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I understand that matrices represent linear maps and all of the exercises I have done have been representing a specific example linear map with a matrix.

It got me wondering: can a unique matrix represent 2 different linear maps (i.e., can 2 different linear maps be represented by the same matrix)? If it is possible, could someone give me an example?

Thank you!

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    $\begingroup$ This is an excellent question $\endgroup$
    – Sambo
    Jun 25 at 14:54
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    $\begingroup$ Yes, of course if you use different bases for the matrix. $\endgroup$ Jun 25 at 15:40

1 Answer 1

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Let $V$ and $V'$ be finite-dimensional vector spaces (it may be the case that $V=V'$). A linear map $F\colon V\to W$ is completely specified by the assignment of three pieces of information: an "input" basis $\mathcal{B}$ of $V$, an "output" basis $\mathcal{B}'$ of $V'$, and a matrix $A$ with appropriate dimension. Changing one of these, while letting the other two fixed, may result in the prescription of a different linear map (there are exceptions, such as the null map $F\equiv 0$).

For example, consider the map $F\colon \mathbb R^2\to \mathbb R^2$ given by $$ F(x, y)=(x+y, x-y).$$ Taking $\mathcal B=\mathcal B'=\{(1,0), (0, 1)\}$, this map is associated to the matrix $\begin{bmatrix} 1 & 1 \\ 1& -1\end{bmatrix}$. However, changing $\mathcal{B}'$ to $\{(1, 1), (1, -1)\}$, the associated matrix is now $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$.

ADDENDUM. Here I wrote an example of two matrices associated to the same linear map. The same blue print can produce two linear maps associated to the same matrix.

A RELATED PROBLEM. An interesting classic problem is the following. Let a finite-dimensional vector space $V\ne \{0\}$ be fixed. Are there linear maps $F\colon V\to V$ such that their associated matrix is the same with respect to all bases? Note that here we only consider maps of $V$ into itself, and we require the same basis in input and in output.

(The answer is that such maps are precisely the scalar maps, that is, the ones of the form $F(v)=\lambda v$ for all $v\in V$.)

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