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I'm learning linear algebra, and I understand that a linear transformation is represented by a matrix. So the following $T: \Bbb R^n \rightarrow \Bbb R^n $ can be represented as $ \mathbf Ax = b $, where A is square.

I also understand that you can find A from x and b by using them in a system of equations with the basis vectors.

However, how do you to find $ \mathbf A$ if both x and b are rectangular matrices?

The only instruction I found suggests inverting $x$, which is impossible since the matrix is not possible in a non square matrix? Is it only possible to find $ \mathbf A $ if $x$ and $b$ are both square?

For example suppose I have the linear transformation:

$$ \mathbf A\begin{bmatrix} 0.6 && 0.2 && 0.2 \\ 0.5 && 0.3 && 0.1 \end{bmatrix} = \begin{bmatrix} 0.9 && 0.05 && 0.05 \\ 1 && 0 && 0 \end{bmatrix} $$

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  • $\begingroup$ Try looking up row reduction/gaussian elimination $\endgroup$
    – Shuri2060
    Jul 30, 2017 at 15:51
  • $\begingroup$ I saw it but I'm not sure how it help. Can use row reduction to turn which rows are linear independent and then use a matrix with only those? $\endgroup$
    – JMenezes
    Jul 31, 2017 at 5:35
  • $\begingroup$ The solutions to a system after being row reduced are the same as the original system. $\endgroup$
    – Shuri2060
    Jul 31, 2017 at 14:37
  • $\begingroup$ @Shuri2060 How do you suggest to do row reduction in this case? $\endgroup$
    – mfl
    Jul 31, 2017 at 18:37
  • $\begingroup$ It's basically mfl's first method, using that system of equations $\endgroup$
    – Shuri2060
    Jul 31, 2017 at 23:15

1 Answer 1

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First of all note that $$A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}.$$

So you need to solve $$ \begin{bmatrix}a&b\\ c&d\end{bmatrix}\begin{bmatrix} 0.6 & 0.2 & 0.2 \\ 0.5 & 0.3 & 0.1 \end{bmatrix} = \begin{bmatrix} 0.9 & 0.05 & 0.05 \\ 1 & 0 & 0 \end{bmatrix}.$$ That is,

$$\begin{cases}0.6a+0.5b= 0.9\\ 0.2a+0.3b=0.05\\0.2a+0.1b=0.05\\0.6c+0.5d= 0.9\\ 0.2c+0.3d=0.05\\0.2c+0.1d=0.05\end{cases}$$

In this case there is no solution. How we know it? Substract to the first equation the second one and the double of the third one. We get a contradiction.

Edit

Another way to solve it is as follows. If we have

$$ A\begin{bmatrix} 0.6 & 0.2 & 0.2 \\ 0.5 & 0.3 & 0.1 \end{bmatrix} = \begin{bmatrix} 0.9 & 0.05 & 0.05 \\ 1 & 0 & 0 \end{bmatrix}$$ then $$A\begin{bmatrix} 0.6 & 0.2 & 0.2 \\ 0.5 & 0.3 & 0.1 \end{bmatrix} \begin{bmatrix} 0.6 & 0.5\\ 0.2 & 0.3 \\ 0.2& 0.1 \end{bmatrix}= \begin{bmatrix} 0.9 & 0.05 & 0.05 \\ 1 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0.6 & 0.5\\ 0.2 & 0.3 \\ 0.2& 0.1 \end{bmatrix}.$$

That is

$$A\begin{bmatrix}0.44 & 0.38 \\ 0.38&0.35\end{bmatrix}=\begin{bmatrix} 0.56&0.47\\0.6&0.5\end{bmatrix}.$$ Thus, if $A$ exists, it must be

$$A=\begin{bmatrix} 0.56&0.47\\0.6&0.5\end{bmatrix}\begin{bmatrix}0.44 & 0.38 \\ 0.38&0.35\end{bmatrix}^{-1}.$$ Now, we have to check if this is a solution of the original system or not.

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  • $\begingroup$ Thanks for the answer, but is there a way to represent that by letters? More specifically, if I want to find A write a equation where only A is on the left hand side? $\endgroup$
    – JMenezes
    Jul 31, 2017 at 5:32

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