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This is a task from a test exam you can find here (in German): http://docdro.id/QRtdXkM

Is the following statement true or false?

Every real homogeneous linear system of equation that has more than one solution, has infinite solutions.

I think the statement is true because a linear system of equations can only have either one solution, no solution or infinite solutions. This statement clearly says "more than one solution $\rightarrow$ infinite solutions" which is true.

Is it really correct like that or there is some special case which can make this statement false?

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    $\begingroup$ The statement is correct but you need to prove it mathematically. How do you know that " a linear system of equations can only have either one solution, no solution or infinite solutions"? $\endgroup$
    – Ofek Gillon
    Mar 4, 2017 at 10:59
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    $\begingroup$ @CarstenS: The question is clearly using "infinite" as a numeral, essentially as a shorthand for "an infinite number of". Whether you consider such use of the word "infinite" to be proper English or not, it's certainly clear enough in context. If the question had instead said "... has five solutions", you wouldn't object on the basis that 5 is not necessarily a solution, would you? $\endgroup$
    – Ilmari Karonen
    Mar 4, 2017 at 13:56
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    $\begingroup$ @IlmariKaronen the thing is that five is a numeral and infinite is not. Exact language is important. The OP has already received helpful answers, so my comment will not harm him, but maybe he will think for a moment about why the original problem said “unendlich viele” and not “unendliche”. $\endgroup$
    – Carsten S
    Mar 4, 2017 at 14:02
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    $\begingroup$ And another remark to the OP: Would the answer have been different if the question had not explicitly said real? $\endgroup$
    – Carsten S
    Mar 4, 2017 at 14:04
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    $\begingroup$ @cnmesr: You made a simple English language error, writing "infinite solutions" where you clearly meant "infinitely many solutions". The rest of the discussion is simply whether this is a serious problem (as Carsten S seems to claim) or rather nitpicking (as Ilmari Karonen and myself think). $\endgroup$
    – celtschk
    Mar 4, 2017 at 15:28

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Indeed, this is even true for non-homogeneous linear systems. Consider the system $Ax=b$, and assume $x_0$ and $x_1$ are solutions. Then for any $x_\lambda = (1-\lambda)x_0+\lambda x_1$ you get $$Ax_\lambda = A((1-\lambda)x_0 + \lambda x_1) = (1-\lambda)A x_0 + \lambda A x_1 = (1-\lambda) b + \lambda b = b$$ Therefore $x_\lambda$ is also a solution, thus you get infinitely many (indeed even uncountably many) solutions.

The homogeneous system is just the special case for $b=0$. Since $x=0$ is always a solution of a homogeneous linear system, for those you can even write the condition as:

If any real homogeneous linear system of equations has a non-zero solution, it has infinitely many.

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  • $\begingroup$ The last sentence is very nice to know, surely also the rest of your answer, now I understood, thanks! $\endgroup$
    – cnmesr
    Mar 4, 2017 at 12:05
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    $\begingroup$ @cnsmer: note that the zerovector always forms a solution to a homogeneous system of equations. Therefore, you can only have either 1 (in case A is invertable) or infinitely many (in case A is singular) solutions. $\endgroup$
    – Student
    Mar 4, 2017 at 12:28
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    $\begingroup$ @cnmesr The clam that there are infinitely many solutions is not true generally, e.g. if the coefficient field is a finite field such as $\,\Bbb Z_2 = $ integers mod $2$. Generally vector spaces are defined over any field of scalars. $\endgroup$
    – Bill Dubuque
    Mar 4, 2017 at 14:41
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    $\begingroup$ @BillDubuque: But then it would not be a real homogeneous linear system of equations, would it? $\endgroup$
    – celtschk
    Mar 4, 2017 at 14:53
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    $\begingroup$ @ClassicEndingMusic: I guess you refer to the exam linked in the question. It is labelled "weiß nicht" which is German for "don't know". According to the rules stated on the exam, if you cross the wrong answer, you get a point deducted, but if you cross "don't know" your points are not affected. The idea probably being to discourage guessing the answer. $\endgroup$
    – celtschk
    Mar 4, 2017 at 17:32
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Hint: if the homogeneous system has two distinct solutions, then one of them, call it $v$, is nonzero. What can you say about $\alpha v$ for a scalar $\alpha$?

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Yes indeed this is true. The way that this is best seen is by noting that if a homogeneous system of equations has more than one solution, then the matrix corresponding to the system of equations, $A$, has a non-trivial kernel.

This means that $\exists \vec{v},\vec{u} \in \ker(A): A\vec{v} = 0$ and $A\vec{u} = 0$. Now it is worth noting that we can take any linear combination of these vectors $\lambda \vec{v} + \mu \vec{u}$ for $\lambda, \mu \in \mathbb{R}$ and we have $A(\lambda \vec{v} + \mu \vec{u}) = \lambda A\vec{v} + \mu A\vec{u} = 0$, so $\lambda \vec{v} + \mu \vec{u}$ also solves this system of equations.

Therefore we have an infinite number of solutions if we have more than one unique solution.

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Every linear comination over $\mathbb{R}$ of two solutions of a homogeneous linear system will be solution to the homogeneous linear system. One of the solutions in necessarly not the zero vector and then it has infinte non trivial combinations which are solutions. If you write it as a matrix, $A \in _{\mathbb{R}_{n \times m}}$, represents the homogeneous linear equation system, andfor the vectors $a,b \in \mathbb{R}^{m}$ such that $Aa=0,Ab=0$.

From that you get that for every $\alpha, \beta \in \mathbb{R}$, exists:

$A(\alpha a+\beta b)=\alpha Aa+\beta Ab=0$.

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