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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.