Let $A$ be non-zero matrix such that $a_{ij}=0$ $\forall i\ge j$. $D$ be a diagonal matrix with distinct diagonal entries. Now I want to show that $D$ is similar to $D+A$.

Then how can I show that this does not hold without "Distinct diagonal entries" assumption?

My try:

I am thinking in terms of "change of basis". So $D$ represents a linear transformation which has distinct eigen vectors with distinct eigen values. Now $A$ is also nilpotent. I can't assemble these facts to get through the problem.

Can anyone help or suggest me anything? Thanks in advance.

  • $\begingroup$ The matrices $D=\operatorname{diag}(1,1)$ and $D+A=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ cannot be similar because $D$ has two linearly independent eigenvectors while $D+A$ doesn't. For different elements in the diagonal, compute eigenvectors of $D+A$ for each entry in the diagonal. Show that because their eigenvalues are different, then they are linearly independent. Finally, the matrix of $D+A$ in the basis formed by those eigenvectors, put in the same order as their corresponding eigenvalues appear in the diagonal, is $D$. $\endgroup$ – logarithm May 5 at 18:29
  • $\begingroup$ $D+A$ has the same (distinct) eigenvalues as $D$. What do you get when diagonalizing $D+A$? $\endgroup$ – A.Γ. May 5 at 18:49
  • $\begingroup$ I get back $D$ again. Ok i got it,thanks! $\endgroup$ – Soumyadip Sarkar May 5 at 19:00

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