The proof of SVD solver formula, $C^T C = V \Sigma^T \Sigma V^T$

This formula could be used to compute matrix multiplication transpose

$${\displaystyle (\mathbf {AB} )^{\mathsf {T}}=\mathbf {B} ^{\mathsf {T}}\mathbf {A} ^{\mathsf {T}}}$$

This formula is used to elaborate matrix multiplication associativity

$${\displaystyle (\mathbf {AB} )\mathbf {C} =\mathbf {A} (\mathbf {BC} )}$$

This formula is used to compute svd of a matrix.

$$C = U \Sigma V^T$$

this MIT course puts them together and gives (equation_1)

$$C^T C = V \Sigma^T \Sigma V^T \tag 1$$

what is the detailed proof for equation_1?

first, what is the detailed procedure of $$C^TC = (U \Sigma V^T)^TC$$

how can I transfer this to $$U^TU$$?

$$C^T = V \Sigma^T U^T$$

• $U$ is real orthogonal. Thus $U^TU=I$. Commented Jun 24, 2019 at 8:37
• Remember, $U$ is chosen to be such that $U^T U = I$ (identity matrix). Using this and your first two equations, you should be able to prove the result. (Note that with associativity, you can compute any two adjacent matrices' product first, and with the transpose rule, it extends to multiple matrices: $$(ABC\ldots XYZ)^T = Z^TY^T X^T \ldots C^T B^T A^T.$$) Commented Jun 24, 2019 at 8:37
• You haven’t numbered your equations. Which one is “equation_1?” (Use \tag to number them.)
– amd
Commented Jun 24, 2019 at 23:07

Using your first two equations, you can get

$$(ABC)^T = ((AB)C)^T = C^T(AB)^T = C^T B^T A^T$$

and then

$$C^T = V \Sigma^T U^T$$

and then

$$C^T C = V \Sigma^T U^T C = V \Sigma^T U^T U \Sigma V^T$$

since

$$U^TU = I$$

eventually you get

$$C^T C = V \Sigma^T U^T U \Sigma V^T = V \Sigma^T \Sigma V^T$$