# Similar matrices if only difference is diagonalizable/non-diagonalizable

I am new to linear algebra, and am just looking for some feedback regarding the following solution:

True or false?

1.$$\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ and $$\begin{pmatrix}2&0\\0&-2\end{pmatrix}$$ are similar.

2.$$\begin{pmatrix}3&1&1\\0&1&0\\0&0&1\end{pmatrix}$$ and $$\begin{pmatrix}1&1&0\\0&1&0\\0&0&3\end{pmatrix}$$ are similar.

My answer to 1. is: false, because (amongst others) the determinants are different. My answer to 2. is: false, because, despite the fact that the determinants, trace and rank are the same, the top matrix is diagonalizable, whereas the bottom one is not. Therefore they cannot be similar.

• Yes..you are right. – Sunny Rathore Jan 19 at 13:38
• @sunnyrathore thank you! – dalta Jan 19 at 13:39