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I'm working through an introductory linear algebra textbook and one exercise gives the system

$2x+3y+5z+2w=0$

$-5x+6y-17z-3w=0$

$7x-4y+3z+13w=0$

And asks why, without doing any calculations, it has infinitely many solutions. Now, a previous exercise gives the same system without the fourth column and asks why, without any calculation, you can tell it's consistent, and I realized that it's because it has the trivial solution (0,0,0). But I'm struggling to see how that implies that this new system has infinitely many solutions.

I did some research and found that if an underdetermined linear system has a solution then it has infinitely many, but the explanations of this seem to talk about rank and other things that I'm not familiar with.

So if someone could please explain why you can just tell without doing any calculation why this system has infinitely many solutions (I'm guessing it has something to do with the previous problem that's the same just without that fourth column of variables) from the authors perspective (i.e. they're only assuming we have algebra 2 at this early point in the book) it would be much appreciated.

Liam Cooney
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    The system you gave is homogeneous (all constants on the right are 0) and involves 4 variables, but there's only 3 equations. You must have exactly as many (non-redundant) equations as variables in order to isolate a unique solution. Since you have fewer equations than variables, the system must be under-constrained, which means that it doesn't put enough conditions in to restrict the system to a single solution. About redundancy: This is what in linear algebra is called linear dependence. An equations that can be reconstructed from the others in the system is in some sense "redundant". – Iwillnotexist Idonotexist Jan 11 '16 at 00:39

3 Answers3

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You can translate each equation as a dot product: e.g. the first line is $(2,3,5,2)\cdot(x,y,z,w)=0$.

Now, we are in a $4$ dimensional space (we are looking for the vector $(x,y,z,w)$), and we have $3$ vectors in there (made from the coefficients), so these span at most $3$ dimensions, so there is at least $1$ dimension that is orthogonal to it.

So, there is a whole line or plane (or space), containing the origin, that solves the equation system.

Berci
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  • I'm beginning to understand this, but I have very poor spatial skills so might you be willing to be a by more explicit? Also, I'm thinking there is something even simpler the authors are looking for – Liam Cooney Jan 10 '16 at 20:44
  • In $2$ and $3$ dimensions, I guess, you already know the dot product. It produces $0$ for two vectors iff they are orthogonal to each other. For higher dimensions (where vectors have more coordinates than $3$), this aspect is taken as the definition of orthogonality, and it works well. In the 3d space, if you have a line or a plane, there is a plane or line that is orthogonal to it. The same story goes on in higher dimensions, the dimensions are summed. – Berci Jan 10 '16 at 20:55
  • I'm still not really getting it. Maybe I just need to better understand the dot product – Liam Cooney Jan 10 '16 at 21:07
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Note that the system without the fourth column is not only consistent but also determined ( the rows are linearly independent), this means that also the system: $$ \begin{cases} 2x+3y+5z=-2\\ -5x+6y-17z=3\\ 7x-4y+3z=-13 \end{cases} $$ is detrmined, i.e has one solution $(x,y,z)=(a,b,c)$.

Now, your system is

$$ \begin{cases} 2x+3y+5z=-2w\\ -5x+6y-17z=3w\\ 7x-4y+3z=-13w \end{cases} $$ so, by linearity, has the infinitely many solutions $(x,y,z)=(aw,bw,cw) \quad \forall w \in \mathbb{R}$

Emilio Novati
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  • I'm not sure if you're confused about what I meant or I'm not understanding your explanation. Where is the first system of eqns coming from? My first system is the same eqns on the left but all zeros on the right. So how does those eqns being set equal to zero having the trivial solution (0,0,0) also imply that system has a solution? – Liam Cooney Jan 10 '16 at 21:43
  • What is confusing? – Emilio Novati Jan 10 '16 at 21:45
  • Does that clarify my confusion? – Liam Cooney Jan 10 '16 at 22:00
  • Now I understand. And, yes, my answer was confusing, so I better specify it. – Emilio Novati Jan 10 '16 at 22:04
  • I'm not familiar with linear independence. What does it mean? I understand the explanation from there but I'm still struggling to see how we now that new matrix has a solution. Is it as simple as just verifying that it does by noting that the lines are not parallel? – Liam Cooney Jan 10 '16 at 22:08
  • It means that the system without the fourth column has only one solution . – Emilio Novati Jan 10 '16 at 22:11
  • Okay so since it is determined, then obviously we can add any constant term to it and it will still only have one solution thus your new system has a solution? :) – Liam Cooney Jan 10 '16 at 22:15
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Obviously , $x=y=z=w=0$ is a solution. The rank of matrix $A$ is at most $3$, but we have $4$ columns.

If the solution would be unique, the rank of $A$ would have to be equal to the number of columns, which is not the case. Hence, there are infinite many solutions.

Peter
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  • This is one of the results I found in my "research" but as I mentioned, I have not yet learned about rank and so I'm avoiding that explanation. Also, this explanation obviously requires a proof which is likely to if my current grasp. I'm still thinking there's something easier that the authors are trying to get us to see – Liam Cooney Jan 10 '16 at 20:41