Inverse matrix

Let $A∈RN×NA \in \R^{N \times N}$ represent a square matrix,

$$A−1Am=emA−1(A1A2…AM)=(e1e2…eM)\begin{align*} A^{-1} A_m &= e_m \\ A^{-1} \begin{pmatrix} A_1 & A_2 & \dots & A_M \end{pmatrix} &= \begin{pmatrix} e_1 & e_2 & \dots & e_M \end{pmatrix} \end{align*}$$

$$\begin{align*} \end{align*}$$

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$$A−1(∑n=1NxnAn)=x,x∈RN\begin{align*} A^{-1} \left( \sum_{n=1}^N x_n A_n \right) = x ,\quad x \in \R^{N} \end{align*}$$

This is because

$$Ax=∑n=1NxnAnA−1Ax=A−1(∑n=1NxnAn)x=A−1(∑n=1NxnAn)\begin{align*} A x = \sum_{n=1}^N x_n A_n \\ A^{-1} A x = A^{-1} \left( \sum_{n=1}^N x_n A_n \right) \\ x = A^{-1} \left( \sum_{n=1}^N x_n A_n \right) \\ \end{align*}$$

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If $A−1A^{-1}$ exists, then any $y∈RNy \in \R^N$ can be expressed as a linear combination $y=x1A1+⋯+xNAN=Axy = x_1 A_1 + \dots + x_N A_N = A x$.

The weights $x={ xn }n=1Nx = \set{x_n}_{n=1}^N$ can be obtained using $A−1yA^{-1} y$.

This is because

$$y=Ax,A−1y=A−1Ax,A−1y=x.\begin{align*} y = A x ,\\ A^{-1} y = A^{-1} A x ,\\ A^{-1} y = x . \\ \end{align*}$$