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Pseudo Inverse

Problem Definition

This problem appears in an optimization process or controller design including a coordinate transformation of a state.
A problem is to get x\bm{x} that satisfy
Ax=y,\begin{align} \bm{A}\bm{x}=\bm{y}, \end{align}
where
A∈Rm×n, x∈Rn, y∈Rm.\begin{align} \bm{A}\in \mathbb{R}^{m\times n}, \ \bm{x}\in \mathbb{R}^{n}, \ \bm{y}\in \mathbb{R}^{m}. \end{align}
How to solve this depends on the size of A\bm{A}.

Determined system

When A\bm{A} is a positive-definite, m=nm=n. In the case, there exist the inverse matrix of A\bm{A}. A solution is uniquely sought as
x^=A−1y.\begin{align} \hat{\bm{x}}=\bm{A}^{-1}\bm{y}. \end{align}

Overdetermined system

When m>nm>n. There is no solution. As an alternative, a vector that satisfy the following equation is provided.
x^=arg min⁡x 12∥Ax−y∥22.\begin{align} \hat{\bm{x}}=\argmin_{\bm{x}} \ \frac{1}{2}\| \bm{A}\bm{x}-\bm{y} \|^2_2. \end{align}
A solution is sought as
∂∂x(12∥Ax−y∥22)=AT(Ax−y)=0⇔ATAx−ATy=0\begin{align} \frac{\partial}{\partial \bm{x}} \left(\frac{1}{2}\| \bm{A}\bm{x}-\bm{y} \|^2_2\right) &= \bm{A}^{\mathrm T}(\bm{A}\bm{x}-\bm{y}) = 0 \\ \Leftrightarrow \bm{A}^{\mathrm T}\bm{A}\bm{x}-\bm{A}^{\mathrm T}\bm{y} &= 0\end{align}
∴x^=(ATA)−1ATy.\begin{align} \therefore \hat{\bm{x}} = \left(\bm{A}^{\mathrm T}\bm{A}\right)^{-1}\bm{A}^{\mathrm T}\bm{y}.\end{align}

Underdetermined system

When m<nm<n. There are many solutions. A generalized solution is provided using the sum of a vectors
x^=xm+xnullxm≡arg min⁡x 12∥x∥22  s. t. Ax=yxnull≡{x ∣ Ax=0}∈Ker(A)\begin{align} \hat{\bm{x}}&=\bm{x}_{\rm m}+\bm{x}_{\rm null} \\ \bm{x}_{\rm m} &\equiv \argmin_{\bm{x}} \ \frac{1}{2}\|\bm{x}\|^2_2\ \ {\rm s.\ t.\ }\bm{A}\bm{x}=\bm{y} \\ \bm{x}_{\rm null} &\equiv \{ \bm{x}\ |\ \bm{A}\bm{x} = 0 \} \in {\rm Ker}(\bm{A}) \end{align}
These vectors are sought as
L(x,λ)=12xTx+λ(Ax−y){∂∂xL(xm,λ)=xm−ATλ=0∂∂λL(xm,λ)=Axm−y=0⇔{xm=ATλAATλ=y⇔{xm=AT(AAT)−1yλ=(AAT)−1y\begin{align} &\mathcal{L}(\bm{x}, \lambda)=\frac{1}{2}\bm{x}^{\mathrm T}\bm{x}+\lambda(\bm{A}\bm{x}-\bm{y})\\ &\left\{\begin{matrix}\frac{\partial}{\partial \bm{x}}\mathcal{L}(\bm{x}_{\rm m}, \lambda)=\bm{x}_{\rm m}-\bm{A}^{\mathrm T}\lambda=0\\ \frac{\partial}{\partial \lambda}\mathcal{L}(\bm{x}_{\rm m}, \lambda)=\bm{A}\bm{x}_{\rm m}-\bm{y}=0\end{matrix}\right.\\ \Leftrightarrow&\left\{\begin{matrix}\bm{x}_{\rm m}=\bm{A}^{\mathrm T}\lambda\\\bm{A}\bm{A}^{\mathrm T}\lambda=\bm{y}\\\end{matrix}\right.\\ \Leftrightarrow&\left\{\begin{matrix}\bm{x}_{\rm m}=\bm{A}^{\mathrm T}\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1}\bm{y}\\ \lambda=\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1}\bm{y}\end{matrix}\right. \end{align}
xnull=(I−AT(AAT)−1A)k.\begin{align} \bm{x}_{\rm null} &= (I-\bm{A}^{\mathrm T}\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1}\bm{A})k. \end{align}
Then, the generalized form is obtained as
x^=AT(AAT)−1y+(I−AT(AAT)−1A)k.\begin{align} \hat{\bm{x}} &=\bm{A}^{\mathrm T}\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1}\bm{y}+(I-\bm{A}^{\mathrm T}\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1}\bm{A})k. \end{align}

Pseudo Inverse and solution

A pseudo inverse matrix is given as
A+={A−1(m=n)(ATA)−1AT(m>n)AT(AAT)−1(m<n).\begin{align} \bm{A}^{+}&=\left\{\begin{matrix} \bm{A}^{-1} & (m=n) \\ \left(\bm{A}^{\mathrm T}\bm{A}\right)^{-1}\bm{A}^{\mathrm T} & (m>n) \\ \bm{A}^{\mathrm T}\left(\bm{A}\bm{A}^{\mathrm T}\right)^{-1} & (m<n) \end{matrix}\right..\end{align}
This matrix provide a solution as
x^={A+y(m=n)A+y(m>n)A+y+(I−A+A)k(m<n)\begin{align} \hat{\bm{x}}&=\left\{\begin{matrix} \bm{A}^{+}\bm{y} & (m=n) \\ \bm{A}^{+}\bm{y} & (m>n) \\ \bm{A}^{+}\bm{y} + (I - \bm{A}^{+}\bm{A})k & (m<n) \end{matrix}\right.\end{align}

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