# Change to standard basis

Let $A=\{a_1, a_2, a_3\}$ be a basis, of which each vector is aligned to a cartesian axis. Given a vector $v_{\langle A\rangle}$, how can I get transform it to the standard, canonical basis so it becomes $v_{\langle I\rangle}$?

I must apologize if this question is silly, I've completely forgotten my linear algebra classes (which makes me want to take them again).

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First express the basis vectors $\{a_1 , a_2, a_3\}$ in the standard basis: $$a_{1, \langle I \rangle} = \pmatrix{a_{1,1} \\ a_{1,2} \\ a_{1,3} } \\ a_{2, \langle I \rangle} = \pmatrix{a_{2,1} \\ a_{2,2} \\ a_{2,3} } \\ a_{3, \langle I \rangle} = \pmatrix{a_{3,1} \\ a_{3,2} \\ a_{3,3} } \\$$ Form the matrix $$A = \pmatrix{a_{1,1} & a_{2,1} & a_{3,1} \\ a_{1,2} & a_{2,2} & a_{3,2} \\ a_{1,3} & a_{2,3} & a_{3,3}}$$ Now, given any $v_{\langle A \rangle} = (v_1 , v_2, v_3),$ we have $$v_{\langle I \rangle} = A \pmatrix{v_1 \\ v_2 \\v_3}.$$

TL;DR $v_{\langle I \rangle}= (v_1, v_2, v_3)_{\langle A \rangle} = v_1 a_{1, \langle I \rangle} + v_2 a_{2, \langle I \rangle} + v_3 a_{3, \langle I \rangle}.$

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I spent way too long messing with the matrix implementation, while the TL;DR was much simpler and faster. Thanks, although now I feel bad and want to take lin. alg. again. – Lazlo Jul 2 '12 at 17:31
@LazloBonin If you can't enrol in a class, you can go through Gilbert Strang's videos here for example – user2468 Jul 2 '12 at 17:56