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Let S be a set of n vectors in an n-dimensional vector space V .
(a) If S is independent, then S is a basis for V .
(b) If S spans V , then S is a basis for V .
kia ap kr chuki hein prove??
ap extended day k chakr my kun ho? abi koshis kry
malika sister mujy ye chez bi confuse kr rhi hy k is ka answer points my ho ga ya nahi.
i think,points mein nhi hoga Coz,
Hint: Here's the geometric interpretation. Any two vectors in R^3 make a plane.
Add any vector which is not in that plane and you get a basis set (because the new vector is not a linear combination of the first two).
For example, the vectors { (1, 0, 0) , (0, 1, 0) , (0, 0, -1) } are a basis for R3
different from the standard basis, because the last one has -1 instead of 1.
Now, I would like to check that they are linearly independent. This means that
I have to imagine that someone did SOME combination of the three which equals
zero. Write it like that:
a*(1, 0, 0) + b*(0, 1, 0) + c*(0, 0, -1) = (0, 0, 0),
where a,b,c are real numbers. The important thing here is that YOU DON'T choose
the numbers a, b, c. If you choose them, then you're doing nothing. What you
have to do is to show that they MUST be all zero; that is, you have to show that
there is NO OTHER POSSIBILITY for them.
That is the meaning of linearly independent. So how do you go about this?
Simple, just make the computation on the left side above, and you get:
(a, b, -c) = (0, 0, 0).
And from this, its obvious that you must have a = 0, b = 0, c = 0.
Hence that is a basis, and its different from the standard basis.
ye solution hy ???
no example hai
finally i think linearly independent prove krna hai and
result geometrically show krna hai
see theorem
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