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Math Help - Find matrix U

  1. #1
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    Find matrix U

    Third part of the same question...the only part i cant seem to understand at all....

    A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

    The characteristic equation is f(x)=(3-x)(x-4)(x-1)

    The eigenvalues for the characteristic equation are 3, 4 and 1?

    I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix
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  2. #2
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    Hello,

    Quote Originally Posted by deragon999 View Post
    Third part of the same question...the only part i cant seem to understand at all....

    A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

    The characteristic equation is f(x)=(3-x)(x-4)(x-1)

    The eigenvalues for the characteristic equation are 3, 4 and 1?

    I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix
    Find the eigenspaces/vectors for each eigenvalue (3, 4, 1). Do you know how to do it ?
    Then, find the matrix transition between the canonical basis (assuming that M was written according to this basis) and the new basis, formed by the eigenvectors.
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    Quote Originally Posted by deragon999 View Post
    Third part of the same question...the only part i cant seem to understand at all....

    A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

    The characteristic equation is f(x)=(3-x)(x-4)(x-1)

    The eigenvalues for the characteristic equation are 3, 4 and 1?

    I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix
    We need U^{\,T}MU to be diagonal.

    You are looking for an orthogonal matrix U (meaning that U^{-1} = U^{\,T}) that diagonalizes M. Find all linearly independent eigenvectors of M. Since M is symmetric, any pair of eigenvectors that correspond to different eigenvalues will be orthogonal to each other. So, take your 3 eigenvectors, and normalize them to get an orthonormal set, which you can then place in the columns of U to diagonalize M. Note that if your matrix had had an eigenvalue of multiplicity k greater than 1, you would need to apply the Gram-Schmidt orthonormalization process on the k linearly independent vectors corresponding to that eigenvalue in order to ensure that your set is orthonormal.

    If you do the above, the columns (and rows) of U will form an orthonormal set, so U will be orthogonal, and U^{-1}MU = U^{\,T}MU = D,\text{ where }D\text{ is diagonal}.
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  4. #4
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    From eigenvectors onwards

    I have found the the three eigenvectors(i hope) which are:

    v1=[1;0;-1] v2=[1;1;1] v3=[-1;2;-1]

    I have used the following method on them: (<-- is the pivot row)

    1 0 -1 <--
    1 1 1
    -1 2 -1
    0 1 2 <--
    0 2 -2
    0 0 -6<--

    Which, when the pivot rows are written as columns of a U, gives U as being:[1,0,0;0,1,0;-1,2,-6]

    Which when applied to (U^t)*M*U this matrix, U, finds D to be:[6,-6,-18;-6,18,-42;-18,-42,108]
    Which satisfies the question "Find a matrix(any matrix) U, so that (U^t)*M*U is diagonal"....correct?

    Is this a valid method for working out a matrix, U?
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    Quote Originally Posted by deragon999 View Post
    I have found the the three eigenvectors(i hope) which are:

    v1=[1;0;-1] v2=[1;1;1] v3=[-1;2;-1]

    I have used the following method on them: (<-- is the pivot row)

    1 0 -1 <--
    1 1 1
    -1 2 -1
    0 1 2 <--
    0 2 -2
    0 0 -6<--
    I find exactly the same eigenvectors

    Now you have to normalize them...

    |v1|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}

    -> v'1=v1/|v1|

    |v2|=\sqrt{3}

    |v3|=\sqrt{6}

    Which, when the pivot rows are written as columns of a U, gives U as being:[1,0,0;0,1,0;-1,2,-6]
    Where does it come from ? o.O

    Which when applied to (U^t)*M*U this matrix, U, finds D to be:[6,-6,-18;-6,18,-42;-18,-42,108]
    Which satisfies the question "Find a matrix(any matrix) U, so that (U^t)*M*U is diagonal"....correct?

    Is this a valid method for working out a matrix, U?
    Hm nope ~

    You got U, formed by the normalised eigenvectors in columns (in the order 3-4-1) :

    U=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \\<br />
0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\<br />
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \end{pmatrix}


    D=U^t M U= \begin{pmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}

    (in the same order as you put the eigenvectors in U)


    Hope it helps
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  6. #6
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    Is that correct for the eigenvectors?
    I ask because I get (1, -2, 1) not (-1, 2, -1) for eigenvalue: 1

    which of us is correct?
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  7. #7
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    Quote Originally Posted by Dr Zoidburg View Post
    Is that correct for the eigenvectors?
    I ask because I get (1, -2, 1) not (-1, 2, -1) for eigenvalue: 1

    which of us is correct?
    The two of you :-)

    The eigenvectors form basis, so you don't care wether it is -u or u.
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