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We have $$ \bar{L}\bar{C}=T^{-1}LCT$$ where $T$ is a coordinate change matrix, $L$ is a column vector and $C$ a row vector of compatible dimension.

What if I want to find out $L$? The result should be $L=T^{-1} \bar{L}$ but I can't understand why.

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    What does overline mean? – 2017-01-11
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    @mathreadler it is not a particular linear algebra convention.. $\bar{C}$ is the C matrix in input state output representation of a dynamic systems when the states are taken from the transformed coordinates space... – 2017-01-11
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    you can multiply with $T$ to the left on both sides and then $T^{-1}C^{-1}$ to the right and you will get $L$ on the right hand side. – 2017-01-11
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    Acutally it should be $L = T^{-1} \bar{L} T$ , but you can not derive that from the one equation that you have given. – 2017-01-11
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    @mathreadler: $C$ is a row vector so it does not have a matrix inverse. If you meant a pseudo-inverse it would work though. – 2017-01-11
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    @Diracules: $L$ is a column vector, so it cannot be the result of computing $T^{-1} \bar{L} T$, since $T$ is a matrix. – 2017-01-11
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    Ok, I don't think I understand the notation well enough to understand the problem. – 2017-01-11

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I'm assuming you are given the product $\bar{L} \bar{C}$ and the matrix $T$ from which you want to find $L$, is that right?

In this case, if you are given $\bar{L} \bar{C} = T^{-1} L C T$ then you can compute $L C = T \bar{L} \bar{C} T^{-1}$. From this, there is no unique way to get $L$ since $L C = (\alpha L) \cdot (\frac{1}{\alpha}C)$ for any $\alpha \neq 0$. However, besides for this scaling ambiguity you can find $L$: since $T \bar{L} \bar{C} T^{-1}$ is rank one, take any nonzero column of it and it qualifies as an estimate of $L$ (they are all scaled copies of each other and scaling is not determined uniquely). Moreover, if $T \bar{L} \bar{C} T^{-1}$ is not exactly rank one (e.g., due some observation errors), you can obtain a rank-one approximation via the truncated SVD. In this case, the dominant left singular vector is a viable estimate for $L$.

Of course, if you also know $C$ it's simpler, you then solve for $L$ by multiplying $L C = T \bar{L} \bar{C} T^{-1}$ with $C^T$ from the right to get $$L C C^T = T \bar{L} \bar{C} T^{-1} C^T \quad \Rightarrow \quad {L} = T \bar{L} \cdot \frac{\bar{C} T^{-1} C^T}{C C^T}.$$

Note that the two terms in the fraction are scalar, since $C$ is a row vector.