Mixing
Eigenvalues of A are also eigenvalues of T
$begingroup$
Let $V$ be the set of all $ntimes n$ matrices over a field $F$. Let $A$ be a fixed element of $V$. Define a linear operator $T$ on $V$ by $T(B)=AB$. I am trying to show that if $lambda$ is an eigenvalue of $A$, then $lambda$ is also an eigenvalue of $T$.
So suppose $Av=lambda v$ for some $vneq 0$ in $V$ and $lambdain F$. So I'd like to prove the existence of a matrix $B$ such that $T(B)=AB=lambda A$, or equivalently, show that $T-lambda I_V$ is not invertible (or injective or surjective). But I am not sure how to proceed from here. What can I do?
linear-algebra matrices eigenvalues-eigenvectors linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
$endgroup$
add a comment |
$begingroup$
Let $V$ be the set of all $ntimes n$ matrices over a field $F$. Let $A$ be a fixed element of $V$. Define a linear operator $T$ on $V$ by $T(B)=AB$. I am trying to show that if $lambda$ is an eigenvalue of $A$, then $lambda$ is also an eigenvalue of $T$.
So suppose $Av=lambda v$ for some $vneq 0$ in $V$ and $lambdain F$. So I'd like to prove the existence of a matrix $B$ such that $T(B)=AB=lambda A$, or equivalently, show that $T-lambda I_V$ is not invertible (or injective or surjective). But I am not sure how to proceed from here. What can I do?
linear-algebra matrices eigenvalues-eigenvectors linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
$endgroup$
add a comment |
$begingroup$
Let $V$ be the set of all $ntimes n$ matrices over a field $F$. Let $A$ be a fixed element of $V$. Define a linear operator $T$ on $V$ by $T(B)=AB$. I am trying to show that if $lambda$ is an eigenvalue of $A$, then $lambda$ is also an eigenvalue of $T$.
So suppose $Av=lambda v$ for some $vneq 0$ in $V$ and $lambdain F$. So I'd like to prove the existence of a matrix $B$ such that $T(B)=AB=lambda A$, or equivalently, show that $T-lambda I_V$ is not invertible (or injective or surjective). But I am not sure how to proceed from here. What can I do?
linear-algebra matrices eigenvalues-eigenvectors linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
$endgroup$
Let $V$ be the set of all $ntimes n$ matrices over a field $F$. Let $A$ be a fixed element of $V$. Define a linear operator $T$ on $V$ by $T(B)=AB$. I am trying to show that if $lambda$ is an eigenvalue of $A$, then $lambda$ is also an eigenvalue of $T$.
So suppose $Av=lambda v$ for some $vneq 0$ in $V$ and $lambdain F$. So I'd like to prove the existence of a matrix $B$ such that $T(B)=AB=lambda A$, or equivalently, show that $T-lambda I_V$ is not invertible (or injective or surjective). But I am not sure how to proceed from here. What can I do?
linear-algebra matrices eigenvalues-eigenvectors linear-transformations
linear-algebra matrices eigenvalues-eigenvectors linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
167k22132235
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
asked Oct 16 '18 at 17:38
confusedmath confusedmath
24717
24717
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Take the matrix such that all of its columns are equal to $v$.
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
An eigenvalue of $T$ is some number $lambda$ such that $T(B)=lambda B$. Thus, we need to find a matrix $B$ such that $AB=lambda B$, not $AB=lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[lambda v, lambda v, lambda v, ...]=lambda [v,v,v,...]=lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $lambda$.
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the matrix such that all of its columns are equal to $v$.
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
Take the matrix such that all of its columns are equal to $v$.
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
add a comment |
$begingroup$
Take the matrix such that all of its columns are equal to $v$.
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
Take the matrix such that all of its columns are equal to $v$.
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
answered Oct 16 '18 at 17:41
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
add a comment |
add a comment |
$begingroup$
An eigenvalue of $T$ is some number $lambda$ such that $T(B)=lambda B$. Thus, we need to find a matrix $B$ such that $AB=lambda B$, not $AB=lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[lambda v, lambda v, lambda v, ...]=lambda [v,v,v,...]=lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $lambda$.
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
An eigenvalue of $T$ is some number $lambda$ such that $T(B)=lambda B$. Thus, we need to find a matrix $B$ such that $AB=lambda B$, not $AB=lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[lambda v, lambda v, lambda v, ...]=lambda [v,v,v,...]=lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $lambda$.
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
An eigenvalue of $T$ is some number $lambda$ such that $T(B)=lambda B$. Thus, we need to find a matrix $B$ such that $AB=lambda B$, not $AB=lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[lambda v, lambda v, lambda v, ...]=lambda [v,v,v,...]=lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $lambda$.
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
$endgroup$
An eigenvalue of $T$ is some number $lambda$ such that $T(B)=lambda B$. Thus, we need to find a matrix $B$ such that $AB=lambda B$, not $AB=lambda A$. We can write a matrix as a row vector of column vectors, i.e. $B=[b_1,b_2,b_3...]$. Matrix multiplication acts on those columns independently: $AB=[Ab_1,Ab_,Ab_3...]$. If $b_i=v$ for all $i$, then $AB=[lambda v, lambda v, lambda v, ...]=lambda [v,v,v,...]=lambda B$. So this shows that $[v,v,v,...]$ is an eigenvector with eigenvalue $lambda$.
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
answered Oct 16 '18 at 20:27
AcccumulationAcccumulation
7,1252619
7,1252619
add a comment |
add a comment |
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