Change of (orthonormal) basis.
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As I see it, the author says that $[Tv]_{e} = A[v]_{e}$ in the last paragraph. How do I see that ?
I think I've jusitied the first entry in $[Tv]_{e} = A[v]_{e}$ as follows
begin{align*}
langle text{T}v,e_{1}rangle &= langle text{T}e_{1},e_{1}ranglelangle v,e_{1}rangle + cdots + langle text{T}e_{n},e_{1}ranglelangle v,e_{n}rangle\
&=langlelangle text{T}e_{1},e_{1}rangle v,e_{1}ranglerangle\
&=langle text{T}v,e_{1}rangle.
end{align*}
Is there any other way to see it, or interpret it ?
linear-algebra
asked Oct 12 '13 at 13:07
New_to_this
162111
add a comment |
up vote
0
down vote
favorite
As I see it, the author says that $[Tv]_{e} = A[v]_{e}$ in the last paragraph. How do I see that ?
I think I've jusitied the first entry in $[Tv]_{e} = A[v]_{e}$ as follows
begin{align*}
langle text{T}v,e_{1}rangle &= langle text{T}e_{1},e_{1}ranglelangle v,e_{1}rangle + cdots + langle text{T}e_{n},e_{1}ranglelangle v,e_{n}rangle\
&=langlelangle text{T}e_{1},e_{1}rangle v,e_{1}ranglerangle\
&=langle text{T}v,e_{1}rangle.
end{align*}
Is there any other way to see it, or interpret it ?
linear-algebra
asked Oct 12 '13 at 13:07
New_to_this
162111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As I see it, the author says that $[Tv]_{e} = A[v]_{e}$ in the last paragraph. How do I see that ?
I think I've jusitied the first entry in $[Tv]_{e} = A[v]_{e}$ as follows
begin{align*}
langle text{T}v,e_{1}rangle &= langle text{T}e_{1},e_{1}ranglelangle v,e_{1}rangle + cdots + langle text{T}e_{n},e_{1}ranglelangle v,e_{n}rangle\
&=langlelangle text{T}e_{1},e_{1}rangle v,e_{1}ranglerangle\
&=langle text{T}v,e_{1}rangle.
end{align*}
Is there any other way to see it, or interpret it ?
linear-algebra
asked Oct 12 '13 at 13:07
New_to_this
162111
As I see it, the author says that $[Tv]_{e} = A[v]_{e}$ in the last paragraph. How do I see that ?
I think I've jusitied the first entry in $[Tv]_{e} = A[v]_{e}$ as follows
begin{align*}
langle text{T}v,e_{1}rangle &= langle text{T}e_{1},e_{1}ranglelangle v,e_{1}rangle + cdots + langle text{T}e_{n},e_{1}ranglelangle v,e_{n}rangle\
&=langlelangle text{T}e_{1},e_{1}rangle v,e_{1}ranglerangle\
&=langle text{T}v,e_{1}rangle.
end{align*}
Is there any other way to see it, or interpret it ?
linear-algebra
linear-algebra
asked Oct 12 '13 at 13:07
New_to_this
162111
asked Oct 12 '13 at 13:07
New_to_this
162111
edited Oct 12 '13 at 13:41
asked Oct 12 '13 at 13:07
New_to_this
162111
asked Oct 12 '13 at 13:07
New_to_this
162111
asked Oct 12 '13 at 13:07
New_to_this
162111
162111
add a comment |
add a comment |
1 Answer
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The idea is this - there is a correspondence between matrices in $mathbb{F}^{ntimes n}$ and linear operators in $mathcal{L}(V,V)$.
To every linear operator $Tin mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $mathcal{B} = {e_i}$ of $V$, and consider the matrix $T_{mathcal{B}}$ whose columns are the vectors ${T(e_i)}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $mathcal{B}'$, then how are the two matrices $T_{mathcal{B}}$ and $T_{mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that
$$
T_{mathcal{B}'} = PT_{mathcal{B}}P^{-1}
$$
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
add a comment |
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1 Answer
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up vote
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The idea is this - there is a correspondence between matrices in $mathbb{F}^{ntimes n}$ and linear operators in $mathcal{L}(V,V)$.
To every linear operator $Tin mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $mathcal{B} = {e_i}$ of $V$, and consider the matrix $T_{mathcal{B}}$ whose columns are the vectors ${T(e_i)}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $mathcal{B}'$, then how are the two matrices $T_{mathcal{B}}$ and $T_{mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that
$$
T_{mathcal{B}'} = PT_{mathcal{B}}P^{-1}
$$
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
add a comment |
up vote
0
down vote
The idea is this - there is a correspondence between matrices in $mathbb{F}^{ntimes n}$ and linear operators in $mathcal{L}(V,V)$.
To every linear operator $Tin mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $mathcal{B} = {e_i}$ of $V$, and consider the matrix $T_{mathcal{B}}$ whose columns are the vectors ${T(e_i)}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $mathcal{B}'$, then how are the two matrices $T_{mathcal{B}}$ and $T_{mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that
$$
T_{mathcal{B}'} = PT_{mathcal{B}}P^{-1}
$$
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
add a comment |
up vote
0
down vote
up vote
0
down vote
The idea is this - there is a correspondence between matrices in $mathbb{F}^{ntimes n}$ and linear operators in $mathcal{L}(V,V)$.
To every linear operator $Tin mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $mathcal{B} = {e_i}$ of $V$, and consider the matrix $T_{mathcal{B}}$ whose columns are the vectors ${T(e_i)}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $mathcal{B}'$, then how are the two matrices $T_{mathcal{B}}$ and $T_{mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that
$$
T_{mathcal{B}'} = PT_{mathcal{B}}P^{-1}
$$
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
The idea is this - there is a correspondence between matrices in $mathbb{F}^{ntimes n}$ and linear operators in $mathcal{L}(V,V)$.
To every linear operator $Tin mathcal{L}(V,V)$ one can associate a matrix as follows : Pick a basis $mathcal{B} = {e_i}$ of $V$, and consider the matrix $T_{mathcal{B}}$ whose columns are the vectors ${T(e_i)}$.
Clearly, this depends on the choice of basis, so the question is : if we choose a different basis $mathcal{B}'$, then how are the two matrices $T_{mathcal{B}}$ and $T_{mathcal{B}'}$ related?
The answer is : There is an invertible matrix $P$ such that
$$
T_{mathcal{B}'} = PT_{mathcal{B}}P^{-1}
$$
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
answered Oct 12 '13 at 13:47
Prahlad Vaidyanathan
25.9k12151
25.9k12151
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
add a comment |
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
Yes I'm familiar with this diagonalization. My first thought was that A actually is the product that you wrote ind the end. I think that the author just make the statement that T$[v]_{e}=$A$[v]_{e}$ without justifying it.
– New_to_this
Oct 12 '13 at 13:55
add a comment |
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