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$T : \Bbb P_n \to \Bbb P_n$ given by $T(f(t)) = \dfrac{df(t)}{dt}$, where $\Bbb P_n$ is space of polynomials with degree $n$.

The matrix representation of this linear transformation (basis as $\{1, .., t^n\}$) would be $$T(\mathbf v) = \begin{bmatrix} 0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ 0&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&n\\ 0&0&0&\cdots&0 \end{bmatrix} \mathbf v$$

where $\bf v$ is $\Bbb R^n$ column vector representation of $f(t)$, like if $f(t) = \sum_{k = 0}^n \alpha_k t^k$ then $v = (\alpha_0, ... , \alpha_n)^T$.

If I am not wrong then the $T$ in matrix representation is not same as $T$ of the original question. The $T$ of the original question is $T : \Bbb P_n \to \Bbb P_n$ whereas the $T$ in the matrix representation is $T : \Bbb R^n \to \Bbb R^n$.

So why we call the $T$ in matrix representation same as the $T$ in original question ?

Is matrix represention of a linear transformation $R : V \to W$ just a way to write the same information as some other linear tranformation $S : \Bbb F^n \to \Bbb F^n$ ?

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  • $\begingroup$ You are not thinking about the bases. You are thinking that the matrix takes in vectors in the standard basis. What is the input basis and output basis? The input and output bases do not have to be the standard $R^n$ bases. $\endgroup$
    – AnlamK
    Oct 1, 2017 at 19:00
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    $\begingroup$ The matrix used to represent the operator $T$ is a different object, but folks often abuse notation knowing that the matrix represents the operator in the context of a specific basis. $\endgroup$
    – copper.hat
    Oct 1, 2017 at 19:00
  • $\begingroup$ @AnlamK The matrix takes $\Bbb R^n$ column vector of coefficients of the polynomial and gives the column vector coefficients of derivative of that polynomial, which under a basis gives the derivative of polynomial. The function $T$ takes a polynomial and gives it derivative. Isn't that two different object ? $\endgroup$
    – user8277998
    Oct 1, 2017 at 19:13
  • $\begingroup$ I see your point (perhaps you are right philosophically) but remember for linear transformations, to describe the linear transformation those coefficients (which basis vector gets mapped to which basis vectors with which coefficients) are all you need. So the matrix representation completely specifies the linear transformation. That’s the reason why people are referring to the $T$ and matrix $T$ as the same. Does this make sense? $\endgroup$
    – AnlamK
    Oct 1, 2017 at 19:24
  • $\begingroup$ @AnlamK Yes it does. $\endgroup$
    – user8277998
    Oct 2, 2017 at 5:39

1 Answer 1

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$T$ is an operator, and whose domain is always same, you have mistaken it to be from $\mathbb{R}^n$ to $\mathbb{R}^n$, while that matrix is also operating on $\mathbb{P}_n$.

Matrix is just the representation of the transformation but to specify the actual transformation it should be operated on same Vector space, $\mathbb{P}_n$ in this case.

Hope it works.

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