1) No matter how the pseudoinverse is constructed, it is always the same? No matter if i use QR or SVD
2) Is the left inverse and the right inverse the same in Moore-Penrose pseudoinverses?
[Math] left inverse == right inverse with Moore–Penrose pseudoinverse
linear algebrapseudoinverse
Related Solutions
The Moore-Penrose inverse of $A\in \mathbb{C}^{m\times n}$, denoted by $A^{+}$, is the unique matrix $X$ satisfying the following four Penrose equations. \begin{eqnarray} (i)~ AXA = A,~ (ii)~ XAX = X, ~ (iii)~ (AX)^* = AX,~ (iv)~ (XA)^* = XA \end{eqnarray}
There are various methods to find out Moore-Penrose inverse of a matrix. For example Rank factorization method, singular value decomposition method, QR decomposition etc. If $Ax = b$ is inconsistent then $x = A^{+}b$ represents the least square solution of minimum 2 norm.
In general $x = A^{+}b$ represents the least square solution to system $Ax = b$. For more detailed information refers to generalized inverse by Ben-Israel http://www.amazon.com/Generalized-inverses-applications-Adi-Ben-Israel/dp/0882759914
The non-uniqueness of SVD can be characterized as follows: suppose that $A = P_0 \Sigma Q_0^T$ is one SVD of $A$. Moreover, suppose that the singular values of $A$ are $s_1$ with multiplicity $k_1$, $s_2$ with multiplicity $k_2$, and so forth, with $s_m = 0$ having multiplicity $k_m = n - r$. That is, we have $$ \Lambda_r = \pmatrix{s_1 I_{k_1} \\ & \ddots \\ && s_{m-1} I_{k_{m-1}}}, \quad \Sigma = \pmatrix{\Lambda_r \\ & 0_{k_m}} $$ Then $A = P\Sigma Q^T$ will be a singular value decomposition of $A$ if and only if there exists an orthogonal matrix $U$ such that $P = P_0 U, Q = Q_0U$, and $U$ is a block-diagonal orthogonal matrix of the form $$ U = \pmatrix{U^{(1)}\\ & \ddots \\ && U^{(m)}} $$ where $U^{(j)}$ is (orthogonal and) of size $k_j \times k_j$.
With that in mind: if you'd like to prove that the pseudoinverse as constructed from SVD is well-defined (that is, uniquely defined regardless of one's choice of SVD), then it suffices to show that for any choice of $U$ of the form prescribed above, we have $$ [Q_0U] \pmatrix{\Lambda_r^{-1} \\ & 0} [P_0U]^T = Q_0 \pmatrix{\Lambda_r^{-1} \\ & 0} P_0 $$ It is straightforward (but in my opinion tedious) to show that this holds if we use the block-structure of $\Lambda_r^{-1}$ and block-matrix multiplication.
Best Answer