ALA240-2009 - October 29th

ALA240-2009 - October 29th - is a linear transformation ²...

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Page 1 of 2 October 29 th MAT 240 Linear Algebra Lecture Reminder : Choosing a basis, V is isomorphic to ° ? Goal : 1. The set ? (V, W) of all linear transformations ± → ² is a V.S. 2. Choosing bases, it is isomorphic to ³ ´?? °µ ´ = dim ±µ ? = dim ( ² ) Extra Claim (from last class final example): If two linear transformations , ` : ? → · agree on a basis of X, they are equal. If ( ? ¸ ) is a basis of X and ∀¸ ¶ ? ¸ µ = ` ? ¸ µ ∈ · then = ` Proof: Pick some element ? ∈ ? , as ( ? ¸ ) as a basis, find scalars ¹ ¸ such that. ? = º ¹ ¸ ? ¸ Now: ¶ ?µ = ( » ¹ ¸ ? ¸ ) = » ¹ ¸ ¶ ? ¸ µ = » ¹ ¸ ` ? ¸ µ = » ¹ ¸ ` ? ¸ µ = `( » ¹ ¸ ? ¸ ) = `( ? ) → ¶ = ` Let ¼ = ( ½ 1, ⋯ ½ ? ) be a basis ( basis is ordered ) of a finite dimension vector space V. ? ∈ ± ? = » ¹ ¸ ½ ¸ Let ¾?¿ ¼ = À ¹ 1 ¹ ? Á = Â ?µ ¼ = ( ½ 1, ⋯ ½ ? ) of V Ã = ( Ä 1 ⋯ Ä ? ) of ° ? Â : ± → ° ? Defined by: ½ ¸ → Ä ¸ Indeed, Â ?µ = Â » ¹ ¸ ½ ¸ µ = » ¹ ¸ Â ½ ¸ µ = » ¹ ¸ Ä ¸ = À ¹ 1 ¹ ? Á = ¾?¿ ¼
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Page 2 of 2 ? : ? → ±
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Unformatted text preview: is a linear transformation, ² = ( ³ 1, ⋯ ³ ´ ) is a basis for V, µ = ( ? 1, ⋯ ? ´ ) is a basis for W ° = ?¶ ² µ = · [ ?¸³ 1 ¹ ] µ ⋯ [ ?¸³ ´ ¹ ] µ º In » 2 ( ¼ ) : ? 2 − 2 ? + 3 ¶ ( ? 2 , ? ,1) = ½ 1 − 2 3 ¾ ? 2 − 2 ? + 3 ¶ (1, ? , ? 2 ) = ½ 3 − 2 1 ¾ ? 2 − 2 ? + 3 ¶ ( ? 2 , ? ,3) = ½ 1 − 2 1 ¾ Coordinates depend on choice of basis! D: » 3 ( ¼ ) → » 2 ( ¼ ) (Differentiation) ² = ( ? 3 , ? 2 , ? , 1) µ = ( ? 2 , ? , 1) ¿¶ ² µ = ¿¶ ( ? 3 , ? 2 , ? ,1) ( ? 2 , ? ,1) = À [ ¿¸? 3 ¹ ] µ [ ¿¸? 2 ¹ ] µ [ ¿¸?¹ ] µ [ ¿¸ 1 ¹ ] µ Á = À [3 ? 2 ] µ [2 ? ] µ [1] µ [0] µ Á =  3 2 1 à ? Ä : ¼ 2 → ¼ 2 ? Ä ¶ Å 1 , Å 2 Å 1 , Å 2 = Æ cos Ä − sin Ä sin Ä cos Ä Ç...
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ALA240-2009 - October 29th - is a linear transformation ²...

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