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If $(f)_BB=Pin mathbb R^ntimes n$, does it make sense to write $(g)_FB=P$?
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Let $f$ a linear application $mathbb Rto mathbb R^n$. Let $B$ and $F$ two basis of $mathbb R^n$. Suppose $(f)_BB=P=(p_ij)_1leq i,jleq ninmathbb R^ntimes n$ and suppose that $P$ is invertible. I want the change of basis $B$ and the basis generated by the column of $P$ denoted $F$. It's of course $(id)_FB$. My question is : Can I write $$(id)_FB=P ?$$
The thing is we have that $$big((id)_FBbig)_ij=p_ij.$$
My problem on writing $(id)_FB=P$ is that $P$ is a linear application that take vectors written in the basis $B$ and gives vectors written in the basis $B$ whereas $(id)_FB$ is an application taking vector written in the basis $B$ and gives vector written in $F$. So $(id)_BF$ and $P$ are definitely not the same application. So I'm a little bit confuse on if I can write $(id)_FB=P$ to say $$big((id)_FBbig)_ij=p_ij,$$
since $(id)_FBneq P$. But in an other way, the notation $(id)_FB=P$ doesn't look wrong since a matrix looks independent of the basis. Indeed, I would interpret a matrix $M$ as "the matrix of a linear application $f$ in a certain basis", but if $g$ is an linear application and $F$ is an other basis, we could also have that $M$ is the basis of $g$ in the basis $F$. In other word
$$(g)_FF: [v]_Flongmapsto (M[v]_F)_F,$$
and thus $$(g)_FF=M,$$
but also $$(f)_BB:[v]_Blongmapsto (M[v]_B)_B,$$
and thus $$(f)_BB=M,$$
but $fneq g$. So maybe a matrix is independent of a basis and an application a priori, and thus the notation $(id)_FB=P$ totally make sense.
I would like to have your opinion on what I just said (you agree or not, and why).
linear-algebra matrices linear-transformations
asked Jul 25 at 16:49
Peter
328112
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up vote
2
down vote
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Let $f$ a linear application $mathbb Rto mathbb R^n$. Let $B$ and $F$ two basis of $mathbb R^n$. Suppose $(f)_BB=P=(p_ij)_1leq i,jleq ninmathbb R^ntimes n$ and suppose that $P$ is invertible. I want the change of basis $B$ and the basis generated by the column of $P$ denoted $F$. It's of course $(id)_FB$. My question is : Can I write $$(id)_FB=P ?$$
The thing is we have that $$big((id)_FBbig)_ij=p_ij.$$
My problem on writing $(id)_FB=P$ is that $P$ is a linear application that take vectors written in the basis $B$ and gives vectors written in the basis $B$ whereas $(id)_FB$ is an application taking vector written in the basis $B$ and gives vector written in $F$. So $(id)_BF$ and $P$ are definitely not the same application. So I'm a little bit confuse on if I can write $(id)_FB=P$ to say $$big((id)_FBbig)_ij=p_ij,$$
since $(id)_FBneq P$. But in an other way, the notation $(id)_FB=P$ doesn't look wrong since a matrix looks independent of the basis. Indeed, I would interpret a matrix $M$ as "the matrix of a linear application $f$ in a certain basis", but if $g$ is an linear application and $F$ is an other basis, we could also have that $M$ is the basis of $g$ in the basis $F$. In other word
$$(g)_FF: [v]_Flongmapsto (M[v]_F)_F,$$
and thus $$(g)_FF=M,$$
but also $$(f)_BB:[v]_Blongmapsto (M[v]_B)_B,$$
and thus $$(f)_BB=M,$$
but $fneq g$. So maybe a matrix is independent of a basis and an application a priori, and thus the notation $(id)_FB=P$ totally make sense.
