# HW4_Sol - Math 104 Fall 2009 Homework 4 solution set We use...

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Math 104 Fall 2009 Homework 4: solution set We use the notations below to ease readability. Matrices are bold capital, vectors are bold lowercase and scalars or entries are not bold. For instance, A is a matrix and a ij its ( i, j )th entry. Likewise x is a vector and x j its j th component. The linear span generated by a group of vecotors v 1 , v 2 , . . . , v n is denoted by span( v 1 , v 2 , ..., v n ). Problem 1 It is true that k E k F = k u k F k v k F . Set u = u 1 . . . u m and v = v 1 . . . v n . Then E = uv * = u 1 ¯ v 1 · · · u 1 ¯ v n . . . . . . . . . u m ¯ v 1 · · · u m ¯ v n . By the definition of Frobenius norm, we have k E k 2 F = X 1 i m, 1 j n | u i | 2 | v j | 2 = X 1 i m | u i | 2 X 1 j n | v j | 2 = k u k 2 F k v k 2 F . It follows that k E k F = k u k F k v k F . Problem 2 Recall that k A k 2 = sup x C n k x k 2 =1 k Ax k 2 . Since U is unitary, we have k Uy k 2 = k y k 2 for any y C m . Thus k UA k 2 = sup x C n : k x k 2 =1 k UAx k 2 = sup x C n : k x k 2 =1 k Ax k 2 = k A k 2 . As for the Frobenius norm, We will use the convenient form k A k F = p tr( A * A ). Since U * U = I , we have k UA k F = p tr(( UA

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