Why linear maps act like matrix multiplication?proof linear maps act like matrix multiplicationlinear...
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Why linear maps act like matrix multiplication?
proof linear maps act like matrix multiplicationlinear transformation of a basis where $T(v_1) = w_1, …, T(v_n) = w_n$Basis in a Linear MapTranspose of the matrix of a linear mapHow to define the matrix of a linear map?Prove that all entries in $M(T, (v_1, … ,v_n), (w_1, … , w_m))$ are zero except for the entries in row j, column jNeed help to understand the uniqueness of linear maps on basis of domain.Linear Maps, Basis of Domain, and MatrixIs the linear map on basis of $V$ a basis of $W$?Prove there exist a basis of $V$ and a basis of $W$ such that all entries of $mathcal{M}(T)$ are 0 except row j and column j
$begingroup$
In Linear Algebra Done Right, it said
Suppose $T in mathcal{L}(V,W)$ and $v in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then $$M(Tv) = M(T)M(v)$$
$M(T)$ is the m-by-n matrix whose entries $A_{j,k}$ are defined by $Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$ suppose $T in mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$.
$M(v)$ is the matrix of vector $v$.
I generally follow the following proof:
Suppose $v = c_1v_1 + ... + c_nv_n$, where $c_1,...,c_n in mathbb{F}$. Thus
$$Tv = c_1Tv_1 +...+c_nTv_n$$
Hence
begin{equation}
begin{split}
M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\
& = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \
& = M(T)M(v)
end{split}
end{equation}
But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $A$ induces a linear map from $mathbb{F}^{n,1}$ to $mathbb{F}^{m,1}$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $M$.
- Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?
- What is meant by suitable relabeling via the isomorphisms given by $M$? Does it just mean $M(T)$ is a isomorphism linear map between $M(v)$ and $M(Tv)$?
matrices linear-transformations
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it said
Suppose $T in mathcal{L}(V,W)$ and $v in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then $$M(Tv) = M(T)M(v)$$
$M(T)$ is the m-by-n matrix whose entries $A_{j,k}$ are defined by $Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$ suppose $T in mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$.
$M(v)$ is the matrix of vector $v$.
I generally follow the following proof:
Suppose $v = c_1v_1 + ... + c_nv_n$, where $c_1,...,c_n in mathbb{F}$. Thus
$$Tv = c_1Tv_1 +...+c_nTv_n$$
Hence
begin{equation}
begin{split}
M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\
& = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \
& = M(T)M(v)
end{split}
end{equation}
But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $A$ induces a linear map from $mathbb{F}^{n,1}$ to $mathbb{F}^{m,1}$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $M$.
- Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?
- What is meant by suitable relabeling via the isomorphisms given by $M$? Does it just mean $M(T)$ is a isomorphism linear map between $M(v)$ and $M(Tv)$?
matrices linear-transformations
$endgroup$
$begingroup$
In your blockquote, you haven't told us what $M$ means.
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
@GerryMyerson revised
$endgroup$
– JOHN
12 hours ago
$begingroup$
Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
$endgroup$
– Gerry Myerson
12 hours ago
1
$begingroup$
@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
$endgroup$
– Theo Bendit
11 hours ago
add a comment |
$begingroup$
In Linear Algebra Done Right, it said
Suppose $T in mathcal{L}(V,W)$ and $v in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then $$M(Tv) = M(T)M(v)$$
$M(T)$ is the m-by-n matrix whose entries $A_{j,k}$ are defined by $Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$ suppose $T in mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$.
$M(v)$ is the matrix of vector $v$.
I generally follow the following proof:
Suppose $v = c_1v_1 + ... + c_nv_n$, where $c_1,...,c_n in mathbb{F}$. Thus
$$Tv = c_1Tv_1 +...+c_nTv_n$$
Hence
begin{equation}
begin{split}
M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\
& = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \
& = M(T)M(v)
end{split}
end{equation}
But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $A$ induces a linear map from $mathbb{F}^{n,1}$ to $mathbb{F}^{m,1}$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $M$.
- Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?
- What is meant by suitable relabeling via the isomorphisms given by $M$? Does it just mean $M(T)$ is a isomorphism linear map between $M(v)$ and $M(Tv)$?
matrices linear-transformations
$endgroup$
In Linear Algebra Done Right, it said
Suppose $T in mathcal{L}(V,W)$ and $v in V$. Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. Then $$M(Tv) = M(T)M(v)$$
$M(T)$ is the m-by-n matrix whose entries $A_{j,k}$ are defined by $Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$ suppose $T in mathcal{L}(V,W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$.
$M(v)$ is the matrix of vector $v$.
I generally follow the following proof:
Suppose $v = c_1v_1 + ... + c_nv_n$, where $c_1,...,c_n in mathbb{F}$. Thus
$$Tv = c_1Tv_1 +...+c_nTv_n$$
Hence
begin{equation}
begin{split}
M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\
& = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \
& = M(T)M(v)
end{split}
end{equation}
But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $A$ induces a linear map from $mathbb{F}^{n,1}$ to $mathbb{F}^{m,1}$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $M$.
- Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?
- What is meant by suitable relabeling via the isomorphisms given by $M$? Does it just mean $M(T)$ is a isomorphism linear map between $M(v)$ and $M(Tv)$?
matrices linear-transformations
matrices linear-transformations
edited 12 hours ago
JOHN
asked 12 hours ago
JOHN JOHN
4589
4589
$begingroup$
In your blockquote, you haven't told us what $M$ means.
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
@GerryMyerson revised
$endgroup$
– JOHN
12 hours ago
$begingroup$
Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
$endgroup$
– Gerry Myerson
12 hours ago
1
$begingroup$
@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
$endgroup$
– Theo Bendit
11 hours ago
add a comment |
$begingroup$
In your blockquote, you haven't told us what $M$ means.
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
@GerryMyerson revised
$endgroup$
– JOHN
12 hours ago
$begingroup$
Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
$endgroup$
– Gerry Myerson
12 hours ago
1
$begingroup$
@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
$endgroup$
– Theo Bendit
11 hours ago
$begingroup$
In your blockquote, you haven't told us what $M$ means.
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
In your blockquote, you haven't told us what $M$ means.
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
@GerryMyerson revised
$endgroup$
– JOHN
12 hours ago
$begingroup$
@GerryMyerson revised
$endgroup$
– JOHN
12 hours ago
$begingroup$
Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
$endgroup$
– Gerry Myerson
12 hours ago
$begingroup$
Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
$endgroup$
– Gerry Myerson
12 hours ago
1
1
$begingroup$
@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
$endgroup$
– Theo Bendit
11 hours ago
$begingroup$
@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
$endgroup$
– Theo Bendit
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n times 1$ column vector. Similarly, $Tv in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m times 1$ coordinate column vector, which are put into an $m times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):
The process of applying $T$ to a vector $v in V$ is the top row of the diagram. However, there's a parallel process happening between $Bbb{F}^n$ and $Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $Bbb{F}^n$ and $W$ and $Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $Bbb{F}^n$ to $Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
$endgroup$
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
add a comment |
$begingroup$
Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.
