# Linear Algebra.similarity

• Mar 21st 2009, 03:03 AM
ypatia
Linear Algebra.similarity
Let the 2x2 matrix $\displaystyle \left&#91;\begin {array}{cc} a & b \\ c & d \end {array} \right&#93;$ How can I found a necessary and sufficient condition in order the matrix A to be similar to the following matrix $\displaystyle \left&#91;\begin {array}{cc} 2 & 1 \\ 0 & 2 \end {array} \right&#93;$

I would be grateful if someone show me a step-by-step solution.
• Mar 21st 2009, 03:33 AM
peteryellow
The matrices are same if a=2, b=1, c=0, and d=2.
• Mar 21st 2009, 03:42 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, 04:00 AM
peteryellow
what is the difference between similar and same???
• Mar 21st 2009, 04:28 AM
SIMILAR MATRICES AND CHANGE OF BASIS
and NO Peteryellow its not true (Itwasntme)
• Mar 21st 2009, 05:51 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), $\displaystyle A=\text{Id}\cdot A \cdot (\text{Id})^{-1}$

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.
• Mar 25th 2009, 11:09 AM
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), $\displaystyle A=\text{Id}\cdot A \cdot (\text{Id})^{-1}$

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.

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, 12:04 PM
Opalg
Quote:

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

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

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

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

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

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

Therefore $\displaystyle \color{red}a+d=4$, $\displaystyle \color{red}bc = (a-2)(d-2)$, $\displaystyle \color{red}bc\leqslant0$ and $\displaystyle \color{red}b,\,c$ 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 $\displaystyle A-2I$ and $\displaystyle \begin{bmatrix}0&1\\0&0\end{bmatrix}$ are similar. So the conditions in red are the necessary and sufficient conditions for A to be similar to $\displaystyle \begin{bmatrix}2&1\\0&2\end{bmatrix}$.