v2010.10.26 - Convex Optimization

628 APPENDIX A. LINEAR ALGEBRAThe (right-)eigenvectors of a diagonalizable matrix A∈ R m×m are linearlyindependent if and only if the left-eigenvectors are. So, matrix A hasa representation in terms of its right- and left-eigenvectors; from thediagonalization (1547), assuming 0 eigenvalues are ordered last,A =m∑λ i s i wi T =i=1k∑≤ mi=1λ i ≠0λ i s i w T i (1586)From the linearly independent dyads theorem (B.1.1.0.2), the dyads {s i w T i }must be independent because each set of eigenvectors are; hence rankA = k ,the number of nonzero eigenvalues. Complex eigenvectors and eigenvaluesare common for real matrices, and must come in complex conjugate pairs forthe summation to remain real. Assume that conjugate pairs of eigenvaluesappear in sequence. Given any particular conjugate pair from (1586), we getthe partial summationλ i s i w T i + λ ∗ i s ∗ iw ∗Ti = 2re(λ i s i w T i )= 2 ( res i re(λ i w T i ) − im s i im(λ i w T i ) ) (1587)where A.18 λ ∗ i λ i+1 , s ∗ i s i+1 , and w ∗ i w i+1 . Then (1586) isequivalently writtenA = 2 ∑ iλ ∈ Cλ i ≠0res 2i re(λ 2i w T 2i) − im s 2i im(λ 2i w T 2i) + ∑ jλ ∈ Rλ j ≠0λ j s j w T j (1588)The summation (1588) shows: A is a linear combination of real and imaginaryparts of its (right-)eigenvectors corresponding to nonzero eigenvalues. Thek vectors {re s i ∈ R m , ims i ∈ R m | λ i ≠0, i∈{1... m}} must therefore spanthe range of diagonalizable matrix A .The argument is similar regarding span of the left-eigenvectors. A.7.40 trace and matrix productFor X,A∈ R M×N+ (39)tr(X T A) = 0 ⇔ X ◦ A = A ◦ X = 0 (1589)A.18 Complex conjugate of w is denoted w ∗ . Conjugate transpose is denoted w H = w ∗T .