I would like to have your opinion on what I just said (you agree or not, and why).
linear-algebra matrices linear-transformations
asked Jul 25 at 16:49
Peter
328112
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f$ a linear application $mathbb Rto mathbb R^n$. Let $B$ and $F$ two basis of $mathbb R^n$. Suppose $(f)_BB=P=(p_ij)_1leq i,jleq ninmathbb R^ntimes n$ and suppose that $P$ is invertible. I want the change of basis $B$ and the basis generated by the column of $P$ denoted $F$. It's of course $(id)_FB$. My question is : Can I write $$(id)_FB=P ?$$
The thing is we have that $$big((id)_FBbig)_ij=p_ij.$$
My problem on writing $(id)_FB=P$ is that $P$ is a linear application that take vectors written in the basis $B$ and gives vectors written in the basis $B$ whereas $(id)_FB$ is an application taking vector written in the basis $B$ and gives vector written in $F$. So $(id)_BF$ and $P$ are definitely not the same application. So I'm a little bit confuse on if I can write $(id)_FB=P$ to say $$big((id)_FBbig)_ij=p_ij,$$
since $(id)_FBneq P$. But in an other way, the notation $(id)_FB=P$ doesn't look wrong since a matrix looks independent of the basis. Indeed, I would interpret a matrix $M$ as "the matrix of a linear application $f$ in a certain basis", but if $g$ is an linear application and $F$ is an other basis, we could also have that $M$ is the basis of $g$ in the basis $F$. In other word
$$(g)_FF: [v]_Flongmapsto (M[v]_F)_F,$$
and thus $$(g)_FF=M,$$
but also $$(f)_BB:[v]_Blongmapsto (M[v]_B)_B,$$
and thus $$(f)_BB=M,$$
but $fneq g$. So maybe a matrix is independent of a basis and an application a priori, and thus the notation $(id)_FB=P$ totally make sense.
I would like to have your opinion on what I just said (you agree or not, and why).
linear-algebra matrices linear-transformations
asked Jul 25 at 16:49
Peter
328112
Let $f$ a linear application $mathbb Rto mathbb R^n$. Let $B$ and $F$ two basis of $mathbb R^n$. Suppose $(f)_BB=P=(p_ij)_1leq i,jleq ninmathbb R^ntimes n$ and suppose that $P$ is invertible. I want the change of basis $B$ and the basis generated by the column of $P$ denoted $F$. It's of course $(id)_FB$. My question is : Can I write $$(id)_FB=P ?$$
The thing is we have that $$big((id)_FBbig)_ij=p_ij.$$
My problem on writing $(id)_FB=P$ is that $P$ is a linear application that take vectors written in the basis $B$ and gives vectors written in the basis $B$ whereas $(id)_FB$ is an application taking vector written in the basis $B$ and gives vector written in $F$. So $(id)_BF$ and $P$ are definitely not the same application. So I'm a little bit confuse on if I can write $(id)_FB=P$ to say $$big((id)_FBbig)_ij=p_ij,$$
since $(id)_FBneq P$. But in an other way, the notation $(id)_FB=P$ doesn't look wrong since a matrix looks independent of the basis. Indeed, I would interpret a matrix $M$ as "the matrix of a linear application $f$ in a certain basis", but if $g$ is an linear application and $F$ is an other basis, we could also have that $M$ is the basis of $g$ in the basis $F$. In other word
$$(g)_FF: [v]_Flongmapsto (M[v]_F)_F,$$
and thus $$(g)_FF=M,$$
but also $$(f)_BB:[v]_Blongmapsto (M[v]_B)_B,$$
and thus $$(f)_BB=M,$$
but $fneq g$. So maybe a matrix is independent of a basis and an application a priori, and thus the notation $(id)_FB=P$ totally make sense.
I would like to have your opinion on what I just said (you agree or not, and why).
linear-algebra matrices linear-transformations
asked Jul 25 at 16:49
Peter
328112
edited Jul 26 at 9:11
asked Jul 25 at 16:49
Peter
328112
asked Jul 25 at 16:49
Peter
328112
asked Jul 25 at 16:49
Peter
328112
328112
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add a comment |Â
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