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I'm stuck with a question.

The question is: What is the change of base matrix $P_{C\leftarrow B}$ from the base $B$ of $V=Span_c\{cos(x),sin(x)\}$ to the base $C$ of $V$ if $B=(\exp(ix), \exp(-ix))$ and $C=(\cos(x), \sin(x))$?

I have no idea how to handle this situation. I thought the concept was as follows: $C=P(C\leftarrow B) B$ But I can't see how I would apply this.

1 Answers 1

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Hint: Forget definitions for now. How would you rewrite $b_1 \exp(ix) + b_2 \exp(-ix)$ as a linear combination of $\cos(x)$ and $\sin(x)$?

If you can write $$ b_1 \exp(ix) + b_2 \exp(-ix) = c_1 \cos(x) + c_2 \sin(x) $$ then your change of base matrix is the matrix satisfying $$ M\pmatrix{b_1\\b_2} = \pmatrix{c_1\\ c_2} $$

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    Oh wow! Thank you very much! I thought I had to satisfy M(exp(ix),exp(-ix))=(cos(x),sin(x)). But looking at your answer, it makes sense now! Which makes the matrix M: (1 1, i -i) Because exp(ix)= cos(x)+i sin(x) and exp(-ix)= cos(x)- i sin(x). Thanks again! – 2017-02-05
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    You got it. When in doubt, check the definitions – 2017-02-05