Actually, I doubt it. Saying that you do not understand a particular symbol doesn't mean using it isn't the best way to do a problem.
Certainly, you can give examples that might be clearer but don't fool yourself into thinking that that is the same as a proof!
Suppose {V1, V2, V3} is a spanning set for a vector space V and {u1, u2, u3, u4} is a collection of vectors in V. (Here, I'm taking n= 3, m= 4> 3.)
Now, suppose you have some numbers, a, b, c, d, so that au1+ bu2+ cu3+ du4= 0.
Since V1, V2, V3 span V, we can write each of u1, u2, u3, u4 using those:
u1= p1V1+ p2V2+ p3V3 for some numbers p1, p2, p3
u2= q1V1+ q3V2+ q3V3 for some numbers q1, q2, q3
u3= r1V1+ r2V2+ r3V3 for some numbers r1, r2, r3
u4= s1V1+ s2V2+ s3V3 for some numbers s1, s2, s3
Now put those into the equation au1+ bu2+ cu3+ du4= 0:
a(p1V1+ p2V2+ p3V3)+ b(q1V1+ q2V2+ q3V3)+c(r1V1+ r2V2+ r3V3+d(s1V1+ s2V2+ s3V3)= 0.
we can write that as
(ap1+ bq1+ cr1+ ds1)V1+ (ap2+ bq2+cr2+ds2)V2+ (ap3+bq3+cr3+ds3)V3= 0
Certainly one way that can be true (not necessarily the only way- we are not assuming V1, V2, V3 are independent themselves) is if each coefficient is 0:
ap1+ bq1+ cr1+ ds1= 0
ap2+ bq2+ cr2+ ds2= 0
ap3+ bq3+ cr3+ ds3= 0
That's three homogenous equations for 4 unknowns, a, b, c, d. Certainly a= b= c= d= 0 is one solution but its easy to see that there are others. For example, take d= 0, c= 1 Then we have ap2+ bq2= -r2, ap3+ bq3= -r3 and can solve those two equations for non-zero a and b.
That's an example to illustrate how the general proof works. If you want to prove it is true for any n, you HAVE to use some kind of general notation. I recommend the you rewrite this example using summation notation to get a better idea of how summation notation works!