Why linear maps act like matrix multiplication?proof linear maps act like matrix multiplicationlinear...

Is there really no realistic way for a skeleton monster to move around without magic?

What would happen to a modern skyscraper if it rains micro blackholes?

Is it legal for company to use my work email to pretend I still work there?

What is the command to reset a PC without deleting any files

Is it possible to do 50 km distance without any previous training?

Is the month field really deprecated?

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

Is the language {<p,n> | p and n are natural numbers and there's no prime number in [p,p+n]} belongs to NP class?

Why are weather verbs 曇る and 晴れる treated differently in this sentence?

Draw simple lines in Inkscape

How is it possible for user to changed after storage was encrypted? (on OS X, Android)

Why has Russell's definition of numbers using equivalence classes been finally abandonned? ( If it has actually been abandonned).

Can a German sentence have two subjects?

How to add power-LED to my small amplifier?

Is there any sparring that doesn't involve punches to the head?

How can I hide my bitcoin transactions to protect anonymity from others?

Why CLRS example on residual networks does not follows its formula?

Why was the small council so happy for Tyrion to become the Master of Coin?

Email Account under attack (really) - anything I can do?

XeLaTeX and pdfLaTeX ignore hyphenation

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

What defenses are there against being summoned by the Gate spell?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

How does one intimidate enemies without having the capacity for violence?



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













4












$begingroup$


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 $M$.




  1. Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?

  2. 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 $M(Tv)$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















4












$begingroup$


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 $M$.




  1. Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?

  2. 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 $M(Tv)$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















4












4








4





$begingroup$


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 $M$.




  1. Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?

  2. 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 $M(Tv)$?










share|cite|improve this question











$endgroup$




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 $M$.




  1. Is the shape of $M(Tv)$ m by 1, $M(T)$ m by n, and $M(v)$ n by 1?

  2. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















13












$begingroup$

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 :( ):



enter image description here



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 $M(T)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very nice graph!
    $endgroup$
    – JOHN
    9 hours ago






  • 2




    $begingroup$
    Upvote for the commutative paint diagram!
    $endgroup$
    – Jannik Pitt
    5 hours ago



















3












$begingroup$


  1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.


  2. 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.






share|cite|improve this answer









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177868%2fwhy-linear-maps-act-like-matrix-multiplication%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13












    $begingroup$

    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 :( ):



    enter image description here



    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 $M(T)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Very nice graph!
      $endgroup$
      – JOHN
      9 hours ago






    • 2




      $begingroup$
      Upvote for the commutative paint diagram!
      $endgroup$
      – Jannik Pitt
      5 hours ago
















    13












    $begingroup$

    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 :( ):



    enter image description here



    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 $M(T)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Very nice graph!
      $endgroup$
      – JOHN
      9 hours ago






    • 2




      $begingroup$
      Upvote for the commutative paint diagram!
      $endgroup$
      – Jannik Pitt
      5 hours ago














    13












    13








    13





    $begingroup$

    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 :( ):



    enter image description here



    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 $M(T)$.






    share|cite|improve this answer









    $endgroup$



    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 :( ):



    enter image description here



    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 $M(T)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    3












    $begingroup$


    1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.


    2. 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.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$


      1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.


      2. 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.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$


        1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.


        2. 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.






        share|cite|improve this answer









        $endgroup$




        1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.


        2. 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 11 hours ago









        MelodyMelody

        1,00912




        1,00912






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177868%2fwhy-linear-maps-act-like-matrix-multiplication%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

            Puerta de Hutt Referencias Enlaces externos Menú de navegación15°58′00″S 5°42′00″O /...

            Castillo d'Acher Características Menú de navegación