When trying to find a basis, I got (-3, -11, 1, 0 ) & (3, 1, 0, 10)

When i saw an example using a different technique to me, (3, 1, 0, 10) & (3, 0, 1, 11) was the basis?

Is that possible, generally and in this example? Thanks

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- May 13th 2011, 09:08 AMsj247Is a Basis unique?
When trying to find a basis, I got (-3, -11, 1, 0 ) & (3, 1, 0, 10)

When i saw an example using a different technique to me, (3, 1, 0, 10) & (3, 0, 1, 11) was the basis?

Is that possible, generally and in this example? Thanks - May 13th 2011, 09:15 AMspiral
A basis is not unique.

As long as your basis spans the vector space, and the elements in the basis are linearly independent, then your basis will work; in general, we usually try to find the smallest or simplest basis that will work. - May 14th 2011, 02:24 AMDeveno
all bases have the same "size" (literally if V is finite-dimensional).

in this case, however, one of the bases must be incorrect: (3,0,1,11) is not in span{(-3,-11,1,0),(3,1,0,10)}.

if it were, we would have (3,0,1,11) = a(-3,-11,1,0) + b(3,1,0,10), giving us:

-3a + 3b =3

-11a + b = 0

a = 1

10b = 11 --> b = 11/10. but -11 + 11/10 is not 0, so we cannot find any such a and b. - May 14th 2011, 06:40 AMHallsofIvy
"Ackbeet: Deveno beat me"

Do you have that on a rubber stamp?(Rofl) - May 14th 2011, 08:06 AMspiralQuote:

-3a + 3b =3

-11a + b = 0

a = 1

10b = 11 --> b = 11/10. but -11 + 11/10 is not 0, so we cannot find any such a and b.

I got a = 1/10, b = 11/10 which seems to work out. - May 14th 2011, 02:09 PMDeveno
the 3rd coordinate of (3,0,1,11) is 1. the 3rd coordinate of a(-3,-11,1,0) + b(3,1,0,10) = (-3a+3b,-11a+b,a,10b) is a.

hence, if the two are to be equal, a MUST EQUAL 1. the 4 equations

-3a+3b = 3

-11a+b = 0

a=1

10b=11

must all have a SIMULTANEOUS solution for (3,1,0,11) to be in the span.