v2004.06.19 - Convex Optimization

320 APPENDIX A. LINEAR ALGEBRANow suppose A has more than m−rank(A) eigenvalues equal to 0. Thenthere are more than m−rank(A) linearly independent eigenvectors associatedwith 0 eigenvalues, and each of those eigenvectors must be in N(A). Thusthere are more than m−rank(A) linearly independent vectors in N(A) ; acontradiction.Therefore diagonalizable A has rank(A) nonzero eigenvalues and exactlym−rank(A) eigenvalues equal to 0 whose corresponding eigenvectors spanN(A).By similar argument, the left-eigenvectors corresponding to 0 eigenvaluesspan N(A T ).Next we show when A is diagonalizable, the real and imaginary parts ofits eigenvectors (corresponding to nonzero eigenvalues) span R(A):The right-eigenvectors of a diagonalizable matrix A∈ R m×m are linearlyindependent if and only if the left-eigenvectors are. So, matrix A has a representationin terms of its right and left-eigenvectors; from the diagonalization(913), 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 (940)From the linearly independent dyads theorem (§B.1.1.0.1), 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 (940), we getthe partial summationλ i s i w T i + λ ∗ i s ∗ iw ∗Ti = 2Re(λ i s i w T i )= 2 ( Re s i Re(λ i w T i ) − Im s i Im(λ i w T i ) ) (941)where A.16 λ ∗ iwrittenA = 2 ∑ iλ ∈ Cλ i ≠0∆=λ i+1 , s ∗ ∆i = s i+1 , and wi∗∆= w i+1 . Then (940) is equivalentlyRe s 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 (942)A.16 The complex conjugate of w is denoted w ∗ , while its conjugate transpose is denotedby w H = w ∗T .