Prove there exists $x neq 0$ such that $(I - M)x = 0$, where $(I - M)^-1$ does not exist. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving that $A$ is singular iff it's not injectiveCan we say that there exist an integer $n$ such $A+nB$ invertible?Does there exists a vector v such that $Avneq 0$ but $A^2v=0$existence of $lambda$ in $V^*$ in an $n$ dimensional vector spaceA matrix $M$ is similar to [$T$]$_B$ $iff$ there exists a basis $B'$ such that $M =$ [$T$]$_B'$Does there exist a basis for the set of $2times 2$ matrices such that all basis elements are invertible?Prove that there exists a vector $v$ such that $|v|=sqrt2$ and $|Tv|=5$Prove that there exists a vector $u ∈ W$ such that $u · v = 0$ and $u · u = 1$.Prove that exist some base of V such that does not have any vector from subspaceProve that there exists a non-zero vector $v in V$ such that $v perp U$Suppose $S$ is angle preserving, show that there is a constant such that $S=cT$ for some orthogonal map $T$
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Prove there exists $x neq 0$ such that $(I - M)x = 0$, where $(I - M)^-1$ does not exist.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving that $A$ is singular iff it's not injectiveCan we say that there exist an integer $n$ such $A+nB$ invertible?Does there exists a vector v such that $Avneq 0$ but $A^2v=0$existence of $lambda$ in $V^*$ in an $n$ dimensional vector spaceA matrix $M$ is similar to [$T$]$_B$ $iff$ there exists a basis $B'$ such that $M =$ [$T$]$_B'$Does there exist a basis for the set of $2times 2$ matrices such that all basis elements are invertible?Prove that there exists a vector $v$ such that $|v|=sqrt2$ and $|Tv|=5$Prove that there exists a vector $u ∈ W$ such that $u · v = 0$ and $u · u = 1$.Prove that exist some base of V such that does not have any vector from subspaceProve that there exists a non-zero vector $v in V$ such that $v perp U$Suppose $S$ is angle preserving, show that there is a constant such that $S=cT$ for some orthogonal map $T$
$begingroup$
How can I show that there exists a nonzero vector $x$ such that, provided $(I - M)$ is not invertible, we have $(I - M)x = 0$?
I'm not sure about how to go about this. I think that I'll need to use the fact that $(I - M)$ is not invertible, but I have no idea where.
Any help is appreciated.
linear-algebra
$endgroup$
|
show 1 more comment
$begingroup$
How can I show that there exists a nonzero vector $x$ such that, provided $(I - M)$ is not invertible, we have $(I - M)x = 0$?
I'm not sure about how to go about this. I think that I'll need to use the fact that $(I - M)$ is not invertible, but I have no idea where.
Any help is appreciated.
linear-algebra
$endgroup$
$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
1
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47
|
show 1 more comment
$begingroup$
How can I show that there exists a nonzero vector $x$ such that, provided $(I - M)$ is not invertible, we have $(I - M)x = 0$?
I'm not sure about how to go about this. I think that I'll need to use the fact that $(I - M)$ is not invertible, but I have no idea where.
Any help is appreciated.
linear-algebra
$endgroup$
How can I show that there exists a nonzero vector $x$ such that, provided $(I - M)$ is not invertible, we have $(I - M)x = 0$?
I'm not sure about how to go about this. I think that I'll need to use the fact that $(I - M)$ is not invertible, but I have no idea where.
Any help is appreciated.
linear-algebra
linear-algebra
edited Apr 8 at 13:43
asked Apr 8 at 13:33
user660922
$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
1
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47
|
show 1 more comment
$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
1
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47
$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
1
1
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $uneq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $uneq v$ so $u-vneq 0$.
