Unformatted text preview: corresponding to σ1
Once σ1 , v1 , and v1 are determined, the remainder of SVD is
determined by the action of A on the space orthogonal to v1
Since v1 is unique up to a sign, the orthogonal space is unique deﬁned
and so are the remaining singular values
5 / 20 Matrix properties via SVD
Theorem
The rank of A is r , the number of nonzero singular values. Proof.
The rank of a diagonal matrix is equal to the number of its nonzero
entries, and in SVD, A = U ΣV where U and V are of full rank. Thus,
rank(A) = rank(Σ) = r Theorem
A 2 = σ1 , and A F = 2
2
σ1 + · · · + σr Proof.
As U and V are orthogonal, A = U ΣV , A 2 = Σ 2 . By deﬁnition,
Σ 2 = max x =1 Σx 2 = max{σi } = σ1 . Likewise, A F = Σ F , and
by deﬁnition Σ F = 2
2
σ1 + · · · + σr
6 / 20 Eigenvalue decomposition
From linear algebra, Ax = λx, λ is an eigenvalue, and x is an
eigenvector
For m eigenvectors, λ1 λ2 A[x1 , x2 , . . . , xm ] = [x1 , x2 , . . . , xm ] .. .
λm
and
AX = X Λ
where Λ is an m × m diagonal matrix whose entries are the eigenvalues
of A, and X ∈ IRm×m contains linearly independent eigenvector of A
The eigenvalue decomposition of A
A = X ΛX −1
7 / 20 SVD and eigenvalue decomposition SVD uses two diﬀerent bases (the sets of left and right singular
vectors), whereas the eigenvalue decomposition uses just one
(eigenvectors)
SVD uses orthonormal bases, whereas the eigenvalue decomposition
uses a basis that generically is not orthogonal
Not all matrices have an eig...
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 Spring '14
 Computer Science, Linear Algebra, Singular value decomposition, eigenvalue decomposition, minimum norm, minimum norm solution

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