# Solution 12 Computational Intelligence Lab 2013

Copied and adapted from: Solution 10 Computational Intelligence Lab 2012

# Problem 1

1
a) rank=4, eigenvalues = -1,1,2,2 => spectral norm = 2, nuclear norm = 6
b) rank=2, eigenvalues = 0,3,3 => spectral norm = 3, nuclear norm = 6
2
${\displaystyle L={\begin{pmatrix}1&1&1&1&1\\2&2&2&2&2\\3&3&3&3&3\\1&1&0&0&0\\1&1&0&0&0\end{pmatrix}}}$
${\displaystyle S={\begin{pmatrix}0&1&0&0&0\\0&0&0&-1&0\\0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&0\end{pmatrix}}}$
rank(L) = 2
card(S) = 3
F(L,S) = 2+3 = 5

# Problem 2

1
• 1-norm sums up the absolute value of all elements -> small if only few elements are nonzero
• large matrix with one very large value
• large matrix filled with small values
2
• nuclear norm = sum of singular values -> small if only few singular values are nonzero -> small rank
• rank 1 matrix with a very high singular value
• full rank matrix with all very small singular values (nearly singular matrix)
3
• in general, sparse means number of elements are of order O(sqrt(m*n))