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
$endgroup$
add a comment |
$begingroup$
Let $M:ell_2rightarrow ell_2$ be given by $Me_i=e_i-e_i+1$, where $e_i$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_i=e_i+1$. Note that $(I-M)$ is not invertible as there is no $xinell_2$ such that $(I-M)x=e_1$, but $(I-M)xneq 0$ for all $xneq 0$.
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
$endgroup$
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $uneq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $uneq v$ so $u-vneq 0$.
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
$endgroup$
add a comment |
$begingroup$
Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $uneq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $uneq v$ so $u-vneq 0$.
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
$endgroup$
add a comment |
$begingroup$
Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $uneq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $uneq v$ so $u-vneq 0$.
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
$endgroup$
Assume that $I-M$ is not invertible. Then, $I-M$ is not an injection (aka not 1-to-1) (see here) and so there exists $u$ and $v$ with $uneq v$ such that $(I-M)v = (I-M)u$. Thus $(I-M)(u-v) = 0$ and by assumption $uneq v$ so $u-vneq 0$.
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
edited Apr 8 at 14:01
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
answered Apr 8 at 13:55
Tony S.F.Tony S.F.
3,51421030
3,51421030
add a comment |
add a comment |
$begingroup$
Let $M:ell_2rightarrow ell_2$ be given by $Me_i=e_i-e_i+1$, where $e_i$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_i=e_i+1$. Note that $(I-M)$ is not invertible as there is no $xinell_2$ such that $(I-M)x=e_1$, but $(I-M)xneq 0$ for all $xneq 0$.
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
$endgroup$
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
add a comment |
$begingroup$
Let $M:ell_2rightarrow ell_2$ be given by $Me_i=e_i-e_i+1$, where $e_i$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_i=e_i+1$. Note that $(I-M)$ is not invertible as there is no $xinell_2$ such that $(I-M)x=e_1$, but $(I-M)xneq 0$ for all $xneq 0$.
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
$endgroup$
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
add a comment |
$begingroup$
Let $M:ell_2rightarrow ell_2$ be given by $Me_i=e_i-e_i+1$, where $e_i$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_i=e_i+1$. Note that $(I-M)$ is not invertible as there is no $xinell_2$ such that $(I-M)x=e_1$, but $(I-M)xneq 0$ for all $xneq 0$.
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
$endgroup$
Let $M:ell_2rightarrow ell_2$ be given by $Me_i=e_i-e_i+1$, where $e_i$ is the standard unit vector. Then $(I-M)$ is given by $(I-M)e_i=e_i+1$. Note that $(I-M)$ is not invertible as there is no $xinell_2$ such that $(I-M)x=e_1$, but $(I-M)xneq 0$ for all $xneq 0$.
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
answered Apr 8 at 13:47
Floris ClaassensFloris Claassens
1,47429
1,47429
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
add a comment |
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
$begingroup$
Note that this kind of construction only works in infinite dimensional space, for finite dimensional spaces you can follow the suggestions in the comments on your question.
$endgroup$
– Floris Claassens
Apr 8 at 13:49
add a comment |
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$begingroup$
You can't. If $I-M$ is invertible, it is bijective, hence as $(I-M)0=0$ there is no non-zero vector that gets mapped to 0 due to injectivity.
$endgroup$
– Floris Claassens
Apr 8 at 13:40
$begingroup$
This is false : if $(I-M)$ is invertible, then multiplying $(I-M)x=0$ by $(I-M)^-1$ gives $x=0$.
$endgroup$
– Segipp
Apr 8 at 13:40
$begingroup$
Sorry - I know that $(I - M)$ is singular. So it is NOT invertible.
$endgroup$
– user660922
Apr 8 at 13:43
1
$begingroup$
If $(I-M)$ is not invertible, then $ker(I-M)neq 0$. Try use this.
$endgroup$
– Matheus Nunes
Apr 8 at 13:45
$begingroup$
So since ker$(I - M) neq 0$, there exists some $v$ such that $Mv = 0$? Is that right?
$endgroup$
– user660922
Apr 8 at 13:47