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Does the statement "having a linear map that maps several vectors to one" make sense?

The question involved actually asks us to find a transformation matrix T(X) such that T(X)=Xa, where X is a nxn matrix and a is a column vector. The way I interpreted this is that T(X) is a map that takes in three column vectors in X and returns the single a. Is this correct?

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    Traditionally, the map $T: \mathbb{R}^{n} \to \mathbb{R}^{m}$ where $x \mapsto Ax = y$, here $x$ is an $n \times 1$ column vector and $A$ is an $m \times n$ matrix, and carrying out the matrix multiplication $y$ is then an $m \times 1$ column vector. – 2017-02-19
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    @JohnSmith, yes, that's why the question arises in the first place – 2017-02-19

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I believe you're asking about the domain space $\mathcal{M}^{n \times n}$, the space of all $n \times n$ matrices. Then, trying to construct a $T$, where $T : \mathcal{M}^{n \times n} \to \mathbb{R}^{n} $ under $A \mapsto Ax$.

Is your question asking you to verify that this is actually a linear transformation? If so,

Take $A, B \in \mathcal{M}^{n \times n}$ and then you have: $$T(A + B) = (A + B)x = Ax + Bx = T(A) + T(B)$$ For any $\lambda \in \mathbb{F}$, the field under the vector space, $$T(\lambda A) = (\lambda A)x = \lambda Ax = \lambda T(A)$$

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    So its fine to map a matrix to a vector, right? I guess that is how I should have framed the question in the first place. – 2017-02-19
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    @RadialApps, yes, we can map a matrix to a vector. – 2017-02-19