Eigenvalues of A are also eigenvalues of T

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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







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    up vote
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    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







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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







      share|cite|improve this question















      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-algebra matrices eigenvalues-eigenvectors






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      edited 29 mins ago









      José Carlos Santos

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      asked 52 mins ago









      confusedmath

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          Take the matrix such that all of its columns are equal to $v$.






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            1 Answer
            1






            active

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            active

            oldest

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            up vote
            5
            down vote



            accepted










            Take the matrix such that all of its columns are equal to $v$.






            share|cite|improve this answer
























              up vote
              5
              down vote



              accepted










              Take the matrix such that all of its columns are equal to $v$.






              share|cite|improve this answer






















                up vote
                5
                down vote



                accepted







                up vote
                5
                down vote



                accepted






                Take the matrix such that all of its columns are equal to $v$.






                share|cite|improve this answer












                Take the matrix such that all of its columns are equal to $v$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 49 mins ago









                José Carlos Santos

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