v2010.10.26 - Convex Optimization

A.7. ZEROS 627(Transpose.)Likewise, for any matrix A∈ R m×nrank(A T ) + dim N(A T ) = m (1585)For any square A∈ R m×m , number of 0 eigenvalues is at least equalto dim N(A T )=dim N(A) while all left-eigenvectors (eigenvectors of A T )corresponding to those 0 eigenvalues belong to N(A T ).For diagonalizable A , number of 0 eigenvalues is precisely dim N(A T )while the corresponding left-eigenvectors span N(A T ). Real and imaginaryparts of the left-eigenvectors remaining span R(A T ).⋄Proof. First we show, for a diagonalizable matrix, the number of 0eigenvalues is precisely the dimension of its nullspace while the eigenvectorscorresponding to those 0 eigenvalues span the nullspace:Any diagonalizable matrix A∈ R m×m must possess a complete set oflinearly independent eigenvectors. If A is full-rank (invertible), then allm=rank(A) eigenvalues are nonzero. [331,5.1]Suppose rank(A)< m . Then dim N(A) = m−rank(A). Thus there isa set of m−rank(A) linearly independent vectors spanning N(A). Eachof those can be an eigenvector associated with a 0 eigenvalue becauseA is diagonalizable ⇔ ∃ m linearly independent eigenvectors. [331,5.2]Eigenvectors of a real matrix corresponding to 0 eigenvalues must be real. A.17Thus A has at least m−rank(A) eigenvalues equal to 0.Now suppose A has more than m−rank(A) eigenvalues equal to 0.Then there are more than m−rank(A) linearly independent eigenvectorsassociated with 0 eigenvalues, and each of those eigenvectors must be inN(A). Thus there are more than m−rank(A) linearly independent vectorsin N(A) ; a contradiction.Diagonalizable A therefore has rank(A) nonzero eigenvalues and exactlym−rank(A) eigenvalues equal to 0 whose corresponding eigenvectorsspan N(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) :A.17 Proof. Let ∗ denote complex conjugation. Suppose A=A ∗ and As i =0. Thens i = s ∗ i ⇒ As i =As ∗ i ⇒ As ∗ i =0. Conversely, As∗ i =0 ⇒ As i=As ∗ i ⇒ s i= s ∗ i .