroup$
– Atsina
Dec 7 '18 at 5:23