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Identifying Nullity (Linear AlgebrA)
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[Q11
Q2
I understand Nullity in a computational-matrix form context. I.e, identify the nullity of a given arbitrary matrix A. This is done through RREF.
However, in the context of projections and dot products, how do matrices that form dot products/perpendicular matrixes affect the nullity? Are there specifications of the matrix I should be looking for?
Thank You
linear-algebra linear-transformations orthonormal projection-matrices
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
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up vote
-1
down vote
favorite
[Q11
Q2
I understand Nullity in a computational-matrix form context. I.e, identify the nullity of a given arbitrary matrix A. This is done through RREF.
However, in the context of projections and dot products, how do matrices that form dot products/perpendicular matrixes affect the nullity? Are there specifications of the matrix I should be looking for?
Thank You
linear-algebra linear-transformations orthonormal projection-matrices
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
[Q11
Q2
I understand Nullity in a computational-matrix form context. I.e, identify the nullity of a given arbitrary matrix A. This is done through RREF.
However, in the context of projections and dot products, how do matrices that form dot products/perpendicular matrixes affect the nullity? Are there specifications of the matrix I should be looking for?
Thank You
linear-algebra linear-transformations orthonormal projection-matrices
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
[Q11
Q2
I understand Nullity in a computational-matrix form context. I.e, identify the nullity of a given arbitrary matrix A. This is done through RREF.
However, in the context of projections and dot products, how do matrices that form dot products/perpendicular matrixes affect the nullity? Are there specifications of the matrix I should be looking for?
Thank You
linear-algebra linear-transformations orthonormal projection-matrices
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
asked Jul 21 at 11:44
PERTURBATIONFLOW
396
396
The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00
add a comment |Â
The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00
The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00
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The nullity is the dimension of the subspace that is the inverse image of $0$. In the case of matrices the transformation was the matrix multiplying a vector from the left. For Q1 the transformation is the orthogonal projection onto $V$. If you project $mathbbR^5$ onto $V$, which has dimention $3$, the size of the space projected to $0$ can be determined using the rank-nullity theorem $dim(Range(T))+dim(Ker(T))=dim(Domain(T))$, where $T=proj_V$, $Domain(T)=mathbbR^5$, and $Range(T)=V$.
– user574889
Jul 21 at 11:58
For Q2, the same idea applies. One only needs to realize that because $wneq0$, then $Range(T)$ is all of $mathbbR$. In fact, $Tleft(fracr^2wright)=r$ already returns all real values as $r$ moves along all reals.
– user574889
Jul 21 at 11:59
@cactus so for Q1, would nullity be 2 since 3+2 = 5? And likewise, for Q2, would nullity be simply 1? Q1: 2 since we simply use nullity-rank theorem so then 5-3 = 2 Q2: 1 since we span an infinite space of reals?
– PERTURBATIONFLOW
Jul 22 at 2:55
That's right. In Q2 the fact that the range of the transformation is infinite is not what is important, but that that range has dimension $1$, and therefore $dim(Ker(T))=dim(Domain(T))-dim(Range(T))=dim(mathbbR^2)-dim(mathbbR)=2-1=1$.
– user574889
Jul 22 at 12:00