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Math Help - Linear Algebra.similarity

  1. #1
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    Linear Algebra.similarity

    Let the 2x2 matrix <br />
 \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 <br />
 \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.
    Thanks in Advance
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  2. #2
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    The matrices are same if a=2, b=1, c=0, and d=2.
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  3. #3
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    Quote Originally Posted by peteryellow View Post
    The matrices are same if a=2, b=1, c=0, and d=2.
    The matrices must be similar not same
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  4. #4
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    what is the difference between similar and same???
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    SIMILAR MATRICES AND CHANGE OF BASIS
    that link should help
    and NO Peteryellow its not true
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  6. #6
    Super Member PaulRS's Avatar
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    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 <br />
P \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) \cdot P^{ - 1}  = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right)<br />

    Or equivalently: <br />
P \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right) \cdot P<br />
and P is invertible

    Let's work with a generic matrix: <br />
P = \left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right)<br />
with <br />
\det \left( P \right) = a_1  \cdot a_4  - a_2  \cdot a_3  \ne 0<br />
(1) so that it is indeed invertible.

    Now: <br />
\left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right)<br />
<br />
 \Leftrightarrow \left( {\begin{array}{*{20}c}<br />
   {a_1  \cdot a + a_2  \cdot c} & {a_1  \cdot b + a_2  \cdot d}  \\<br />
   {a_3  \cdot a + a_4  \cdot c} & {a_3  \cdot b + a_4  \cdot d}  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   {2 \cdot a_1  + a_3 } & {2 \cdot a_2  + a_4 }  \\<br />
   {2a_3 } & {2a_4 }  \\<br /> <br />
 \end{array} } \right)<br />

    Equivalently: <br />
\left\{ \begin{gathered}<br />
  a_1  \cdot a + a_2  \cdot c = 2 \cdot a_1  + a_3  \hfill \\<br />
  a_1  \cdot b + a_2  \cdot d = 2 \cdot a_2  + a_4  \hfill \\<br />
  a_3  \cdot a + a_4  \cdot c = 2a_3  \hfill \\<br />
  2a_4  = a_3  \cdot b + a_4  \cdot d \hfill \\ <br />
\end{gathered}  \right.<br />

    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.
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  7. #7
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    Quote Originally Posted by PaulRS View Post
    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 <br />
P \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) \cdot P^{ - 1}  = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right)<br />

    Or equivalently: <br />
P \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right) \cdot P<br />
and P is invertible

    Let's work with a generic matrix: <br />
P = \left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right)<br />
with <br />
\det \left( P \right) = a_1  \cdot a_4  - a_2  \cdot a_3  \ne 0<br />
(1) so that it is indeed invertible.

    Now: <br />
\left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br />
   a & b  \\<br />
   c & d  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   2 & 1  \\<br />
   0 & 2  \\<br /> <br />
 \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}<br />
   {a_1 } & {a_2 }  \\<br />
   {a_3 } & {a_4 }  \\<br /> <br />
 \end{array} } \right)<br />
<br />
 \Leftrightarrow \left( {\begin{array}{*{20}c}<br />
   {a_1  \cdot a + a_2  \cdot c} & {a_1  \cdot b + a_2  \cdot d}  \\<br />
   {a_3  \cdot a + a_4  \cdot c} & {a_3  \cdot b + a_4  \cdot d}  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   {2 \cdot a_1  + a_3 } & {2 \cdot a_2  + a_4 }  \\<br />
   {2a_3 } & {2a_4 }  \\<br /> <br />
 \end{array} } \right)<br />

    Equivalently: <br />
\left\{ \begin{gathered}<br />
  a_1  \cdot a + a_2  \cdot c = 2 \cdot a_1  + a_3  \hfill \\<br />
  a_1  \cdot b + a_2  \cdot d = 2 \cdot a_2  + a_4  \hfill \\<br />
  a_3  \cdot a + a_4  \cdot c = 2a_3  \hfill \\<br />
  2a_4  = a_3  \cdot b + a_4  \cdot d \hfill \\ <br />
\end{gathered}  \right.<br />

    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
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  8. #8
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    Quote Originally Posted by ypatia View Post
    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}.
    Last edited by Opalg; March 25th 2009 at 01:20 PM. Reason: made some corrections
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