MATH-M 303 Lecture Notes - Lecture 18: Linear Combination, Coordinate Vector, Isomorphism
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Since (cid:1828) spans (cid:1848), relation above holds. Since (cid:1828) linearly independent, all scalars must be 0, so (cid:1855)=(cid:1856) for (cid:882) ; so representation of (cid:2206) must be unique. For (cid:2206)=(cid:4666)(cid:2869),(cid:2870)(cid:4667) (cid:2870), what is [(cid:2206)]: (cid:2206)=(cid:2869)(cid:2778)+(cid:2870)(cid:2779) by definition of (cid:2206) given here and by the fact we know any vector can be expressed, so [(cid:2206)]=(cid:2206) Given a general basis (cid:1828) for and vector (cid:2206), how do we find [(cid:2206)]: ex. Need scalars (cid:1855)(cid:2869),(cid:1855)(cid:2870) such that (cid:2206)=(cid:1855)(cid:2869)(cid:2778)+(cid:1855)(cid:2870)(cid:2779)=[(cid:2778) (cid:2779)][(cid:1855)(cid:2869)(cid:1855)(cid:2870)] Solve equation for (cid:1855)(cid:2869),(cid:1855)(cid:2870) with augmented matrix [(cid:2778) (cid:2779)|(cid:2206)] (cid:883)|(cid:886)(cid:887)] ~ [(cid:883) (cid:882)(cid:882) (cid:883)|(cid:885)(cid:884)] =(cid:2206: found that, for basis for , [(cid:2206)]=(cid:2206), where =[(cid:2778) (cid:2779) ] and (cid:2206) . Converts (cid:1828)-coordinates into standard coordinates by multiplication. Since (cid:1828) is a basis, we know invertible (columns span , linearly independent) Can rewrite equation as [(cid:2206)]= (cid:2869)(cid:2206: recall coordinate map (cid:1846): given by (cid:1846)(cid:4666)(cid:2206)(cid:4667)=[(cid:2206)] (cid:1846) is linear with standard matrix (cid:2869: since invertible, (cid:1846) invertible- holds more generally.
