Unformatted text preview: envalue decomposition, but all matrices
have a SVD 8 / 20 Matrix properties via SVD (cont’d)
Theorem
The nonzero singular values of A are the square roots of the nonzero
eigenvalues of AA or A A (they have the same nonzero eigenvalues). Proof.
From deﬁnition,
AA = (U ΣV )(U ΣV ) = U ΣV V ΣU 2
2
= U diag(σ1 , . . . , σp ) U Theorem
For A ∈ IRm×m ,  det(A) = m
i =1 σi Proof.
m det(A) = det(U ΣV ) = det(U )det(Σ)det(V ) = det(Σ) = σi
i =1
9 / 20 Lowrank approximation
Theorem
(EckartYoung 1936) Let A = U ΣV = U diag(σ1 , . . . , σr , 0, . . . , 0)V .
For any ν with 0 ≤ ν ≤ r , Aν = ν=1 σi ui vi ,
i
A − Aν 2 =
min
A − B 2 = σν +1
rank(B )≤ν Proof.
Suppose there is some B with rank(B ) ≤ ν such that
A − B 2 < A − Aν 2 = σν +1 . Then there exists an (n − ν )dimensional
subspace W ∈ IRn such that w ∈ W ⇒ B w = 0. Then
Aw 2 = (A − B )w 2 ≤ A − B 2 w 2 < σν +1 w 2
Thus W is a (n − ν )dimensional subspace where Aw < σν +1 w . But
there is a (ν + 1)dimensional subspace where Aw ≥ σν +1 w , namely
the space spanned by the ﬁrst ν + 1 right singular vector of A. Since the
sum of the dimensions of these two spaces exceeds n, there must be a
nonzero vector lying in both, and this is a contradiction.
10 / 20 Lowrank approximat...
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 Spring '14
 Computer Science, Linear Algebra, Singular value decomposition, eigenvalue decomposition, minimum norm, minimum norm solution

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