The 'T' means transposed.

Consider the four non-zero COLUMN vectors in R^2:

x=[x1, x2]; y=[y1, y2]; w=[w1, w2]; z=[z1, z2]

Let xTw=0, yTz=0, A=xyT, B=wzT

Determine:

a) The 2-norm of A

b) The singular value decomposition (SVD) of A

c) The Range space and Nullspace of A

d) The pseudo-inverse de A

e) The singular value decomposition (SVD) of A+B

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PART A:

, whereby is the conjugate transpose of .

(Note that since each of the entries in ).

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EDIT: I feel like I owe more of an explanation for why in this problem. is called the conjugate transpose matrix of because each entry in is the complex conjugate (remember from algebra?) of its respective entry (i.e., same location in the matrix) in . In our case, sinceR, the entries in and are identical; .

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So,

Now, we compute our characteristic polynomial, which will give us what we need to know in terms of eigenvalues.

Sooooooooooo,

The greatest of these two roots, is of course, , since it absolutely must be positive by virtue of the fact that both x and y are given as non-zero column vectors, and addition and multiplication of positive real numbers will never give a negative result.

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So FINALLY, we compute the 2-norm of :

This problem is the pinnacle of tedium.

-Andy