Originally Posted by

**PaulRS** Peteryellow: Please re-read the first post. What is being asked is a necessary and sufficient condition.

ADARSH: Any matrix is similar to itself -in fact similarity is an equivalence relation-, that is what Peteryellow meant ( what I interpret, because he certainly didn't say that), $\displaystyle A=\text{Id}\cdot A \cdot (\text{Id})^{-1}$

As for the question:

The 2 matrixes are similar if and only if there exists a non-singular Matrix $\displaystyle P$ such that $\displaystyle

P \cdot \left( {\begin{array}{*{20}c}

a & b \\

c & d \\

\end{array} } \right) \cdot P^{ - 1} = \left( {\begin{array}{*{20}c}

2 & 1 \\

0 & 2 \\

\end{array} } \right)

$

Or equivalently: $\displaystyle

P \cdot \left( {\begin{array}{*{20}c}

a & b \\

c & d \\

\end{array} } \right) = \left( {\begin{array}{*{20}c}

2 & 1 \\

0 & 2 \\

\end{array} } \right) \cdot P

$ and $\displaystyle P$ **is invertible**

Let's work with a generic matrix: $\displaystyle

P = \left( {\begin{array}{*{20}c}

{a_1 } & {a_2 } \\

{a_3 } & {a_4 } \\

\end{array} } \right)

$ with $\displaystyle

\det \left( P \right) = a_1 \cdot a_4 - a_2 \cdot a_3 \ne 0

$ (1) so that it is indeed invertible.

Now: $\displaystyle

\left( {\begin{array}{*{20}c}

{a_1 } & {a_2 } \\

{a_3 } & {a_4 } \\

\end{array} } \right) \cdot \left( {\begin{array}{*{20}c}

a & b \\

c & d \\

\end{array} } \right) = \left( {\begin{array}{*{20}c}

2 & 1 \\

0 & 2 \\

\end{array} } \right) \cdot \left( {\begin{array}{*{20}c}

{a_1 } & {a_2 } \\

{a_3 } & {a_4 } \\

\end{array} } \right)

$ $\displaystyle

\Leftrightarrow \left( {\begin{array}{*{20}c}

{a_1 \cdot a + a_2 \cdot c} & {a_1 \cdot b + a_2 \cdot d} \\

{a_3 \cdot a + a_4 \cdot c} & {a_3 \cdot b + a_4 \cdot d} \\

\end{array} } \right) = \left( {\begin{array}{*{20}c}

{2 \cdot a_1 + a_3 } & {2 \cdot a_2 + a_4 } \\

{2a_3 } & {2a_4 } \\

\end{array} } \right)

$

Equivalently: $\displaystyle

\left\{ \begin{gathered}

a_1 \cdot a + a_2 \cdot c = 2 \cdot a_1 + a_3 \hfill \\

a_1 \cdot b + a_2 \cdot d = 2 \cdot a_2 + a_4 \hfill \\

a_3 \cdot a + a_4 \cdot c = 2a_3 \hfill \\

2a_4 = a_3 \cdot b + a_4 \cdot d \hfill \\

\end{gathered} \right.

$

Solve for $\displaystyle a_1,a_2,...,a_4$ (see under what conditions on a, b, c, d this is possible), and then what other conditions you have to add for (1) to hold. And then you are done.