Sorry for the long title, English is not my first language and I couldn't find a shorter description of the question.
I need help proving / disproving this question:
A is a matrix, B is the matrix acquired from A after doing a set of elementary operations. A is diagonalizable, prove or disprove B is diagonalizable.
My attempt is as following:
If a is diagonalizable then there is a diagonal matrix D and an inversible matrix P that follows this equation $\mathrm{D=P^{-1}AP}$. Also, you can show B as a multiplication of multiple elementary matrices followed by A ($\mathrm{B=\phi_k(I)\phi_{k-1}(I)...\phi_{1}(I)A}$), because every elementary matrix is inversible, we can multiply by the inverse matrices from the left and receive this equation: $\mathrm{A=\phi_k(I)^{-1}\phi_{k-1}(I)^{-1}...\phi_{1}(I)^{-1}B}$, after combining the two equations we get $\mathrm{D=P^{-1}\phi_k(I)^{-1}\phi_{k-1}(I)^{-1}...\phi_{1}(I)^{-1}BP}$ but I couldn't find a way to continue my answer.
Thanks.
Consider \begin{align*} A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}. \end{align*}
It is easy to verify that $A$ is diagonal, $B$ is obtained by adding the first column of $A$ to its second column, but $B$ is not diagonalizable (it is a Jordan block).