How Can You Prove a Spanning Set Theorem Without Summation?

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SUMMARY

The theorem states that if {V1, V2,...Vn} is a spanning set for a vector space V, then any collection of m vectors in V, where m>n, is linearly dependent. The discussion provides a proof without summation notation by expressing vectors in terms of the spanning set and demonstrating that the coefficients must satisfy a system of homogeneous equations. This illustrates the linear dependence of the vectors when m exceeds n, confirming the theorem's validity. The proof emphasizes the necessity of general notation for broader applications.

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loli12
Can someone help me prove this theorem without using the summation sign? because the proof in my book uses the summation method to prove this and i have trouble understanding the those signs. Thanks a lot!

Theorem: If {V1, V2,...Vn} is a spanning set for a vector space V, then any colletion of m vectors in V, where m>n, is linearly dependent.
 
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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!
 
Thanks a lot. I will try using the summation notation with it!
 

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