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1. - 1.71. Our proof of the Cauchy-Schwarz inequality, Theorem 1.13, used that when  $U$ is a unit vector,  $0 \leq ||V−(U·V)U||^2 = ||V||2 ||V||^2  −(U·V)^2$. Therefore if $U$ is a unit vector and equality holds, then $V = (U · V)U$. Show that equality occurs in the Cauchy Schwarz inequality for two arbitrary vectors $V$ and $W$ only if one of the vectors is a multiple (perhaps zero) of the other vector.  Answer: 
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$$V = (U · V)U = (\frac{W}{||W||}· V)·\frac{W}{||W||}= (\frac{W·V}{||W||^2})·W$$  As $(\frac{W·V}{||W||^2})$ is a constant, so $V$ is a multiple of $W$.  2. -2.19. 2.19.  Suppose C is an n by n matrix with orthonormal columns. Use Theorem 2.2 to show that  $$||CX|| \leq \sqrt{n} ||X||$$  Use the Pythagorean theorem and the result of Problem 2.17 to show that in fact 
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(2)By Problem 2.17, $CX=x_1C_1 +x_2C_2 +\dots+x_nC_n$.  To find the norm of RHS, we need to apply the Pythagorean theorem (we need that C has orthonormal columns) to get  $$||x_1C_1 + x_2C_2 + \dots + x_nC_n||^2 = ||x_1C_1||^2 + ||x_2C_2||^2 + \dots + ||x_nC_n||^2$$  We Now we  can put the pieces together: $$||CX||^2 = ||x_1C_1 + x_2C_2 + \dots · · ·  + x_nC_n||^2$$ $$= ||x_1C_1||^2+||x_2C_2||^2+ \dots+ ||x_nC_n||^2 $$  $$=x_1^2||C_1||^2+x_2^2||C_2||^2+\dots+x_n^2||C_n||^2$$  $$=x_1^2+x_2^2+\dots+x_n^2$$  $$= ||X||^2.$$ = ||X||2.  Since norms are nonnegative, we can conclude that $||CX|| = ||X||$.