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Linear combination

  • #1
162
0
Given the vectors
v1=(1, 1) ^t
v2=(3, -1)^t

setting up the matrix gives det≠0, thus any vector in R^n can be written as a linear combination of v1 and v2.

This is where I'm getting confused.
If the numbers in the matrix were changed so det=0, can you still right any vector in R^n as a linear combination of v1 and v2?
If det=0, this would yield a free variable. In the examples in the book, they say you can write a vector as a linear combination of other vectors even if a free variable exists.
 

Answers and Replies

  • #2
162
0
I guess this leads to my next question:
If v1 and v2 are linearly independent vectors
in R^n and v3 cannot be written as a scalar
multiple of v1, then v1, v2, and v3 are linearly
independent.
Why is this statement false?



Never mind, I thought the statement said "and v3 cannot be written as a scalar multiple of v1&v2" instead of just v1. If it was v1 and v2, then it would be linearly independent.
 
  • #3
jbunniii
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Given the vectors
v1=(1, 1) ^t
v2=(3, -1)^t

setting up the matrix gives det≠0, thus any vector in R^n can be written as a linear combination of v1 and v2.

This is where I'm getting confused.
If the numbers in the matrix were changed so det=0, can you still right any vector in R^n as a linear combination of v1 and v2?
If det=0, this would yield a free variable. In the examples in the book, they say you can write a vector as a linear combination of other vectors even if a free variable exists.
det=0 if and only if both columns are a scalar multiple of a particular vector, call it v. If that is the case, then any linear combination of the columns will also be a scalar multiple of v. Therefore any vector that is not a scalar multiple of v cannot be expressed as a linear combination of the columns.
 

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