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Given a set of column vectors $v_1, v_2,...,v_t$ is there a way to calculate a unique transition matrix?

In other words, is there one and only one matrix $A$ such that $Av_{i} = v_{i+1}$?

Additionally, is it possible to find the "transition matrix of best fit" for $v_1, v_2,...v_t$? In other words, the matrix $A$ that minimizes the error between a predicted vector and the actual?

I'm asking because I'm trying to use Markov chains to predict a series of column vectors and I'm curious if there's a more "vectorized" (i.e. linear algebra based) implementation of calculating the transition matrix.

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    I'm assuming you actually want $Av_i = v_{i+1}$, and not just $Av_1 = v_2$. It depends on the $v_i$; for a trivial example, if $v_i=\mathbf{0}$ but $v_{i+1}\neq\mathbf{0}$, then no matrix $A$ with $Av_i=v_{i+1}$ exists. More generally, if $v_i\in\mathrm{span}(v_1,\ldots,v_{i-1})$, $v_i=\alpha_1v_1+\cdots+\alpha_{i-1}v_{i-1}$, then the relations $Av_1=v_2$, $Av_2=v_3$, etc. force $v_{i+1}=Av_i = \alpha_1v_2+\cdots+\alpha_{i-1}v_i$. These conditions are necessary; if the $v_i$ satisfy the conditions, then $A$ exists; uniqueness only follows if the $v_i$ span the vector space.2012-01-30
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    If I misread your problem and you actually want just $Av_1=v_2$, then if $v_1\neq\mathbf{0}$ there are generally *lots* of matrices $A$: complete $\{v_1\}$ to a basis $v_1,u_2,\ldots,u_n$, pick your favorite vectors $w_2,\ldots,w_n$; then there is a matrix $A$ such that $Av_1=v_2$ and $Au_j=w_j$ for $j=2,\ldots,n$. Clearly, this gives lots of liberty in choosing $A$ if you only subject it to the condition $Av_1=v_2$.2012-01-30
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    Yep, that was a typo, thanks. What you said about span makes sense, however, for my particular dataset I can't guarantee that will be true.2012-01-30

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