Results 1 to 7 of 7

Thread: Find matrix U

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    72

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    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 $\displaystyle U^{\,T}MU$ to be diagonal.

    You are looking for an orthogonal matrix $\displaystyle U$ (meaning that $\displaystyle U^{-1} = U^{\,T}$) that diagonalizes $\displaystyle M$. Find all linearly independent eigenvectors of $\displaystyle M$. Since $\displaystyle M$ 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 $\displaystyle U$ to diagonalize $\displaystyle M$. Note that if your matrix had had an eigenvalue of multiplicity $\displaystyle k$ greater than 1, you would need to apply the Gram-Schmidt orthonormalization process on the $\displaystyle k$ 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 $\displaystyle U$ will form an orthonormal set, so $\displaystyle U$ will be orthogonal, and $\displaystyle U^{-1}MU = U^{\,T}MU = D,\text{ where }D\text{ is diagonal}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    Posts
    72

    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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...

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

    -> v'1=v1/|v1|

    $\displaystyle |v2|=\sqrt{3}$

    $\displaystyle |v3|=\sqrt{6}$

    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) :

    $\displaystyle U=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \\
    0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\
    \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \end{pmatrix}$


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

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


    Hope it helps
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    77
    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?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] How do you find matrix this kind of matrix?
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Feb 4th 2010, 04:22 AM
  2. Find the Matrix E, such that
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Oct 18th 2009, 10:58 AM
  3. Replies: 3
    Last Post: Jun 15th 2009, 02:16 PM
  4. Find the matrix
    Posted in the Advanced Algebra Forum
    Replies: 10
    Last Post: May 13th 2009, 06:40 AM
  5. Find a matrix
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Jun 21st 2008, 02:28 PM

Search Tags


/mathhelpforum @mathhelpforum