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    This is driving me mad!

    I have to find the eigenvectors of the matrix A=[1, 0 : 1, 2 ]

    So first I find the eigenvalues by solving the det (A - kI) = 0
    Where k = lamda , the eigenvalue and I is the identity matrix.

    So I end up with [ 1-k , 0 : 1 , 2-k] which gives the characteristic polynomial as:
    k^2 -2k +2 = 0

    So k=2 ,1

    Then To find eigenvectors I solve (A-kI)v = 0

    Where v is the eigenvector.

    So if I sub my eigenvalues back into the A-kI matrix and then solve I SHOULD get my eigenvectors. BUT my matrix becomes:

    for k=1

    [0,0 : 1,1] [v1:v2] = [0:0]

    Giving:

    0x + 0y =0

    which to me gives the eigenvector [1:1] not to sure why just thats what others have looked like but the actual answers are:

    [0:1] and 1/sqrt(2) *[1:-1]

    HOW?
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    (Original post by Carlo08)
    This is driving me mad!

    I have to find the eigenvectors of the matrix A=[1, 0 : 1, 2 ]

    So first I find the eigenvalues by solving the det (A - kI) = 0
    Where k = lamda , the eigenvalue and I is the identity matrix.

    So I end up with [ 1-k , 0 : 1 , 2-k] which gives the characteristic polynomial as:
    k^2 -2k +2 = 0

    So k=2 ,1

    Then To find eigenvectors I solve (A-kI)v = 0

    Where v is the eigenvector.

    So if I sub my eigenvalues back into the A-kI matrix and then solve I SHOULD get my eigenvectors. BUT my matrix becomes:

    for k=1

    [0,0 : 1,1] [v1:v2] = [0:0]

    Giving:

    0x + 0y =0

    which to me gives the eigenvector [1:1] not to sure why just thats what others have looked like but the actual answers are:

    [0:1] and 1/sqrt(2) *[1:-1]

    HOW?
     \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}

    Eigenvectors that satisfy the equation are   \begin{pmatrix} 0 \\ y \end{pmatrix} and  k  \begin{pmatrix} 1 \\ 1 \end{pmatrix} where k is an arbitrary constant. Of course,   \begin{pmatrix} 0 \\ 0 \end{pmatrix} is a trivial solution.
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    (Original post by Clarity Incognito)
     \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \end{pmatrix}

    Eigenvectors that satisfy the equation are   \begin{pmatrix} 0 \\ y \end{pmatrix} and  k  \begin{pmatrix} 1 \\ 1 \end{pmatrix} where k is an arbitrary constant. Of course,   \begin{pmatrix} 0 \\ 0 \end{pmatrix} is a trivial solution.
    I'm sorry I have no idea what you mean here.
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    (Original post by Carlo08)
    I'm sorry I have no idea what you mean here.
    (A-kI)v=0

    Expanding and rearranging we have:

    Av = kv where k is an eigenvalue and v is the eigenvector corresponding to the eigenvalue.

    v =   \begin{pmatrix} x \\ y \end{pmatrix} where we want to know what x and y are right?

    We get the two equations:

    x = 2x
    x+2y=2y

    Solve, you'll find that x must be 0 and y can be anything.

    Do the same for

    x = x

    x+ 2y = y ( y=-x)

    If you chose x=800 then y=-800 or   800 \begin{pmatrix} 1 \\ -1 \end{pmatrix}


     \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} and   \begin{pmatrix} 0 \\ 1 \end{pmatrix} are both normalised eigenvectors, that might be specified in the question.
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    (Original post by Clarity Incognito)
    (A-kI)v=0

    Expanding and rearranging we have:

    Av = kv where k is an eigenvalue and v is the eigenvector corresponding to the eigenvalue.

    v =   \begin{pmatrix} x \\ y \end{pmatrix} where we want to know what x and y are right?

    We get the two equations:

    x = 2x
    x+2y=2y

    Solve, you'll find that x must be 0 and y can be anything.

    Do the same for

    x = x

    x+ 2y = y ( y=-x)

    If you chose x=800 then y=-800 or   800 \begin{pmatrix} 1 \\ -1 \end{pmatrix}


     \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} and   \begin{pmatrix} 0 \\ 1 \end{pmatrix} are both normalised eigenvectors, that might be specified in the question.
    Did you get different eigenvalues than I did, or did you just pick two for example? Mine where 2 and 1 which for k=1 gave me the equations : 0x+0y = x and x+y=y
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    (Original post by Carlo08)
    Did you get different eigenvalues than I did, or did you just pick two for example? Mine where 2 and 1 which for k=1 gave me the equations : 0x+0y = x and x+y=y
    Yeah, I got k=2 and 1 as well

    I might be taking your matrix A to be different though

    A =   \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}

    or

    A =   \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} ? Are you multiplying matrix A with   \begin{pmatrix} x \\ y \end{pmatrix} correctly? where are the two 0s coming from in the first equation?
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    (Original post by Clarity Incognito)
    Yeah, I got k=2 and 1 as well

    I might be taking your matrix A to be different though

    A =   \begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}

    or

    A =   \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} ? Are you multiplying matrix A with   \begin{pmatrix} x \\ y \end{pmatrix} correctly? where are the two 0s coming from in the first equation?
    $A=\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix} $

    and then when I put the value 1 back into the matrix as k I get
    \begin{pmatrix} 1-1 & 0 \\ 1 & 2-1 \end{pmatrix}

    giving:

