v2007.09.13 - Convex Optimization

A.7. ZEROS 511For diagonalizable matrix A (A.5), the number of 0 eigenvalues isprecisely dim N(A) while the corresponding eigenvectors span N(A). Thereal and imaginary parts of the eigenvectors remaining span R(A).(TRANSPOSE.)Likewise, for any matrix A∈ R m×nrank(A T ) + dim N(A T ) = m (1366)For any square A∈ R m×m , the number of 0 eigenvalues is at least equalto dim N(A T ) = dim N(A) while the left-eigenvectors (eigenvectors of A T )corresponding to those 0 eigenvalues belong to N(A T ).For diagonalizable A , the number of 0 eigenvalues is preciselydim N(A T ) while the corresponding left-eigenvectors span N(A T ). The realand imaginary parts 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. [247,5.1]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. [247,5.2]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.dim N(A) = 2, λ(A) = [0 0 0 1] T ; three eigenvectors in the nullspace but only two areindependent. The right-hand side of (1365) is tight for nonzero matrices; e.g., (B.1) dyaduv T ∈ R m×m has m 0-eigenvalues when u ∈ v ⊥ .A.17 Let ∗ denote complex conjugation. Suppose A=A ∗ and As i = 0. Then s i = s ∗ i ⇒As i =As ∗ i ⇒ As ∗ i =0. Conversely, As∗ i =0 ⇒ As i=As ∗ i ⇒ s i= s ∗ i .