Mixing
When is the nullspace unique in this case?
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 '18 at 15:56
Iamanon
1157
add a comment |
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 '18 at 15:56
Iamanon
1157
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17
add a comment |
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 '18 at 15:56
Iamanon
1157
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
linear-algebra matrices projection projection-matrices
asked Nov 21 '18 at 15:56
Iamanon
1157
asked Nov 21 '18 at 15:56
Iamanon
1157
asked Nov 21 '18 at 15:56
Iamanon
1157
asked Nov 21 '18 at 15:56
Iamanon
1157
asked Nov 21 '18 at 15:56
Iamanon
1157
1157
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17
add a comment |
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17
1
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
1
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17
add a comment |
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1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 '18 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 '18 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 '18 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 '18 at 16:17