1. ## Linear Algebra.similarity

Let the 2x2 matrix $
\left[\begin {array}{cc} a & b \\ c & d \end {array} \right]$
How can I found a necessary and sufficient condition in order the matrix A to be similar to the following matrix $
\left[\begin {array}{cc} 2 & 1 \\ 0 & 2 \end {array} \right]$

I would be grateful if someone show me a step-by-step solution.

2. The matrices are same if a=2, b=1, c=0, and d=2.

3. Originally Posted by peteryellow
The matrices are same if a=2, b=1, c=0, and d=2.
The matrices must be similar not same

4. what is the difference between similar and same???

5. SIMILAR MATRICES AND CHANGE OF BASIS
and NO Peteryellow its not true

6. 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), $A=\text{Id}\cdot A \cdot (\text{Id})^{-1}$

As for the question:

The 2 matrixes are similar if and only if there exists a non-singular Matrix $P$ such that $
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: $
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 $P$ is invertible

Let's work with a generic matrix: $
P = \left( {\begin{array}{*{20}c}
{a_1 } & {a_2 } \\
{a_3 } & {a_4 } \\

\end{array} } \right)
$
with $
\det \left( P \right) = a_1 \cdot a_4 - a_2 \cdot a_3 \ne 0
$
(1) so that it is indeed invertible.

Now: $
\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)
$
$
\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: $
\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 $a_1,a_2,...,a_4$ (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.

7. Originally Posted by PaulRS

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), $A=\text{Id}\cdot A \cdot (\text{Id})^{-1}$

As for the question:

The 2 matrixes are similar if and only if there exists a non-singular Matrix $P$ such that $
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: $
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 $P$ is invertible

Let's work with a generic matrix: $
P = \left( {\begin{array}{*{20}c}
{a_1 } & {a_2 } \\
{a_3 } & {a_4 } \\

\end{array} } \right)
$
with $
\det \left( P \right) = a_1 \cdot a_4 - a_2 \cdot a_3 \ne 0
$
(1) so that it is indeed invertible.

Now: $
\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)
$
$
\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: $
\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 $a_1,a_2,...,a_4$ (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.
I try to solve the system but I cannot found a condition as you describe before. Is there any other way to solve this exercise or could anyone help me to solve the above complicate system?? Thanks anyone in advance

8. Originally Posted by ypatia
Let A be the 2x2 matrix $\left[\begin {array}{cc} a & b \\ c & d \end {array} \right]$ How can I found a necessary and sufficient condition in order the matrix A to be similar to the following matrix $\left[\begin {array}{cc} 2 & 1 \\ 0 & 2 \end {array} \right]$

I would be grateful if someone show me a step-by-step solution.
Here's a solution that cuts out some of the algebra.

If A is similar to $\begin{bmatrix}2&1\\0&2\end{bmatrix}$ then $A - 2I$ will be similar to $\begin{bmatrix}2&1\\0&2\end{bmatrix}-2I = \begin{bmatrix}0&1\\0&0\end{bmatrix}$.

So suppose that $P = \begin{bmatrix}w&x\\y&z\end{bmatrix}$ is an invertible matrix such that $\begin{bmatrix}a-2&b\\c&d-2\end{bmatrix} = P^{-1}\begin{bmatrix}0&1\\0&0\end{bmatrix}P$.

The inverse of P is given by $P^{-1} = \delta^{-1}\begin{bmatrix}z&-x\\-y&w\end{bmatrix}$, where $\delta = wz-xy\ (\ne0)$ is the determinant of P. Then

$\begin{bmatrix}a-2&b\\c&d-2\end{bmatrix} = \delta^{-1}\begin{bmatrix}z&-x\\-y&w\end{bmatrix} \begin{bmatrix}0&1\\0&0\end{bmatrix} \begin{bmatrix}w&x\\y&z\end{bmatrix} = \delta^{-1}\begin{bmatrix}yz&z^2\\-y^2&-yz\end{bmatrix}$.

Comparing coefficients, you see that $a-2 = -(d-2) = \delta^{-1}yz$ and $bc = -\delta^{-2}y^2z^2$.

Therefore $\color{red}a+d=4$, $\color{red}bc = (a-2)(d-2)$, $\color{red}bc\leqslant0$ and $\color{red}b,\,c$ are not both 0. Conversely, if a,b,c,d satisfy those conditions then you can reconstruct w,x,y,z so that the matrices $A-2I$ and $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ are similar. So the conditions in red are the necessary and sufficient conditions for A to be similar to $\begin{bmatrix}2&1\\0&2\end{bmatrix}$.