Proving that $Awin langle wrangle implies A$ is of the form $lambda I_n$
$begingroup$
If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
$A=
begin{bmatrix}
lambda & & \
& ddots & a_{ij}\
& & lambda
end{bmatrix}$ is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.
But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.
How to show that $Awnotin langle wrangle$?
Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?
For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$
linear-algebra group-theory linear-groups general-linear-group
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
$A=
begin{bmatrix}
lambda & & \
& ddots & a_{ij}\
& & lambda
end{bmatrix}$ is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.
But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.
How to show that $Awnotin langle wrangle$?
Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?
For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$
linear-algebra group-theory linear-groups general-linear-group
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
$A=
begin{bmatrix}
lambda & & \
& ddots & a_{ij}\
& & lambda
end{bmatrix}$ is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.
But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.
How to show that $Awnotin langle wrangle$?
Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?
For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$
linear-algebra group-theory linear-groups general-linear-group
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
$endgroup$
If $w=begin{bmatrix} w_1\ vdots\w_n end{bmatrix}$ is a vector in $K^n$ for a field $K$ and
$A=
begin{bmatrix}
lambda & & \
& ddots & a_{ij}\
& & lambda
end{bmatrix}$ is a $ntimes n $ matrix with entries on $K$ and $lambdain K^{*}$ then $Aw=lambda w+e_i a_{ij} w_jnotinlangle wrangle$ if $ine j$. And so $langle A wranglene langle wrangle$ where $langle wrangle$ is the $1$ dimensional vector space generated by $w$.
But if $A$ is a general matrix in $GL_n(K)$ that is not of the form $lambda I_n$ then $Aw=sumlimits_{i=1}^nsumlimits_{j=1}^n a_{ij}w_je_i$.
How to show that $Awnotin langle wrangle$?
Can we decompose $A$ into $sum A_{ij}$ where each $A_{ij}$ has a simple form like above (so $A_{ij}wnotin langle wrangle$) and say that $Aw=underbrace{sum A_{ij}w}_{notin langle wrangle} notin langle wrangle$?
For context the question asks to prove that $AW=W~~forall Win{text{sub-vector-spaces of } K^n text{ with dimension 1}} iff A=lambda I_n$ for some $lambdain K^{*}$
linear-algebra group-theory linear-groups general-linear-group
linear-algebra group-theory linear-groups general-linear-group
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
edited Jan 8 at 7:37
Martin Sleziak
44.8k10119273
44.8k10119273
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
asked Jan 5 at 10:39
John CataldoJohn Cataldo
1,1881316
1,1881316
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
$endgroup$
add a comment |
$begingroup$
The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
$endgroup$
The condition $Awinlangle wrangle$, applied to elements of the standard basis of $K^n$, implies that $A$ is a diagonal matrix. Now, for elements $w_1neq w_2$ of the basis, being linearly independent, $Aw_1=lambda_1 w_1$ and $Aw_2=lambda_2 w_2$ together with $A(w_1+w_2)=lambda(w_1+w_2)$ imply $lambda_1=lambda_2(=lambda)$.
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
answered Jan 5 at 11:14
metamorphymetamorphy
3,7021621
3,7021621
add a comment |
add a comment |
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