Matrix compression

1. Symmetric matrix compression

$$\left[\begin{array}{ccccc}a& x& x& x& x\\ x& b& x& x& x\\ x& x& c& x& x\\ x& x& x& d& x\\ x& x& x& x& e\end{array}\right]$$

$$\left[\begin{array}{ccccc}a& & & & \\ x& b& & & \\ x& x& c& & \\ x& x& x& d& \\ x& x& x& x& e\end{array}\right]$$

In a two-dimensional array $a_{ij}$ Subscript in one-dimensional array :
$$k=i(i-1)/2+j-1$$

For symmetric matrices, we can save only the lower triangle （ Including elements on the diagonal ）, Or upper triangular element

The number of storage units required $n\times n$ Turned into $(n(n+1)/2)$

2. Upper and lower triangular matrix compression

2.1 Lower triangular symmetric matrix compression

about
$$\left[\begin{array}{ccccc}{a}_{11}& & & & \\ a_{21}& a_{22}& & & \\ a_{31}& a_{32}& a_{33}& & \\ a_{41}& a_{42}& a_{43}& a_{44}& \\ a_{51}& a_{52}& a_{53}& a_{54}& a_{55}\end{array}\right]$$
take
$$\left[\begin{array}{ccccc}{a}_{11}& & & & \\ ...& & & & \\ a_{i1}& ...& a_{ij}& ...& \\ ...& & & & \\ a_{n1}& ...& a_{nj}& ...& a_{nn}\end{array}\right]$$
Turn into
$$\left[\begin{array}{cccccccccccc}{a}_{11}& ...& a_{i1}& ...& a_{ij}& ...& a_{n1}& ...& a_{nj}& ...& a_{nn}& \end{array}\right]$$
Subscripts in the original matrix $i,j$, Corresponds to the subscript in the array $k$

2.2 Upper triangular symmetric matrix compression

about
$$\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ & a_{22}& a_{23}& a_{24}\\ & & a_{33}& a_{34}\\ & & & a_{44}\end{array}\right]$$
take
$$\left[\begin{array}{ccccc}{a}_{11}& ...& a_{1j}& ...& a_{1n}\\ & & ...& & ...\\ & & a_{ij}& ...& a_{in}\\ & & & & ...\\ & & & & a_{nn}\end{array}\right]$$
Turn into
$$\left[\begin{array}{cccccccccccc}{a}_{11}& ...& a_{1j}& ...& a_{1n}& ...& a_{ij}& ...& a_{in}& ...& a_{nn}& \end{array}\right]$$
Subscripts in the original matrix $i,j$, Corresponds to the subscript in the array $k$

namely :
$$k=(i-1)(2n-i+2)/2+j-i$$

2. Diagonal matrix compression

The matrix
$$\left[\begin{array}{ccccc}{a}_{11}& a_{12}& 0& 0& 0\\ a_{21}& a_{22}& a_{23}& 0& 0\\ 0& a_{32}& a_{33}& a_{34}& 0\\ 0& 0& a_{43}& a_{44}& a_{45}\\ 0& 0& 0& a_{54}& a_{55}\end{array}\right]$$
Compressed into a one-dimensional matrix
$$\left[\begin{array}{ccccccccccccc}{a}_{11}& a_{12}& a_{21}& a_{22}& a_{23}& a_{32}& a_{33}& a_{34}& a_{43}& a_{44}& a_{45}& a_{54}& a_{55}\end{array}\right]$$
original $n\times n$ Storage units are now just $3n-2$ individual

Subscripts in the original matrix $i,j$, Corresponds to the subscript in the array $k$

namely
$$k=2i+j-3$$
It is known that $k$:
$$\{\begin{array}{l}i=\lfloor (k+1)/3\rfloor +1\\ j=k-2i+3\end{array}$$