Lecture 7 - A Start into Chapter 2

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Starting Chapter 2

Suppose v→1,...,v→n∈V. If we scale each one of them:

α1v→1,...,αnv→n∈V

And:

α1v→1+...+αnv→n∈V

Because both of closure under vector addition and scalar multiplication. We give a special name to this, calling it a linear combination:

Linear Combination

Given vectors v→1,...,v→n∈V, a linear combination of these vectors is a vector of the form:

α1v→1+...+αnv→n∈V

Where each αi∈F.

Note that subspaces are thus closed under linear combinations.

Let's look at some examples:

Example

Is (7,4,0) a linear combination of v→1=(0,1,0), v→2=(1,−1,0)?

\begin{proof}
Notice that:

(7,4,0)=11(0,1,0)+7(1,−1,0)=11v→1+7v→2

So yes it's a linear combination.
\end{proof}

Example

Is the function 2sin⁡(x+π4) a linear combination of sin⁡(x) and cos⁡(x)?

\begin{proof}
It is a linear combination since, using sin⁡(α+β)=sin⁡(α)cos⁡(β)+cos⁡(α)sin⁡(β):

2sin⁡(x+π4)=2(sin⁡(x)cos⁡(π/4)+cos⁡(x)sin⁡(π/4))=sin⁡(x)+cos⁡(x)

So yes it is a linear combination. (you can note that in this case V=RR).
\end{proof}

Example

Is ex a L.C. of sin⁡(x) and cos⁡(x)?

\begin{proof}
The answer is no! Notice that any linear combination of sin,cos must live within some rance [a,b] for any a,b∈R. But ex∈[0,∞] which breaks this pattern, so it can't be a linear combination of the two vectors in question.

Also, we can show that vectors in our subspace are periodic, while ex definitely is not.
\end{proof}