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We know the standard form of expressing a system of linear equations in $n$ variables in $n$ equations.

$$A_{n \times n} \cdot X_{n \times 1} = B_{n \times 1}$$

Where $A$ is the coefficient matrix, $X$ is the unknown variables (each variable maybe a real or complex number) matrix and $B$ is the constants matrix.

Now, arriving at the question, let's say I have a set of $n$ vectors $X$, each of length $n$ and another set of $n+1$ vectors $Y$, each of length $n$. (Clearly, $size(X) < size(Y) $). All elements of $X$ are independent. The same goes for $Y$.

Now I want to express each element in $Y$, as a linear expression of all the elements in $X$. So that would be something like below, for some coefficient matrix $K$.

$$ K_{(n+1)\times n} \cdot X_{n \times n} = Y_{(n+1)\times n} $$

My question is, is it necessary that $K$ must always exist i.e. there must be a way to express all elements in $Y$ as a linear combination of $X$.

I think the answer is No. The reasoning is that, for a system of linear equations where number of equations is strictly greater than the number of unknowns, there does not exist a solution at all. In the above case, we have $n$ unknowns from $X$ and $n+1$ equations, each one for each element in $Y$.

Is my answer and reasoning correct?

Edit Given all vectors in $X$ are linearly independent and all vectors in $Y$ are linearly independent. Also given that if out of $n+1$ vectors in $Y$, each of the $n$ vectors from $Y$ can be expressed as a linear sum of the $n$ vectors in $X$, then is it or is it not possible for the $n+1^{th}$ vector in $Y$ to always be expressed in terms of the vectors in $X$?

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Since the vectors $X$ span n-dimensional space each vector $Y$ can be expressed as a linear combination of them. This is true for as many $Y$s as you like.

Note that only $n$ of your vectors $Y$ can be linearly independent.

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  • $\begingroup$ So there will be exactly one vector in $Y$ which cannot be expressed in terms of vectors in $X$? $\endgroup$ – RandomPerfectHashFunction Aug 26 '19 at 13:22
  • $\begingroup$ No, every $Y$ can be expressed. You might find it helpful to think about the example where all the $Y$s are the same - its easy to express all of them in terms of the $X$s. $\endgroup$ – S. Dolan Aug 26 '19 at 13:29
  • $\begingroup$ I think my question is not clear. Let me modify it. $\endgroup$ – RandomPerfectHashFunction Aug 26 '19 at 13:31
  • $\begingroup$ You could consider this example. $X=$ \begin{pmatrix}1&0\\0&1\end{pmatrix} and let $Y$ be any nx2 matrix. Then, for $K=Y$, we have $KX=Y$. Does this help? $\endgroup$ – S. Dolan Aug 26 '19 at 13:46
  • $\begingroup$ Yeah your case is perfectly valid for matrices with atomic elements like real or complex numbers. But my matrix elements are vectors and that is my doubt. Is such a case possible? $\endgroup$ – RandomPerfectHashFunction Aug 26 '19 at 13:54

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