    \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}

    This is correct right?
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    (Original post by Carlo08)
     A=\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}

    and then when I put the value 1 back into the matrix as k I get
    \begin{pmatrix} 1-1 & 0 \\ 1 & 2-1 \end{pmatrix}

    giving:

    \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}

    This is correct right?
    You've got to add latex] or tex] at the start of latex statements and when you're done, finish with /latex] or /tex] (all the latex commands start with "[")

    Yeah, I see what you've done now, so the equations you have now are

    \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =0
    giving x +y = 0 i.e. x = -y. You can now choose any value for x or y and the corresponding y or x value. The easiest being 1 giving  v_1 = \begin{pmatrix} 1\\ -1 \end{pmatrix} They have then normalised this vector.

    For the eigenvalue k=2 we have \begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix} giving -x=0 and x=0 and no conditions for y i.e. y can be anything but x must be 0. And so we have  v_2 = \begin{pmatrix} 0 \\ y \end{pmatrix} The most straightforward being y=1 as y=0 gives the trivial zero vector. With y =1, this is already a normalised eigenvector.
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    (Original post by Clarity Incognito)
    You've got to add latex] or tex] at the start of latex statements and when you're done, finish with /latex] or /tex] (all the latex commands start with "[")

    Yeah, I see what you've done now, so the equations you have now are

    \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =0
    giving x +y = 0 i.e. x = -y. You can now choose any value for x or y and the corresponding y or x value. The easiest being 1 giving  v_1 = \begin{pmatrix} 1\\ -1 \end{pmatrix} They have then normalised this vector.

    For the eigenvalue k=2 we have \begin{pmatrix} -1 & 0 \\ 1 & 0 \end{pmatrix} giving -x=0 and x=0 and no conditions for y i.e. y can be anything but x must be 0. And so we have  v_2 = \begin{pmatrix} 0 \\ y \end{pmatrix} The most straightforward being y=1 as y=0 gives the trivial zero vector. With y =1, this is already a normalised eigenvector.

    Ah thank you, your explanation of the first answer makes perfect sense to me. So basically I want to simplify the equations I get into x=y and then just give a value to x and see what the corresponding y value is. How does this work if I have 2 equations, or will they both be multiples of eachother? ie x=y and 4x=4y kind of thing?
    The answer given in my notes for the first one is, as you showed me [1:-1] however its multiplied by 1/sqrt(2) why is this .... and for k=2 it has been given as [0:1] as you said, thank you
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    (Original post by Carlo08)
    Ah thank you, your explanation of the first answer makes perfect sense to me. So basically I want to simplify the equations I get into x=y and then just give a value to x and see what the corresponding y value is. How does this work if I have 2 equations, or will they both be multiples of eachother? ie x=y and 4x=4y kind of thing?
    The answer given in my notes for the first one is, as you showed me [1:-1] however its multiplied by 1/sqrt(2) why is this .... and for k=2 it has been given as [0:1] as you said, thank you
    Yeah, if there are two equations then you just solve for x and y like you would for standard simultaneous equations, if one of the equations was x=y then yes the other could be a multiple of it.

    It's multiplied by 1/sqrt2 to 'normalise' the eigenvector. A normalised vector is a vector of unit magnitude. You normalise a vector  \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{pmatrix} by dividing each element by the magnitude of the vector, which is  \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 + ... + (x_n)^2}

    So the eigenvector  \begin{pmatrix} 1 \\ -1 \end{pmatrix} has a magnitude of  \sqrt{(1)^2 + (-1)^2} Dividing each element of the eigenvector by this you get  \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}
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    (Original post by Clarity Incognito)
    Yeah, if there are two equations then you just solve for x and y like you would for standard simultaneous equations, if one of the equations was x=y then yes the other could be a multiple of it.

    It's multiplied by 1/sqrt2 to 'normalise' the eigenvector. A normalised vector is a vector of unit magnitude. You normalise a vector  \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{pmatrix} by dividing each element by the magnitude of the vector, which is  \sqrt{(x_1)^2 + (x_2)^2 + (x_3)^2 + ... + (x_n)^2}

    So the eigenvector  \begin{pmatrix} 1 \\ -1 \end{pmatrix} has a magnitude of  \sqrt{(1)^2 + (-1)^2} Dividing each element of the eigenvector by this you get  \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}
    Thank you so so much for all that help! I missed the lecture on it and have been trying to learn it on my own but with your help I not understand it. When would I know to normalise the vector or not? It doesn't say in the question, would I just always do it?

    Thank you
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    (Original post by Carlo08)
    Thank you so so much for all that help! I missed the lecture on it and have been trying to learn it on my own but with your help I not understand it. When would I know to normalise the vector or not? It doesn't say in the question, would I just always do it?

    Thank you
    No worries! You don't always have to normalise the vector. I assumed that they specified it in the question. They are both eigenvectors nonetheless so that's the important bit. The reason you'd want to use normalised eigenvectors is to diagonalise the matrix A. This will give a matrix os the same size as A, which will be a diagonal matrix with the eigenvalues along the leading diagonal.
 
 
 
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