$A$ is an $m\times n$ matrix. If $v$ is an $n\times 1$ vector and it is an element of a subspace of dimension $$d \le n $$, then must vectors of the form $Av$ of dimension $m\times 1$ be elements of a subspace of dimension $$e \le d $$?
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$\begingroup$ What do you mean by lives in a subspace? $\endgroup$– Jürgen SukumaranCommented Jun 14, 2016 at 22:31
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$\begingroup$ Something feels a bit weird here: All vectors $v \in V$ live in subspaces of dimensions $1$ through $\dim V$. I imagine alternative wording could be something like, "Suppose $W$ is a subspace of $V$ with $\dim W = d \le n$, and suppose the image $\{Av : v \in W\}$ of $W$ under $A$ has dimension $e$. Is $e \le d$?" $\endgroup$– pjs36Commented Jun 14, 2016 at 22:31
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1$\begingroup$ I think you more clearly restated my question. This may help. $\endgroup$– AlexCommented Jun 14, 2016 at 22:57
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$\begingroup$ @Alex I hope it does! If you get the answer on your own, be sure to write it up as an answer to the question. $\endgroup$– pjs36Commented Jun 15, 2016 at 1:12
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1 Answer
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I will make some assumptions, but I think this is the basic idea:
v is an element of a subspace of dimension d, so $$v= \alpha_1 \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n\end{pmatrix} + \alpha_2 \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n\end{pmatrix} + \cdots + \alpha_d \begin{pmatrix} d_1 \\ d_2 \\ \vdots \\ d_n\end{pmatrix}$$
Since $$A = \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \\ a_{m1} & & a_{mn} \end{pmatrix}$$
Then $$Av = \begin{pmatrix} a_{11} (\alpha_1a_1+\alpha_2b_1+\cdots+\alpha_dd_1) + \cdots + a_{1n} (\alpha_1a_n+\alpha_2b_n+\cdots+\alpha_dd_n) \\ \vdots \\ a_{m1} (\alpha_1a_1+\alpha_2b_1+\cdots+\alpha_dd_1) + \cdots + a_{mn} (\alpha_1a_n+\alpha_2b_n+\cdots+\alpha_dd_n) \end{pmatrix}$$
$$ = \begin{pmatrix} \alpha_1c_{11}+\alpha_2c_{12}+\cdots+\alpha_dc_{1d} \\ \vdots \\ \alpha_1c_{m1}+\alpha_2c_{m2}+\cdots+\alpha_dc_{md}\end{pmatrix}$$
$$ = \alpha_1 \begin{pmatrix} c_{11} \\ \vdots \\ c_{m1} \end{pmatrix} + \cdots + \alpha_d \begin{pmatrix} c_{m1} \\ \vdots \\ c_{md} \end{pmatrix} $$
This vector is an element of a d dimensional subspace if the vectors $$\begin{pmatrix} c_{11} \\ \vdots \\ c_{m1} \end{pmatrix}, \cdots, \begin{pmatrix} c_{1d} \\ \vdots \\ c_{md} \end{pmatrix} $$ are linearly independent. The vector may be an element of a less than d dimensional subspace if the vectors $$\begin{pmatrix} c_{11} \\ \vdots \\ c_{m1} \end{pmatrix}, \cdots, \begin{pmatrix} c_{1d} \\ \vdots \\ c_{md} \end{pmatrix} $$ are linearly dependent. Also, this must be the case when $$n < d$$