Proper method to write vectors which span $mathrmker(T)$ The Next CEO of Stack OverflowShow that the span of two vectors is T-invariantBasis of a subspace (What does it mean for vectors to span a subspace?)Number of vectors in a set & span of a setFind $mathrmKer(T)$ and $mathrmIm(T)$ of the following linear transformation with basesDescribe the span of these vectors as the span of as few vectors as possible.The dimension of the intersection of two vector subspacesFinding $T:Vto V$ By The Span Of $ker(T)$ And $operatornameIm(T)$Are the span of these vectors the same?Assume the set u1…u4 is an orthogonal basis for R4. Write v as the sum of two vectors, one in span u1 and the other in span u2, u3, u4Span of vectors with more entries in each vector than the amount of vectors
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Proper method to write vectors which span $mathrmker(T)$
The Next CEO of Stack OverflowShow that the span of two vectors is T-invariantBasis of a subspace (What does it mean for vectors to span a subspace?)Number of vectors in a set & span of a setFind $mathrmKer(T)$ and $mathrmIm(T)$ of the following linear transformation with basesDescribe the span of these vectors as the span of as few vectors as possible.The dimension of the intersection of two vector subspacesFinding $T:Vto V$ By The Span Of $ker(T)$ And $operatornameIm(T)$Are the span of these vectors the same?Assume the set u1…u4 is an orthogonal basis for R4. Write v as the sum of two vectors, one in span u1 and the other in span u2, u3, u4Span of vectors with more entries in each vector than the amount of vectors
$begingroup$
On my linear algebra exam, I was required to find the vectors which span $mathrmker(T)$, where the linear transformation $mathrmT$ was given. The official answers are $beginbmatrix
0 & 1\
-1 & 0
endbmatrix$
or
$beginbmatrix
0 & -1\
1 & 0
endbmatrix$.
I wrote my answer as
$
beginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbR
$
I did not get credit for this problem, as the official solutions says we had to write the vectors which span $mathrmker(T)$ and not the set $mathrmker(T)$.
I realise my answer is not completely right, as when $mathrma = 0$ the vector no longer spans $mathrmker(T)$. But other than that, is there something else I'm missing?
If I had written $mathrma in mathbbR - 0$, would my answer then be correct?
Also, what is the difference between $mathrmker(T)$ and the "set" $mathrmker(T)$ ?
linear-algebra linear-transformations
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
On my linear algebra exam, I was required to find the vectors which span $mathrmker(T)$, where the linear transformation $mathrmT$ was given. The official answers are $beginbmatrix
0 & 1\
-1 & 0
endbmatrix$
or
$beginbmatrix
0 & -1\
1 & 0
endbmatrix$.
I wrote my answer as
$
beginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbR
$
I did not get credit for this problem, as the official solutions says we had to write the vectors which span $mathrmker(T)$ and not the set $mathrmker(T)$.
I realise my answer is not completely right, as when $mathrma = 0$ the vector no longer spans $mathrmker(T)$. But other than that, is there something else I'm missing?
If I had written $mathrma in mathbbR - 0$, would my answer then be correct?
Also, what is the difference between $mathrmker(T)$ and the "set" $mathrmker(T)$ ?
linear-algebra linear-transformations
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago
add a comment |
$begingroup$
On my linear algebra exam, I was required to find the vectors which span $mathrmker(T)$, where the linear transformation $mathrmT$ was given. The official answers are $beginbmatrix
0 & 1\
-1 & 0
endbmatrix$
or
$beginbmatrix
0 & -1\
1 & 0
endbmatrix$.
I wrote my answer as
$
beginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbR
$
I did not get credit for this problem, as the official solutions says we had to write the vectors which span $mathrmker(T)$ and not the set $mathrmker(T)$.
I realise my answer is not completely right, as when $mathrma = 0$ the vector no longer spans $mathrmker(T)$. But other than that, is there something else I'm missing?
If I had written $mathrma in mathbbR - 0$, would my answer then be correct?
Also, what is the difference between $mathrmker(T)$ and the "set" $mathrmker(T)$ ?
linear-algebra linear-transformations
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
On my linear algebra exam, I was required to find the vectors which span $mathrmker(T)$, where the linear transformation $mathrmT$ was given. The official answers are $beginbmatrix
0 & 1\
-1 & 0
endbmatrix$
or
$beginbmatrix
0 & -1\
1 & 0
endbmatrix$.
