# Diagonal matrix with distinct diagonal entries is similar to a diagonal plus an lower triangular matrix

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.

• 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 $D$. – logarithm May 5 at 18:29
• $D+A$ has the same (distinct) eigenvalues as $D$. What do you get when diagonalizing $D+A$? – A.Γ. May 5 at 18:49
• I get back $D$ again. Ok i got it,thanks! – Soumyadip Sarkar May 5 at 19:00