v2009.01.01 - Convex Optimization

572 APPENDIX A. LINEAR ALGEBRA
The right-eigenvectors of a diagonalizable matrix A∈ R m×m are linearly
independent if and only if the left-eigenvectors are. So, matrix A has
a representation in terms of its right- and left-eigenvectors; from the
diagonalization (1438), assuming 0 eigenvalues are ordered last,
A =
m∑
λ i s i wi T =
i=1
k∑
≤ m
i=1
λ i ≠0
λ i s i w T i (1472)
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 eigenvalues
are common for real matrices, and must come in complex conjugate pairs for
the summation to remain real. Assume that conjugate pairs of eigenvalues
appear in sequence. Given any particular conjugate pair from (1472), we get
the partial summation
λ i s i w T i + λ ∗ i s ∗ iw ∗T
i = 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 ) ) (1473)
where A.18 λ ∗ i = λ i+1 , s ∗ i
equivalently written
A = 2 ∑ i
λ ∈ C
λ i ≠0
∆
∆
= s i+1 , and w ∗ i
Re s 2i Re(λ 2i w T 2i) − Im s 2i Im(λ 2i w T 2i) + ∑ j
λ ∈ R
λ j ≠0
∆
= w i+1 . Then (1472) is
λ j s j w T j (1474)
The summation (1474) shows: A is a linear combination of real and imaginary
parts of its right-eigenvectors corresponding to nonzero eigenvalues. The
k vectors {Re s i ∈ R m , Im s i ∈ R m | λ i ≠0, i∈{1... m}} must therefore span
the range of diagonalizable matrix A .
The argument is similar regarding the span of the left-eigenvectors.
A.7.4
0 trace and matrix product
For X,A∈ R M×N
+ (32)
tr(X T A) = 0 ⇔ X ◦ A = A ◦ X = 0 (1475)
A.18 The complex conjugate of w is denoted w ∗ , while its conjugate transpose is denoted
by w H = w ∗T .