# Confused about the relation between linear transformations, matrices and basis vectors

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.

• 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. May 31, 2020 at 20:36
• 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 $1$ May 31, 2020 at 20:47
• Thanks for the edit @Roy. May 31, 2020 at 20:48
• "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. May 31, 2020 at 20:52
• 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). May 31, 2020 at 20:56

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.
• Thanks @Sergio! I really appreciate the detail you went into! May 31, 2020 at 21:42
• @ManishBhat You are welcome, happy to be useful. May 31, 2020 at 21:55