Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.1.9 The Matrix Transpose

Jason Glowney

It is not read­i­ly ap­par­ent, but the SVD gives us a very con­cise geo­met­ric in­ter­pre­ta­tion of what the linear trans­pose of a matrix is, where the linear trans­pose of a matrix MM is

M†=VΣU†.\begin{array}{r} M^{\dagger} = V\Sigma U^{\dagger}. \end{array}
(1.1.35)

Proof (1.1.36) - We prove that M†=VΣU†M^{\dagger} = V\Sigma U^{\dagger}, by start­ing with

M=UΣV†,M = U\Sigma V^{\dagger},

and mul­ti­ply both sides on the left by VV

MV=UΣV†V→MV=UΣ,MV = U\Sigma V^{\dagger}V \rightarrow MV = U\Sigma,

take the trans­form of both MVMV and UΣU\Sigma

(MV)†=(UΣ)†→V†M†=Σ†U†.(MV)^{\dagger} = (U\Sigma)^{\dagger} \rightarrow V^{\dagger}M^{\dagger} = \Sigma^{\dagger}U^{\dagger}.

Since Σ\Sigma is di­ag­o­nal with only real com­po­nents, Σ†=Σ\Sigma^{\dagger} = \Sigma, we can then left-mul­ti­ply both sides by VV, prov­ing that

VV†M†=VΣU†→M†=VΣU†.\begin{array}{r} {VV}^{\dagger}M^{\dagger} = V\Sigma U^{\dagger} \rightarrow M^{\dagger} = V\Sigma U^{\dagger}. \end{array}
□\square

The find­ings from the proof reveal the trans­pose of a matrix rep­re­sents the SVD done in the re­verse order, where first is the ro­ta­tion/re­flec­tion by U†U^{\dagger}, then the scal­ing by Σ\Sigma, and the finish is a ro­ta­tion/re­flec­tion by VV. The reader should keep this notion of the matrix trans­pose in mind, as this is not typ­i­cal­ly taught in in­tro­duc­to­ry linear al­ge­bra.