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Let $V$ and $W$ be vector spaces with (standard, orthonormal) bases $\mathcal{V} = (v_1, \dots, v_k)$ and $\mathcal{W} = (w_1, \dots, w_n)$, and let $T : V \to W$ be a linear map. With respect to the bases $\mathcal{V}$ and $\mathcal{W}$, $T$ has matrix representation $G$.

With respect to some other (unknown) bases $\mathcal{V}' = (v_1', \dots, v_k')$ and $\mathcal{W}' = (w_1', \dots, w_n')$ of $V$ and $W$, $T$ has matrix representation $G'$.

Hence $G$ and $G'$ are related by the matrix equation $$ G = I_W G' I_V^{-1}, $$ where $I_V$ is the identity matrix on $V$ with respect to the bases $\mathcal{V}$ and $\mathcal{V}'$, and likewise, $I_W$ is the identity matrix on $W$ with respect to the bases $\mathcal{W}$ and $\mathcal{W}'$.

The problem: compute the matrices $I_W$ and $I_V^{-1}$ given $G$ and $G'$. (equivalently, compute the new bases vectors in $\mathcal{V}'$ and $\mathcal{W}'$ with respect to the old basis vectors, given only the matrix of $T$ in each representation.)

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  • $\begingroup$ What do you mean by orthonormal? Do you have a scalar product, or some kind of bilinear form on these spaces? $\endgroup$ – Jendrik Stelzner Jan 27 '16 at 0:10
  • $\begingroup$ Yes, take $V$ and $W$ to be inner product spaces $\mathbf{F}^k$ and $\mathbf{F}^n$, respectively, where $\mathbf{F}$ is the common field over which $V$ and $W$ are defined. $\endgroup$ – Jon Warneke Jan 27 '16 at 0:16
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I don’t think that this can be solved in general without more information about the bases.

Suppose for instance that $V=W$ and $G=G'=I$. Then $\mathcal V'=\mathcal W'$ can be any basis for $V$.

Addendum: More generally, if $\operatorname{rank} G = r = \operatorname{rank} G'$, then we can find invertible matrices $S$, $S'$, $T$ and $T'$ (i.e., choose bases for $V$ and $W$) such that $$SGT = \pmatrix{I_r&0\\0&0} = S'G'T',$$ so $G=(S^{-1}S')\,G'(T'T^{-1})$. In the construction of these matrices, the bases for $V$ can be chosen arbitrarily, hence $\mathcal V'$ is also arbitrary.

Even if you’re given a basis $\mathcal V'$, that may not be enough to guarantee a unique solution for $\mathcal W'$. If $r<\dim W$, then a part of the basis for $W$ can also be chosen arbitrarily.

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  • $\begingroup$ Right, but then we know $I_V^{-1} = I_W = \mbox{diag}(\underbrace{1, \dots, 1}_{\dim V \mbox{ ones}})$ $\endgroup$ – Jon Warneke Jan 27 '16 at 0:34
  • $\begingroup$ @JonWarneke No. All we know is that $I_WI_V^{-1}=I$, which only lets us conclude that $I_W=I_V$, not that either is the identity, but we already knew that because we took $V=W$. $\endgroup$ – amd Jan 27 '16 at 2:35

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