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Matrix of a linear mapping in the new basis
$begingroup$
Let $N, M$ be a linear spaces over the same field $mathbb{F}$. So ${e_i}$, ${overline{e_i}}$ -- two bases in $N$ and ${e_k'}$, ${overline{e_k}'}$ -- two bases in $M$. Let $f$ be a map between $N$ and $M$, $f: N rightarrow M$, and between bases ${e_i}$ and ${e_k'}$ it will be $A_f$ and between bases ${overline{e}_i}$ and ${overline{e}_k'}$ it will be $overline{A_f}$. Let $B$ be the basis change matrix $B({e_i}) = {overline{e_i}}$ and similar for $C$, $C({e_k'}) = {overline{e_k}'}$.
So in my textbook I have this proof of the identity $overline{A_f} = C^{-1} A_f B$:
$(overline{e}_i)overline{A_f} = f((overline{e}_i)) = f((e_i)B) = (f(e_i))B = (e_i')A_f B = (overline{e_i}')C^{-1}A_f B$
But I do not really understand why and how we can make follow steps: $cdots = f((e_i)B) = (f(e_i))B = (e_i')A_f B = cdots$. Why we can take the matrix $B$ beyond the parentheses? And why $f((e_i)) = (e_i')A_f$ and not $f((e_i)) = (e_i)A_f$?
linear-algebra abstract-algebra change-of-basis
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
$endgroup$
add a comment |
$begingroup$
Let $N, M$ be a linear spaces over the same field $mathbb{F}$. So ${e_i}$, ${overline{e_i}}$ -- two bases in $N$ and ${e_k'}$, ${overline{e_k}'}$ -- two bases in $M$. Let $f$ be a map between $N$ and $M$, $f: N rightarrow M$, and between bases ${e_i}$ and ${e_k'}$ it will be $A_f$ and between bases ${overline{e}_i}$ and ${overline{e}_k'}$ it will be $overline{A_f}$. Let $B$ be the basis change matrix $B({e_i}) = {overline{e_i}}$ and similar for $C$, $C({e_k'}) = {overline{e_k}'}$.
So in my textbook I have this proof of the identity $overline{A_f} = C^{-1} A_f B$:
$(overline{e}_i)overline{A_f} = f((overline{e}_i)) = f((e_i)B) = (f(e_i))B = (e_i')A_f B = (overline{e_i}')C^{-1}A_f B$
But I do not really understand why and how we can make follow steps: $cdots = f((e_i)B) = (f(e_i))B = (e_i')A_f B = cdots$. Why we can take the matrix $B$ beyond the parentheses? And why $f((e_i)) = (e_i')A_f$ and not $f((e_i)) = (e_i)A_f$?
linear-algebra abstract-algebra change-of-basis
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
$endgroup$
$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23
add a comment |
$begingroup$
Let $N, M$ be a linear spaces over the same field $mathbb{F}$. So ${e_i}$, ${overline{e_i}}$ -- two bases in $N$ and ${e_k'}$, ${overline{e_k}'}$ -- two bases in $M$. Let $f$ be a map between $N$ and $M$, $f: N rightarrow M$, and between bases ${e_i}$ and ${e_k'}$ it will be $A_f$ and between bases ${overline{e}_i}$ and ${overline{e}_k'}$ it will be $overline{A_f}$. Let $B$ be the basis change matrix $B({e_i}) = {overline{e_i}}$ and similar for $C$, $C({e_k'}) = {overline{e_k}'}$.
So in my textbook I have this proof of the identity $overline{A_f} = C^{-1} A_f B$:
$(overline{e}_i)overline{A_f} = f((overline{e}_i)) = f((e_i)B) = (f(e_i))B = (e_i')A_f B = (overline{e_i}')C^{-1}A_f B$
But I do not really understand why and how we can make follow steps: $cdots = f((e_i)B) = (f(e_i))B = (e_i')A_f B = cdots$. Why we can take the matrix $B$ beyond the parentheses? And why $f((e_i)) = (e_i')A_f$ and not $f((e_i)) = (e_i)A_f$?
linear-algebra abstract-algebra change-of-basis
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
$endgroup$
Let $N, M$ be a linear spaces over the same field $mathbb{F}$. So ${e_i}$, ${overline{e_i}}$ -- two bases in $N$ and ${e_k'}$, ${overline{e_k}'}$ -- two bases in $M$. Let $f$ be a map between $N$ and $M$, $f: N rightarrow M$, and between bases ${e_i}$ and ${e_k'}$ it will be $A_f$ and between bases ${overline{e}_i}$ and ${overline{e}_k'}$ it will be $overline{A_f}$. Let $B$ be the basis change matrix $B({e_i}) = {overline{e_i}}$ and similar for $C$, $C({e_k'}) = {overline{e_k}'}$.
So in my textbook I have this proof of the identity $overline{A_f} = C^{-1} A_f B$:
$(overline{e}_i)overline{A_f} = f((overline{e}_i)) = f((e_i)B) = (f(e_i))B = (e_i')A_f B = (overline{e_i}')C^{-1}A_f B$
But I do not really understand why and how we can make follow steps: $cdots = f((e_i)B) = (f(e_i))B = (e_i')A_f B = cdots$. Why we can take the matrix $B$ beyond the parentheses? And why $f((e_i)) = (e_i')A_f$ and not $f((e_i)) = (e_i)A_f$?
linear-algebra abstract-algebra change-of-basis
linear-algebra abstract-algebra change-of-basis
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
edited Feb 3 at 15:00
J. W. Tanner
4,7871420
4,7871420
asked Feb 3 at 3:33
Just do itJust do it
19618
asked Feb 3 at 3:33
Just do itJust do it
19618
asked Feb 3 at 3:33
Just do itJust do it
19618
19618
$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23
add a comment |
$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23
$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23
add a comment |
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$begingroup$
I don't like your notation, and can't go no further. Perhaps try identifying the linear transformations with their matrix, to simplify your question.
$endgroup$
– I Said Roll Up n Smoke Adjoint
Feb 3 at 16:46
$begingroup$
@HermitwithAdjoint I simply quoted my textbook and wrote out the notation from it. What do you mean when you suggest replacing functions with matrices? A record of the form $A_f$ -- is a matrix of a map $f$
$endgroup$
– Just do it
Feb 4 at 1:23