There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $Vto F^n$, isomorphism from $Tto F^{ntimes m}$, and an isomorphism from $Wto F^m.$
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n times 1$ column vector. Similarly, $Tv in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m times 1$ coordinate column vector, which are put into an $m times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):
The process of applying $T$ to a vector $v in V$ is the top row of the diagram. However, there's a parallel process happening between $Bbb{F}^n$ and $Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $Bbb{F}^n$ and $W$ and $Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $Bbb{F}^n$ to $Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
$endgroup$
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
add a comment |
$begingroup$
In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n times 1$ column vector. Similarly, $Tv in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m times 1$ coordinate column vector, which are put into an $m times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):
The process of applying $T$ to a vector $v in V$ is the top row of the diagram. However, there's a parallel process happening between $Bbb{F}^n$ and $Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $Bbb{F}^n$ and $W$ and $Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $Bbb{F}^n$ to $Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
$endgroup$
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
add a comment |
$begingroup$
In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n times 1$ column vector. Similarly, $Tv in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m times 1$ coordinate column vector, which are put into an $m times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):
The process of applying $T$ to a vector $v in V$ is the top row of the diagram. However, there's a parallel process happening between $Bbb{F}^n$ and $Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $Bbb{F}^n$ and $W$ and $Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $Bbb{F}^n$ to $Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
$endgroup$
In answer to your first question, yes to all three: $v$ is an element of the $n$-dimensional space $V$, so the coordinate vector with respect to the basis will be an $n times 1$ column vector. Similarly, $Tv in W$, which is an $n$-dimensional space, so $M(Tv)$ will be an $m times 1$ column vector. Finally, $M(T)$ is built from transforming the $n$ basis vectors of the domain, forming each an $m times 1$ coordinate column vector, which are put into an $m times n$ matrix.
In answer to your second question, consult the following commutative diagram (made in Paint :( ):
The process of applying $T$ to a vector $v in V$ is the top row of the diagram. However, there's a parallel process happening between $Bbb{F}^n$ and $Bbb{F}^m$, mirroring the same process.
The isomorphism being referred to are the double arrows, taking us between $V$ and $Bbb{F}^n$ and $W$ and $Bbb{F}^m$, by way of coordinate vectors. The coordinate vector map on $V$ is a linear map between $V$ and $Bbb{F}^n$ that is invertible, making it an isomorphism (and similarly for $W$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.
In $V$, we have some abstract vectors, and an abstract linear transformation $T$ that maps vectors in $V$ to vectors in $W$. However, using this isomorphism, we can view $V$ slightly differently as $Bbb{F}^n$, and similarly for $W$, which means $T$ boils down to a linear map from $Bbb{F}^n$ to $Bbb{F}^m$, which can be characterised as matrix multiplication. The matrix, in particular, is $M(T)$.
answered 11 hours ago
Theo BenditTheo Bendit
20.5k12354
20.5k12354
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
add a comment |
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
$begingroup$
Very nice graph!
$endgroup$
– JOHN
9 hours ago
2
2
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
$begingroup$
Upvote for the commutative paint diagram!
$endgroup$
– Jannik Pitt
5 hours ago
add a comment |
$begingroup$
Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.
There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $Vto F^n$, isomorphism from $Tto F^{ntimes m}$, and an isomorphism from $Wto F^m.$
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
$endgroup$
add a comment |
$begingroup$
Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.
There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $Vto F^n$, isomorphism from $Tto F^{ntimes m}$, and an isomorphism from $Wto F^m.$
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
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add a comment |
$begingroup$
Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.
There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $Vto F^n$, isomorphism from $Tto F^{ntimes m}$, and an isomorphism from $Wto F^m.$
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
$endgroup$
Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.
There is a theorem that if $V$ is an $n-$dimensional vector space over a field $F,$ then $V$ is isomorphic to $F^n.$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $M$ is an isomorphism between $M(v)$ and $M(Tv).$ These are particular vectors. The map $M$ actually induces an isomorphism from $Vto F^n$, isomorphism from $Tto F^{ntimes m}$, and an isomorphism from $Wto F^m.$
I actually like the way that this is done. The Author is telling you that you're representation of $T$ by a matrix depends on your choice of basis in $F^n$. A fact that is important to remember.
answered 11 hours ago
MelodyMelody
1,00912
1,00912
add a comment |
add a comment |
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$begingroup$
In your blockquote, you haven't told us what $M$ means.
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– Gerry Myerson
12 hours ago
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@GerryMyerson revised
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– JOHN
12 hours ago
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Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$?
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– Gerry Myerson
12 hours ago
1
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@GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $Bbb{R}^3$, then $M(1, 2, 3) = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}$.
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– Theo Bendit
11 hours ago