I am using the book "Linear Algebra 4th Edition" by Stephen H. Friedberg, Arnold J. Insel, and Lawrence E. Spence. It gives this corollary: "Let T be a linear operator on an n-dimensional vector space V. If T has n distinct eigenvalues, then T is diagonalizable."
I am attaching a scan of my work and I am wondering if a linear operator can be interpreted the same way as a matrix can. If so, does that mean that each of the eigenvalues has to be a different number (i.e. "distinct" eigenvalue)? I got 3, 3, and -1 as the eigenvalues for the first matrix I tried and then computed the eigenvectors which in turn became the column vectors of the matrix Q. I found Q to be non-invertible and got stuck there. Thanks for the help.