Editing Singular value decomposition (section)

=== Singular values, singular vectors, and their relation to the SVD ===
A non-negative real number {{tmath|\sigma}} is a '''[[singular value]]''' for {{tmath|\mathbf M}} if and only if there exist unit-length vectors {{tmath|\mathbf u}} in {{tmath|K^m}} and {{tmath|\mathbf v}} in {{tmath|K^n}} such that

<math display=block>\begin{align}
\mathbf{M v} &= \sigma \mathbf{u}, \\[3mu]
\mathbf M^*\mathbf u &= \sigma \mathbf{v}.
\end{align}</math>

The vectors {{tmath|\mathbf u}} and {{tmath|\mathbf v}} are called '''left-singular''' and '''right-singular vectors''' for {{tmath|\sigma,}} respectively.

In any singular value decomposition

<math display=block>
\mathbf M = \mathbf U \mathbf \Sigma \mathbf V^*
</math>

the diagonal entries of {{tmath|\mathbf \Sigma}} are equal to the singular values of {{tmath|\mathbf M.}} The first {{tmath|p {{=}} \min(m,n)}} columns of {{tmath|\mathbf U}} and {{tmath|\mathbf V}} are, respectively, left- and right-singular vectors for the corresponding singular values.  Consequently, the above theorem implies that:
* An {{tmath|m \times n}} matrix {{tmath|\mathbf M}} has at most {{tmath|p}} distinct singular values.
* It is always possible to find a [[orthogonal basis|unitary basis]] {{tmath|\mathbf U}} for {{tmath|K^m}} with a subset of basis vectors spanning the left-singular vectors of each singular value of {{tmath|\mathbf M.}}
* It is always possible to find a unitary basis {{tmath|\mathbf V}} for {{tmath|K^n}} with a subset of basis vectors spanning the right-singular vectors of each singular value of {{tmath|\mathbf M.}}

A singular value for which we can find two left (or right) singular vectors that are linearly independent is called ''degenerate''.  If {{tmath|\mathbf u_1}} and {{tmath|\mathbf u_2}} are two left-singular vectors which both correspond to the singular value σ, then any normalized linear combination of the two vectors is also a left-singular vector corresponding to the singular value σ.  The similar statement is true for right-singular vectors.  The number of independent left and right-singular vectors coincides, and these singular vectors appear in the same columns of {{tmath|\mathbf U}} and {{tmath|\mathbf V}} corresponding to diagonal elements of {{tmath|\mathbf \Sigma}} all with the same value {{tmath|\sigma.}}

As an exception, the left and right-singular vectors of singular value 0 comprise all unit vectors in the [[cokernel]] and [[Kernel (linear algebra)|kernel]], respectively, of {{tmath|\mathbf M,}} which by the [[rank–nullity theorem]] cannot be the same dimension if {{tmath|m \neq n.}}  Even if all singular values are nonzero, if {{tmath|m > n}} then the cokernel is nontrivial, in which case {{tmath|\mathbf U}} is padded with {{tmath|m - n}} orthogonal vectors from the cokernel.  Conversely, if {{tmath|m < n,}} then {{tmath|\mathbf V}} is padded by {{tmath|n - m}} orthogonal vectors from the kernel.  However, if the singular value of {{tmath|0}} exists, the extra columns of {{tmath|\mathbf U}} or {{tmath|\mathbf V}} already appear as left or right-singular vectors.

Non-degenerate singular values always have unique left- and right-singular vectors, up to multiplication by a unit-phase factor {{tmath|e^{i\varphi} }} (for the real case up to a sign).  Consequently, if all singular values of a square matrix {{tmath|\mathbf M}} are non-degenerate and non-zero, then its singular value decomposition is unique, up to multiplication of a column of {{tmath|\mathbf U}} by a unit-phase factor and simultaneous multiplication of the corresponding column of {{tmath|\mathbf V}} by the same unit-phase factor.
In general, the SVD is unique up to arbitrary unitary transformations applied uniformly to the column vectors of both {{tmath|\mathbf U}} and {{tmath|\mathbf V}} spanning the subspaces of each singular value, and up to arbitrary unitary transformations on vectors of {{tmath|\mathbf U}} and {{tmath|\mathbf V}} spanning the kernel and cokernel, respectively, of {{tmath|\mathbf M.}}