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Math Help - Canonical Basis?

  1. #1
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    Post Canonical Basis?

    Hi there,

    I need to create a matrix given some information, without the matrix it is impossible to continue the assignment - unfortunately, I don't find much relevant information on canonical basis online, well, I have found a few examples, but they are similar to what I'm looking at...


    Find the 2x2-matrix, A, belonging to the linear "map"/"projection", T, with regard to the canonical basis of R^2 (real numbers).


    A small note e(1) means e sub one.

    T(e(1) + e(2)) = e(1) + e(2), T(-e(1)+e(2)) = (1/2)e(1) - (1/2)e(2)

    Does anyone have an idea on this matter?


    Thanks in advance.
    Simon DK.
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  2. #2
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    Quote Originally Posted by sh01by View Post
    Hi there,

    I need to create a matrix given some information, without the matrix it is impossible to continue the assignment - unfortunately, I don't find much relevant information on canonical basis online, well, I have found a few examples, but they are similar to what I'm looking at...


    Find the 2x2-matrix, A, belonging to the linear "map"/"projection", T, with regard to the canonical basis of R^2 (real numbers).


    A small note e(1) means e sub one.

    T(e(1) + e(2)) = e(1) + e(2), T(-e(1)+e(2)) = (1/2)e(1) - (1/2)e(2)

    Does anyone have an idea on this matter?


    Thanks in advance.
    Simon DK.
    are these two separate transformations that we are talking about?

    the first: T(e_1 + e_2) = e_1 + e_2

    and the second: T(-e_1 + e_2) = \frac {1}{2}e_1 - \frac {1}{2}e_2

    ?

    I find it difficult to see how one transformtion would do both operations, but then again, i am a bit rusty with this


    Here's what i am thinking:

    Recall: in \mathbb {R}, e_1 = {1 \choose 0} and e_2 = {0 \choose 1}

    we want T_1 such that T_1 (e_1 + e_2) = e_1 + e_2. Now T_1 must be of the form \left(\begin{array}{cc}a&0\\0&b\end{array}\right)

    So we have: \left(\begin{array}{cc}a&0\\0&b\end{array}\right) \left( {1 \choose 0}+{0 \choose 1} \right) = {1 \choose 0}+{0 \choose 1}

    \Rightarrow \left(\begin{array}{cc}a&0\\0&b\end{array}\right) {1 \choose 1} = {1 \choose 1}

    \Rightarrow a \cdot 1 = 1 \mbox { and } b \cdot 1 = 1 \Rightarrow a = b = 1

    So, T_1 = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)

    the second one is done similarly
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    are these two separate transformations that we are talking about?

    the first: T(e_1 + e_2) = e_1 + e_2

    and the second: T(-e_1 + e_2) = \frac {1}{2}e_1 - \frac {1}{2}e_2

    ?

    I find it difficult to see how one transformtion would do both operations, but then again, i am a bit rusty with this


    Here's what i am thinking:

    Recall: in \mathbb {R}, e_1 = {1 \choose 0} and e_2 = {0 \choose 1}

    we want T_1 such that T_1 (e_1 + e_2) = e_1 + e_2. Now T_1 must be of the form \left(\begin{array}{cc}a&0\\0&b\end{array}\right)

    So we have: \left(\begin{array}{cc}a&0\\0&b\end{array}\right) \left( {1 \choose 0}+{0 \choose 1} \right) = {1 \choose 0}+{0 \choose 1}

    \Rightarrow \left(\begin{array}{cc}a&0\\0&b\end{array}\right) {1 \choose 1} = {1 \choose 1}

    \Rightarrow a \cdot 1 = 1 \mbox { and } b \cdot 1 = 1 \Rightarrow a = b = 1

    So, T_1 = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)

    the second one is done similarly
    Jhevon, one transformation will do both operations. Why did you assume a diagonal matrix? Letting all 4 elements of the matrix be non-zero leads to 4 equations in 4 unknowns. Here are the results.

