# Existence of coefficient matrix for given System of Linear Equations

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 $$X$$?

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.
• So there will be exactly one vector in $Y$ which cannot be expressed in terms of vectors in $X$? – RandomPerfectHashFunction Aug 26 '19 at 13:22
• 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. – S. Dolan Aug 26 '19 at 13:29
• 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? – S. Dolan Aug 26 '19 at 13:46