Thus p w is given by p w x h x w 1 i h w 1 w 1 i w 1

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Thus, P W is given by P W ( ~x ) = h ~x, ~w 1 i h ~w 1 , ~w 1 i ~w 1 + h ~x, ~w 2 i h ~w 2 , ~w 2 i ~w 2 + h ~x, ~w 3 i h ~w 3 , ~w 3 i ~w 3 . Using this formula, we compute P W ( ~e 1 ) = 1 3 ~w 1 - 1 3 ~w 2 + 1 15 ~w 3 = 1 15 11 2 - 6 2 , P W ( ~e 2 ) = 1 3 ~w 1 + 1 3 ~w 2 + 2 15 ~w 3 = 1 15 2 14 3 - 1 , P W ( ~e 3 ) = 0 3 ~w 1 + 1 3 ~w 2 - 1 15 ~w 3 = 1 15 - 6 3 6 3 , P W ( ~e 4 ) = 1 3 ~w 1 + 0 3 ~w 2 - 3 15 ~w 3 = 1 15 2 - 1 3 14 .
Thus, the desired matrix is 1 15 11 2 - 6 2 2 14 3 - 1 - 6 3 6 3 2 - 1 3 14 . 3. Let V = C ([0 , 2]) be the vector space (over R ) of continuous functions from [0 , 2] to R . Define h f, g i = Z 1 0 f ( t ) g ( t ) dt for f, g V. Prove that , ·i is not an inner product on V . Proof . Define f ( t ) = ( 0 if t 1 , t - 1 if t > 1 , so that f is continuous, and hence f ∈ C ([0 , 2]). Clearly f 6 = ~ 0, since f is not everywhere 0 on [0 , 2]. However, h f, f i = Z 1 0 f 2 dt = Z 1 0 0 dt = 0 , contradicting the positivity axiom for inner products. So , ·i is not an inner product. QED 4. Decide whether or not C = 1 0 0 - 1 1 1 1 0 0 is diagonalizable. Answer . The characteristic polynomial is det( C - λI ) = - λ (1 - λ ) 2 ; so the characteristic polynomial splits, and the only eigenvalues are λ = 0 , 1. To finish testing for diagonalizability, we need to check that each eigenspace has the appropriate dimension. The eigenvalue λ = 0 has algebraic multiplicity 1, so its eigenspace is automatically dimension 1, and we don’t need to check explicitly. [But since the first and third rows of C are the same, clearly the eigenspace of 0, i.e., the null space of C , is nontrivial. And if you really want to be specific, it’s Span { [0 1 - 1] } .] For λ = 1, we compute C - I = 0 0 0 - 1 0 1 1 0 - 1 , which has rank 1 and therefore nullity 2, since each row is a multiple of [1 0 - 1]. Thus, the eigenspace of 1 has the proper dimension of 2, and therefore C is diagonalizable. [If you want to be more specific, the eigenspace of 1 is Span { [1 0 1] , [0 1 0] } . So our three eigenvectors { [1 0 1] , [0 1 0] } , [0 1 - 1] } form a basis for F 3 , thus diagonalizing C .] 5. Let A M 3 × 3 ( R ) be symmetric. Suppose that ~v 1 = 1 1 0 and ~v 2 = 1 - 1 2 are eigenvectors of A with eigenvalues λ 1 = 3 and λ 2 = - 2, respectively. Suppose also that det( A ) = 6. Find A . Answer . Call the third eigenvalue λ 3 . [At the moment, it’s conceivable that λ 3 might be a repeat of one of the two we already know.] Since the product of the eigenvalues equals det( A ), we have - 6 λ 3 = 6, and hence λ 3 = - 1. Since A is symmetric, any eigenvector ~v 3 for λ 3 = - 1 must be orthogonal to the other two. That is, setting B = 1 1 0 1 - 1 2 , we must have B~v 3 = ~ 0. Solving this system gives Ker( B ) = Span 1 - 1 - 1 , so we may choose ~v 3 = 1 - 1 - 1 .
Dividing each of our three eigenvectors by its length, we get that Q = 1 / 2 1 / 6 1 / 3 1 / 2 - 1 / 6 - 1 / 3 0 2 / 6 - 1 / 3 is an orthogonal matrix whose columns are the eigenvectors of A . Thus, putting the eigenvalues in a diagonal matrix in the same order, we have A = Q 3 0 0 0 - 2 0 0 0 - 1 Q t = · · · = 1 6 5 13 - 2 13 5 2 - 2 2 - 10 .

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