    \left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {1 \choose 1} = {1 \choose 1}

    \left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}

    JakeD
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JakeD View Post
    Jhevon, one transformation will do both operations. Why did you assume a diagonal matrix? Letting all 4 elements be non-zero leads to 4 equations in 4 unknowns. Here are the results.

    \left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {1 \choose 1} = {1 \choose 1}

    \left(\begin{array}{cc}1/4&3/4\\3/4&1/4\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}
    nice. i don't remember what was going through my head at the time i assumed it was a diagonal matrix.... what you did seems so obvious now
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  5. #5
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    Thanks, please elaborate on the transformation...

    Hi Jhevon and JakeD,

    it is always perfect with quick replies...

    Yes, Jhevon, they are separate - I tried to make a few spaces, but I guess they were removed, when I posted. : o )

    JakeD, I see your solution, and it is right, but...
    How is it you find the matrix? I see how it works, how you multiply the matrix with the information in T(e(1)+e(2)) as seen in your post - the first math-input...




    Suppose, you're looking at the first case... What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...

    I guess what I am asking is - how do set up and equation, so you get the condition from the other information too - and end up with your matrix..?

    Did that make any sense?

    Simon DK
    Last edited by sh01by; June 4th 2007 at 12:04 PM. Reason: It wasn't quite eloquent enough... hehe.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sh01by View Post
    Hi Jhevon and JakeD,

    it is always perfect with quick replies...

    Yes, Jhevon, they are separate - I tried to make a few spaces, but I guess they were removed, when I posted. : o )

    JakeD, I see your solution, and it is right, but...
    How is it you find the matrix? I see how it works, how you multiply the matrix with the information in T(e(1)+e(2)) as seen in your post - the first math-input...




    Suppose, you're looking at the first case... What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...

    I guess what I am asking is - how do set up and equation, so you get the condition from the other information too - and end up with your matrix..?

    Did that make any sense?

    Simon DK
    Do you remember how to do matrix multiplication?

    we wanted to find a matrix \left(\begin{array}{cc}a&b\\c&d\end{array}\right) such that:

    \left(\begin{array}{cc}a&b\\c&d\end{array}\right) {1 \choose 1} = {1 \choose 1}

    and

    \left(\begin{array}{cc}a&b\\c&d\end{array}\right) {-1 \choose 1} = {1/2 \choose -1/2}


    Then combining the matrices on the left of the first equation, we realize that this means:

    -a + b = \frac {1}{2} and -c + d = - \frac {1}{2}

    Combining the matrices on the left of the second equation, we realize that:

    a + b = 1 and c + d = 1


    So we got the system:

    -a + b = \frac {1}{2}............................(1)
    -c + d = - \frac {1}{2}..........................(2)
    a + b = 1 ..............................(3)
    c + d = 1 ...............................(4)

    solving that system gives us the components of the matrix JakeD came up with

    Quote Originally Posted by sh01by View Post
    What I don't get is? Why don't you just choose a matrix of pure 1/2? That would be satisfying...
    he wanted a single matrix that does both operations. had he choosen a matrix with just 1/2's everywhere, we would satisfy the first operation, but not the second
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  7. #7
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    Hi Jhevon and JakeD, thanks.

    Thanks for your replies!

    It was a big help. I think, I'd be able to solve a similar assignment, if it is part of the exam, which I'm taking tomorrow.

    It makes a little more sense now too, because I've been working on other topics in linear algebra for the last days, so it is all coming together - just not fast enough to ace the test tomorrow, but - maybe next time, for now - I'm just looking to pass.

    Have a nice day,
    Simon DK.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sh01by View Post
    Thanks for your replies!

    It was a big help. I think, I'd be able to solve a similar assignment, if it is part of the exam, which I'm taking tomorrow.

    It makes a little more sense now too, because I've been working on other topics in linear algebra for the last days, so it is all coming together - just not fast enough to ace the test tomorrow, but - maybe next time, for now - I'm just looking to pass.

    Have a nice day,
    Simon DK.
    Ok, good luck
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