# Uniqueness of inverse matrix and possibility of $P=PX$

Suppose that there is non-zero vector $P$ of size $1 \times n$.

1) Does there exist some $P$ that $P=PX$ without $X$ being identity matrix?

2) When $AB = BA = I$ and $A$ given, can there be several candidates for $B$? I learned that inverse is unique, but just to make sure.

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1) Says that $P$ is an eigenvector for $X$ with eigenvalue $1$. Here is a reference: en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors – Jonas Meyer Dec 10 '12 at 4:05
OK, I get it. @JonasMeyer But we usually do the following: $P^{-1}P = P^{-1}PX$ which would mean $I = X$! That seems nonsensical, but we solve the equation this way.. How am I conused at here? – DDR Dec 10 '12 at 4:18
DDR: What is $P^{-1}$? What is the inverse of a $1\times n$ vector? Think about this, and see Michael Hardy's answer. Note that there does not exist a matrix $A$ such that $AP$ is the $3\times 3$ identity matrix, one reason being that $AP$ has rank at most $1$. – Jonas Meyer Dec 10 '12 at 4:21

$$\pmatrix{1&1\cr}\pmatrix{1/3&2/3\cr2/3&1/3\cr}=\pmatrix{1&1\cr}$$
You can't have $AB=BA=I$ unless $A$ is the unique inverse of $B$ and vice-versa, but you can have $AB=I= I_k = (k\times k\text{ identity matrix})$ and $BA=$ an $n\times n$ matrix of rank $k$, which behaves like an identity matrix in that $BAu=u$ whenever $u$ is in the column space of $B$, but behaves like a zero matrix in that $BAu=0$ whenever $u$ is orthogonal to the column space of $B$. This can happen only if $k<n$.