Basic math of the SVD

From standard linear algebra we know that a square matrix $\boldsymbol{X}$ can be diagonalized if and only it is a so-called normal matrix, that is if $\boldsymbol{X}\in {\mathbb{R}}^{n\times n}$ we have $\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X}$ or if $\boldsymbol{X}\in {\mathbb{C}}^{n\times n}$ we have $\boldsymbol{X}\boldsymbol{X}^{\dagger}=\boldsymbol{X}^{\dagger}\boldsymbol{X}$. The matrix has then a set of eigenpairs

$$(\lambda_1,\boldsymbol{u}_1),\dots, (\lambda_n,\boldsymbol{u}_n),$$

and the eigenvalues are given by the diagonal matrix

$$\boldsymbol{\Sigma}=\mathrm{Diag}(\lambda_1, \dots,\lambda_n).$$

The matrix $\boldsymbol{X}$ can be written in terms of an orthogonal/unitary transformation $\boldsymbol{U}$

$$\boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma}\boldsymbol{V}^T,$$

with $\boldsymbol{U}\boldsymbol{U}^T=\boldsymbol{I}$ or $\boldsymbol{U}\boldsymbol{U}^{\dagger}=\boldsymbol{I}$.

Not all square matrices are diagonalizable. A matrix like the one discussed above

$$\boldsymbol{X} = \begin{bmatrix} 1& -1 \\ 1& -1\\ \end{bmatrix}$$

is not diagonalizable, it is a so-called defective matrix. It is easy to see that the condition $\boldsymbol{X}\boldsymbol{X}^T=\boldsymbol{X}^T\boldsymbol{X}$ is not fulfilled.

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