Vextendsingle vextendsingle vextendsingle

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Unformatted text preview: vextendsingle vextendsingle vextendsingle vextendsingle 4 1 vextendsingle vextendsingle vextendsingle vextendsingle = 2 × 4 − 2 − 3 × 4 = 2 . Consequently, the statement is TRUE . 010 10.0 points Simplify the expression (2 u + 4 v ) · (4 u + 3 v ) − bardbl u − 3 v bardbl 2 for vectors u , v in R n . 1. 7 bardbl u bardbl 2 + 28 u · v − 3 bardbl v bardbl 2 2. 7 bardbl u bardbl 2 + 28 u · v + 3 bardbl v bardbl 2 correct 3. 7 bardbl u bardbl 2 − 28 u · v − 3 bardbl v bardbl 2 4. 7 bardbl u bardbl 2 + 4 u · v + 3 bardbl v bardbl 2 5. 7 bardbl u bardbl 2 − 4 u · v − 3 bardbl v bardbl 2 6. 7 bardbl u bardbl 2 − 28 u · v + 3 bardbl v bardbl 2 Explanation: By linearity, (2 u + 4 v ) · (4 u + 3 v ) = 2 u · (4 u + 3 v ) + 4 v · (4 u + 3 v ) = 8 u · u + 6 u · v + 16 v · u + 12 v · v , while bardbl u − 3 v bardbl 2 = ( u − 3 v ) · ( u − 3 v ) = u · ( u − 3 v ) − 3 v · ( u − 3 v ) = u · u − 3 u · v − 3 v · u + 9 v · v . But v · u = u · v , u · u = bardbl u bardbl 2 , v · v = bardbl v bardbl 2 . So after expansion the expression becomes 8 bardbl u bardbl 2 + 22 u · v + 12 bardbl v bardbl 2 − ( bardbl u bardbl 2 − 6 u · v + 9 bardbl v bardbl 2 ) = 7 bardbl u bardbl 2 + 28 u · v + 3 bardbl v bardbl 2 . 011 10.0 points For each y in R n and each subspace W of R n , the vector y − proj W y is in W ⊥ . True or False? 1. FALSE 2. TRUE correct Explanation: Each y can be written uniquely in the form y = hatwide y + z where hatwide y is the projection of y onto W and z is in W ⊥ . But then y − proj W y = y − hatwide y = z is in W ⊥ . Consequently, the statement is TRUE . 012 10.0 points By diagonalizing the matrix A = bracketleftbigg 3 4 − 2 − 3 bracketrightbigg , Version 058 – EXAM03 – gilbert – (56780) 6 compute f ( A ) for the polynomial f ( x ) = 4 x 3 − x 2 + 2 x + 3 . 1. f ( A ) = bracketleftbigg − 20 − 24 − 12 − 16 bracketrightbigg 2. f ( A ) = bracketleftbigg 20 24 12 16 bracketrightbigg 3. f ( A ) = bracketleftbigg − 20 − 24 12 16 bracketrightbigg 4. f ( A ) = bracketleftbigg 20 24 − 12 − 16 bracketrightbigg correct Explanation: If A can be diagonalized by A = PDP-1 = P bracketleftbigg d 1 d 2 bracketrightbigg P-1 , then f ( A ) = Pf ( D ) P-1 = P bracketleftbigg f ( d 1 ) f ( d 2 ) bracketrightbigg P-1 . Now A can be diagonalized if we can find an eigenbasis of R 2 of eigenvectors v 1 , v 2 of A corresponding to eigenvalues λ 1 , λ 2 , for then: A = P bracketleftbigg λ 1 λ 2 bracketrightbigg P-1 , P = [ v 1 v 2 ] . But det[ A − λI ] = vextendsingle vextendsingle vextendsingle vextendsingle 3 − λ 4 − 2 − 3 − λ vextendsingle vextendsingle vextendsingle vextendsingle = 8 − (3 − λ )(3 + λ ) = λ 2 − 1 = 0 , i.e. , λ 1 = 1 and λ 2 = − 1. Corresponding eigenvectors are v 1 = bracketleftbigg 2 − 1 bracketrightbigg , v 2 = bracketleftbigg 1 − 1 bracketrightbigg , so P = bracketleftbigg 2 1 − 1 − 1 bracketrightbigg , P-1 = bracketleftbigg...
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