1
$\begingroup$

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.

  • 2
    1) Says that $P$ is an eigenvector for $X$ with eigenvalue $1$. Here is a reference: http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors2012-12-10
  • 0
    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?2012-12-10
  • 0
    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 $1$.2012-12-10

2 Answers 2