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I was watching 3blue1brown's video series on linear algebra. My understanding till now is :-

  1. A linear transformation takes in a vector and outputs another vector.
  2. The above statement is equivalent to multiplying a unique matrix to the given vector.
  3. 3b1b shows the linear transformation using a new coordinate system, and shows that $\hat{i}$ and $\hat{j}$ change.
  4. When he discusses change of basis, he states that it helps us move between different coordinate systems.
  5. 3b1b also states that a matrix implicitly assumes coordinate systems, as it represents the landing spots of basis vectors after linear transformation.
  6. He shows how to transform a rotation matrix in a conventional Cartesian coordinate system, to Jennifer's coordinate system (one where basis vectors are not perpendicular to one another).

Points 4,5 and 6 have really confused me and now I doubt even points 1,2 and 3.

When we write a matrix what basis vectors does it assume? I have never seen any text stating that this assumes a Cartesian coordinate system. I always assumed that it is somehow independent of coordinate systems.

My second question: I thought that a linear transformation doing a 90° counter-clockwise rotation is represented by a unique matrix \begin{equation*} A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \end{equation*} but, as was shown in the video for Jennifer's choice of basis vectors the same 90° counter-clockwise rotation linear transformation is in fact \begin{equation*} B = \begin{pmatrix} 1/3 & -2/3 \\ 5/3 & -1/3 \end{pmatrix}. \end{equation*} It seems like a linear transformation has a one-one mapping to a unique matrix only for a given set of basis vectors. Thus, the same matrix can refer to different linear transformations if we choose a different basis vector. In case, I am correct, could you provide a mathematically rigorous way of writing this down (using math symbols). I feel that I understand concepts better if I can write it in a mathematical form, instead of relying solely on intuition.

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    $\begingroup$ For your first question: If a text does not specify the basis in which it is working, then you can always think that you are working with the canonical basis, namely $\{ e_1, e_2 \}$ in $\mathbb{R}^2$, where $e_1=(1,0)$ and $e_2=(0,1)$. Remember that, if you are working in a vector space $V$ of dimension $2$ and basis $\{ v_1 , v_2 \}$, then it is exactly as working in $\mathbb{R}^2$ with canonical basis. $\endgroup$ May 31 '20 at 20:36
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    $\begingroup$ As a professor told me: "The canonical basis is the most beautiful basis in the realm". This means, the simplest basis that comes in your mind, This is exactly done with $e_1 = (1,0,0 \dots , 0), e_2 = (0,1,0, \dots, 0), \dots, e_n =(0,0,0 \dots , 1)$ in $\mathbb{R}^n$ where $e_i$ is a vector with coordinates equal zero except for that one in position $i$ where it takes the value $1$ $\endgroup$ May 31 '20 at 20:47
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    $\begingroup$ Thanks for the edit @Roy. $\endgroup$
    – Manish
    May 31 '20 at 20:48
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    $\begingroup$ "It seems like a linear transformation has a one-one mapping to a unique matrix only for a given set of basis vectors. Thus, the same matrix can refer to different linear transformations if we choose a different basis vector." This is correct. It sounds like your understanding is fine. It's just that, as others have mentioned, when interpreting a matrix as a linear transformation we assume the basis for $\mathbb{R}^n$ is the standard basis $(1,0, \ldots, 0), (0, 1, 0, \ldots), \ldots$ unless another basis is indicated. $\endgroup$ May 31 '20 at 20:52
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    $\begingroup$ For your second question: I think it's better to reverse the point of view. The same given linear transformation can refer to different matrices if you choose a different basis vector. You can do imagine it: pick a pencil, you say "oh this is a pencil". Now invert your head. Here you say "oh! This is the same pencil but I'm watching it from another point of view". This means: the same transformation (pencil) has a lot of representations, i.e. matrices, one for each change of basis (invert your head). $\endgroup$ May 31 '20 at 20:56
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A vector is an element of a vector space. An element of a vector space can be an $n$-tuple of numbers, a polynomial, a matrix, a function etc.

A linear transformation transforms a vector ($n$-tuple, polynomial, matrix, function, etc.) into another vector ($n$-tuple, polynomial, matrix, function, etc.). A matrix cannot transform a vector into another vector, because you can multiply a matrix by an $n$-tuple, but you can't multiply a matrix by a polynomial, a matrix (well, not always, see below), a function, etc.

A matrix associated to a linear transformation can only multiply $n$-tuples of coordinates respect to a basis, and the results are $n$-tuples of coordinates respect to a basis.

Imagine that your vector space is the set of all symmetric $2\times 2$ matrices, and that your linear transformation is:$$T\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\begin{bmatrix} c & a \\ a & b \end{bmatrix}$$

The simplest basis is: $\left\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right\}$.

Respect to this basis the coordinates of $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$ are $(a,b,c)$, the coordinates of $\begin{bmatrix} c & a \\ a & b \end{bmatrix}$ are $(c,a,b)$.

The matrix associated to $T$ respect to that basis is: $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$.

You can't multiply $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ by $\begin{bmatrix} a & b \\ b & c \end{bmatrix}$, but: $$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}=\begin{bmatrix} c \\ a \\ b \end{bmatrix}$$ i.e. $$\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\text{Coord}\left(\begin{bmatrix} a & b \\ b & c \end{bmatrix}\right)=\text{Coord}\left(\begin{bmatrix} c & a \\ a & b \end{bmatrix}\right)$$ This is why:

  • you always need a basis to associate a matrix to a linear transformation (when the basis is omitted you assume the canonical basis),
  • the matrix associated to a linear transformation is unique respect to a fixed basis,
  • you can also have different bases for the domain and the range of a linear transformation, so the matrix associated to a linear transformation is unique respect to the basis of its domain and the basis of its range,
  • since there are infinite bases, there also are infinite matrices associated to a linear transformation.
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  • $\begingroup$ Thanks @Sergio! I really appreciate the detail you went into! $\endgroup$
    – Manish
    May 31 '20 at 21:42
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    $\begingroup$ @ManishBhat You are welcome, happy to be useful. $\endgroup$
    – Sergio
    May 31 '20 at 21:55

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