Linear Algebra.similarity

**ypatia** · March 21st 2009, 03:03 AM

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.
Thanks in Advance

**peteryellow** · March 21st 2009, 03:33 AM

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

**ypatia** · March 21st 2009, 03:42 AM

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

**peteryellow** · March 21st 2009, 04:00 AM

what is the difference between similar and same???

**ADARSH** · March 21st 2009, 04:28 AM

SIMILAR MATRICES AND CHANGE OF BASIS
that link should help
and NO Peteryellow its not true

**PaulRS** · March 21st 2009, 05:51 AM

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.

**ypatia** · March 25th 2009, 11:09 AM

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

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

**Opalg** · March 25th 2009, 12:04 PM

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.
Thanks in Advance

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

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