I have the following matrix for T: R3->R3 with respect to the standard basis:
\begin{array}{ccc} 1 & 1 & 0 \\ & -1 & 1 \\ & & 1 \end{array}
and was wondering if such a matrix could be an isomorphism, i.e. whether it has an inverse or not (is this the right logic?).
So I'm looking for a $T^{-1}$ such that $TT^{-1}=I_{output space, i.e. R3}$ or 1 1 1?
I was thinking that such an inverse transformation could be of the form
\begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & \\ b_{31} & & \end{array}
whereupon I could solve for the $b$ variables after doing $TT^{-1}$, but I have more variables than equations, so this seems wrong.
Tips appreciated!
You can very easily fill in values such that the matrix has an inverse, for example, $$\pmatrix{1&1&0\cr \color{red}{0}&-1&1\cr \color{red}{0}&\color{red}{0}&1\cr} \quad\hbox{or}\quad \pmatrix{1&1&0\cr \color{red}{1}&-1&1\cr \color{red}{1}&\color{red}{1}&1\cr}\ .$$