# Linear Algebra.similarity

• Mar 21st 2009, 04:03 AM
ypatia
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.
• Mar 21st 2009, 04:33 AM
peteryellow
The matrices are same if a=2, b=1, c=0, and d=2.
• Mar 21st 2009, 04:42 AM
ypatia
Quote:

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
• Mar 21st 2009, 05:00 AM
peteryellow
what is the difference between similar and same???
• Mar 21st 2009, 05:28 AM
SIMILAR MATRICES AND CHANGE OF BASIS
and NO Peteryellow its not true (Itwasntme)
• Mar 21st 2009, 06:51 AM
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.
• Mar 25th 2009, 12:09 PM
ypatia
Quote:

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
• Mar 25th 2009, 01:04 PM
Opalg
Quote:

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 $\begin{bmatrix}2&1\\0&2\end{bmatrix}$.