each vector vec(x) in R^n can be expressed as a sum vec(x)=vec(x)_1 vec(e)_1+...+vec(x)_n vec(e)_n Show that this expression is unique, that is, there does not exist other, different linear combination of vectors vec(e__1, vec(e)_2,...,vec(e)_n which is equal to vec(x)

Karli Kidd

Karli Kidd

Open question

2022-08-21

each vector x R n can be expressed as a sum
x = x 1 e 1 + · · · + x n e n
Show that this expression is unique, that is, there does not exist other, different linear combination of vectors
e 1 , e 2 , . . . , e n
which is equal to x . I've test vector
[ 16 32 ] = 1 [ 1 1 ] + 2 [ 8 16 ]
[ 16 32 ] = 2 [ 2 8 ] + 4 [ 3 4 ]
and wrote it in such different ways. So I don't get what's happening here, is not this against uniqueness? Or if I am wrong then give me some hints how to prove it.

Answer & Explanation

ekhonawy

ekhonawy

Beginner2022-08-22Added 6 answers

This statement means that for a given basis of linearly independent vectors e 1 , , e n there is a unique decomposition of any vector x into a linear combination x = x 1 e 1 + + x n e n . Note that xi are not vectors, but real numbers: coordinates of the vector x in the basis e 1 , , e n
In your example you considered two different bases, so of course the coordinates are different.

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