I wrote my answer as
$
beginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbR
$
I did not get credit for this problem, as the official solutions says we had to write the vectors which span $mathrmker(T)$ and not the set $mathrmker(T)$.
I realise my answer is not completely right, as when $mathrma = 0$ the vector no longer spans $mathrmker(T)$. But other than that, is there something else I'm missing?
If I had written $mathrma in mathbbR - 0$, would my answer then be correct?
Also, what is the difference between $mathrmker(T)$ and the "set" $mathrmker(T)$ ?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 19:20
RootPi
asked Mar 27 at 19:14
RootPiRootPi
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:14
RootPiRootPi
33
asked Mar 27 at 19:14
RootPiRootPi
33
33
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
RootPi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago
add a comment |
$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago
$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By definition, the $kerT$ is a set. It is defined as follows:
$$kerT = xin E: Tx = 0$$
where $T:Eto V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this:
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright$$
However notice that
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright = operatornameSpanleft(beginbmatrix
0 & 1\
-1 & 0
endbmatrixright).$$
Here the $operatornameSpan(v) = lambda v:lambda in mathbbK$, where $mathbbK$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition, the $kerT$ is a set. It is defined as follows:
$$kerT = xin E: Tx = 0$$
where $T:Eto V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this:
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright$$
However notice that
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright = operatornameSpanleft(beginbmatrix
0 & 1\
-1 & 0
endbmatrixright).$$
Here the $operatornameSpan(v) = lambda v:lambda in mathbbK$, where $mathbbK$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
$endgroup$
add a comment |
$begingroup$
By definition, the $kerT$ is a set. It is defined as follows:
$$kerT = xin E: Tx = 0$$
where $T:Eto V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this:
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright$$
However notice that
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright = operatornameSpanleft(beginbmatrix
0 & 1\
-1 & 0
endbmatrixright).$$
Here the $operatornameSpan(v) = lambda v:lambda in mathbbK$, where $mathbbK$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
$endgroup$
add a comment |
$begingroup$
By definition, the $kerT$ is a set. It is defined as follows:
$$kerT = xin E: Tx = 0$$
where $T:Eto V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this:
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright$$
However notice that
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright = operatornameSpanleft(beginbmatrix
0 & 1\
-1 & 0
endbmatrixright).$$
Here the $operatornameSpan(v) = lambda v:lambda in mathbbK$, where $mathbbK$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
$endgroup$
By definition, the $kerT$ is a set. It is defined as follows:
$$kerT = xin E: Tx = 0$$
where $T:Eto V.$
Your answer would have been correct if you were asked to just state the kernel. You would write something like this:
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright$$
However notice that
$$kerT = leftbeginbmatrix
0 & a\
-a & 0
endbmatrix,ain mathbbKright = operatornameSpanleft(beginbmatrix
0 & 1\
-1 & 0
endbmatrixright).$$
Here the $operatornameSpan(v) = lambda v:lambda in mathbbK$, where $mathbbK$ is the field over which you have defined your vector space. So when they ask you for the vectors that span the kernel you have to esentially tell them the vectors whose scalar mutiples form the kernel. You could also find a basis for the kernel and that would be just fine too.
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
answered Mar 27 at 19:27
model_checkermodel_checker
4,41621931
4,41621931
add a comment |
add a comment |
RootPi is a new contributor. Be nice, and check out our Code of Conduct.
RootPi is a new contributor. Be nice, and check out our Code of Conduct.
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RootPi is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
You were asked for some specific vector that generates the space, but that’s not what you gave in your answer. I would’ve given you at least partial credit for that, though.
$endgroup$
– amd
Mar 27 at 20:18
$begingroup$
What's the linear transformation $T$?
$endgroup$
– egreg
Mar 27 at 21:42
$begingroup$
@egreg $mathrmT:mathcalM_22 to mathcalM_22$ defined by $mathrmT(A) = A + A^T$ where $mathcalM_22$ is the set of all $mathrm2 times 2$ matrices.
$endgroup$
– RootPi
2 days ago