Change of Coordinate Matrix Collary Proof












1















Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.




I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$



And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.



Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.



But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.



Hope someone can give me some hints.
Thanks a lot.






linear-algebra matrices linear-transformations







share|cite|improve this question
























  • The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
    – Omnomnomnom
    Nov 26 at 19:17










  • @Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
    – Ann P.
    Nov 27 at 0:00


















1















Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.




I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$



And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.



Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.



But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.



Hope someone can give me some hints.
Thanks a lot.






linear-algebra matrices linear-transformations







share|cite|improve this question
























  • The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
    – Omnomnomnom
    Nov 26 at 19:17










  • @Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
    – Ann P.
    Nov 27 at 0:00
















1












1








1








Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.




I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$



And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.



Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.



But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.



Hope someone can give me some hints.
Thanks a lot.






linear-algebra matrices linear-transformations







share|cite|improve this question
















Let $Ain M_{n times n}(Bbb F)$, and let $gamma$ be the basis for $Bbb F^n$.
Show that if $[L_A]_γ=Q^{-1}AQ$, then the j-th column of Q is the j-th vector in $γ$.




I start by applying the theorem:
$[L_A]_γ = Q^{-1}[L_A]_{α} Q$



And let $[L_A]_α=A$, $α$ is another unknown basis of $Bbb F^n$.



Then we have $Q=[I_n]_{γ→α}$, $Q$ is an $n times n$ matrix.



But I still can't figure out why the $j$-th column of $Q$ is the same as the $j$-th vector in $γ$.



Hope someone can give me some hints.
Thanks a lot.






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






share|cite|improve this question















share|cite|improve this question













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edited Nov 26 at 19:12









Omnomnomnom

126k788176




126k788176










asked Nov 26 at 15:31









Ann P.

133




133












  • The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
    – Omnomnomnom
    Nov 26 at 19:17










  • @Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
    – Ann P.
    Nov 27 at 0:00




















  • The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
    – Omnomnomnom
    Nov 26 at 19:17










  • @Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
    – Ann P.
    Nov 27 at 0:00


















The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17




The statement is not correct as stated: if $[L_A]_{gamma} = Q^{-1}AQ$, then there are many matrices that $Q$ could be. For instance: for any such matrix $Q$ and non-zero $k in Bbb F$, $kQ$ will satisfy $(kQ)^{-1}A(kQ) = [L_A]_{gamma}$. I suspect that you are meant to prove the converse, that is: if $Q$ is the matrix whose columns are the elements of $gamma$, then $[L_A]_gamma = Q^{-1}AQ$.
– Omnomnomnom
Nov 26 at 19:17












@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00






@Omnomnomnom Thanks, I've revisited the original question and found that you're right. English is not my first language, so sometimes it's hard for me to conveyed the question properly. I'll try considering $alpha$ as standard basis!! Thanks again.
– Ann P.
Nov 27 at 0:00












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Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
$$
Q = [I_n]_{gamma to alpha}
$$
We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
$$
[I_n]_{gamma to alpha} e_i =
[I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
$$






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    Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
    $$
    Q = [I_n]_{gamma to alpha}
    $$
    We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
    $$
    [I_n]_{gamma to alpha} e_i =
    [I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
    $$






    share|cite|improve this answer


























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      Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
      $$
      Q = [I_n]_{gamma to alpha}
      $$
      We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
      $$
      [I_n]_{gamma to alpha} e_i =
      [I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
      $$






      share|cite|improve this answer
























        0












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        0






        Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
        $$
        Q = [I_n]_{gamma to alpha}
        $$
        We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
        $$
        [I_n]_{gamma to alpha} e_i =
        [I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
        $$






        share|cite|improve this answer












        Hint: Let $alpha$ denote the standard basis (that is, $e_1 = (1,0,dots,0)$ is the first vector, $e_2 = (0,1,0,dots,0)$ is the second, and so on). Then if we take $Q$ to be the matrix whose $i$th column is the $i$th vector in $gamma$, we have
        $$
        Q = [I_n]_{gamma to alpha}
        $$
        We can that this is the case as follows: if $v_1,dots,v_n$ are the elements of $gamma$, then we note that the $i$th column of $[I_n]_{gamma to alpha}$ is given by
        $$
        [I_n]_{gamma to alpha} e_i =
        [I_n]_{gamma to alpha} [v_i]_{gamma} = [I_n v_i]_{alpha} = [v_i]_{alpha} = v_i
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 19:22









        Omnomnomnom

        126k788176




        126k788176






























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