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Inlinear algebra, asemi-orthogonal matrix is a non-squarematrix withreal entries where: if the number of columns exceeds the number of rows, then the rows areorthonormal vectors; but if the number of rows exceeds the number of columns, then the columns are orthonormal vectors.

Let${\displaystyle A}$ be an${\displaystyle m\times n}$ semi-orthogonal matrix.

• Either${\displaystyle A^{\mathrm{T}}A=I\text{ or }A{A}^{\mathrm{T}}=I.}$^[1][2][3]
• A semi-orthogonal matrix is anisometry. This means that it preserves the norm either in row space, or column space.
• A semi-orthogonal matrix always has full rank.
• A square matrix is semi-orthogonal if and only if it is anorthogonal matrix.
• A real matrix is semi-orthogonal if and only if its non-zerosingular values are all equal to 1.
• A semi-orthogonal matrixA issemi-unitary (eitherA^†A = I orAA^† = I) and either left-invertible or right-invertible (left-invertible if it has more rows than columns, otherwise right invertible).

Consider the${\displaystyle 3\times 2}$ matrix whose columns are orthonormal:Here, its columns are orthonormal. Therefore, it is semi-orthogonal, which is confirmed by:

Consider the${\displaystyle 2\times 3}$ matrix whose rows are orthonormal:Here, its rows are orthonormal. Therefore, it is semi-orthogonal, which is confirmed by:

The following${\displaystyle 3\times 2}$ matrix has orthogonal, but not orthonormal, columns and is therefore not semi-orthogonal:The calculation confirms this:

If a matrix${\displaystyle A}$ is tall or square (${\displaystyle m\ge n}$), its semi-orthogonality implies${\displaystyle A^{T}A=I_n}$. For any vector${\displaystyle x\in {\mathbb{R}}^n}$,${\displaystyle A}$ preserves its norm:${\displaystyle \Vert Ax{\Vert }_2^2=(Ax{)}^T(Ax)=x^T{A}^{T}Ax=x^T{I}_{n}x=\Vert x{\Vert }_2^2}$If a matrix${\displaystyle A}$ is short (), it preserves the norm of vectors in itsrow space.

If${\displaystyle A^{T}A=I_n}$, then the columns of${\displaystyle A}$ are linearly independent, so the rank of${\displaystyle A}$ must be${\displaystyle n}$.If${\displaystyle A{A}^T=I_m}$, then the rows of${\displaystyle A}$ are linearly independent, so the rank of${\displaystyle A}$ must be${\displaystyle m}$.In both cases, the matrix has full rank.

The statement is that a real matrix${\displaystyle A}$ is semi-orthogonal if and only if all of its non-zero singular values are 1.

This follows directly from theSVD,${\displaystyle A=U\mathrm{\Sigma }{V}^T}$.

(${\displaystyle ⟹}$) Assume${\displaystyle A}$ is semi-orthogonal. Then either${\displaystyle A^{T}A=I}$ or${\displaystyle A{A}^T=I}$. The non-zero singular values of${\displaystyle A}$ are the square roots of the non-zero eigenvalues of both${\displaystyle A^{T}A}$ and${\displaystyle A{A}^T}$. Since one of these "Gramian" matrices is anidentity matrix, its eigenvalues are all 1. Thus, the non-zero singular values of${\displaystyle A}$ must be 1.

(${\displaystyle \Leftarrow }$) Assume all non-zero singular values of${\displaystyle A}$ are 1. This forces the block of${\displaystyle \mathrm{\Sigma }}$ containing the non-zero values to be an identity matrix. This structure ensures that either${\displaystyle {\mathrm{\Sigma }}^T\mathrm{\Sigma }=I_n}$ (if${\displaystyle A}$ has full column rank) or${\displaystyle \mathrm{\Sigma }{\mathrm{\Sigma }}^T=I_m}$ (if${\displaystyle A}$ has full row rank). Substituting this into the expressions for${\displaystyle A^{T}A=V({\mathrm{\Sigma }}^T\mathrm{\Sigma })V^T}$ or${\displaystyle A{A}^T=U(\mathrm{\Sigma }{\mathrm{\Sigma }}^T)U^T}$ respectively shows that one of them must simplify to an identity matrix, satisfying the definition of a semi-orthogonal matrix.

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