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Math Help - Is the matrix Diagonalizable

  1. #1
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    Question Is the matrix Diagonalizable

    Hi I have a problem in Diagonalizability concept.

    A= [7 -4 0
    5 -5 0
    6 -6 3]

    A is 3x3 matrix.

    I got eigen values as 3, 3, -1
    Need to find eigen vectors and matrix D, where [D= Q^(-1)AQ]
    D being the diagonal matrix.

    If there is any easy methodology plz suggest

    Need help from eigen vectors onwards.

    Thanks in advance.
    Last edited by sofwhere; August 27th 2009 at 04:24 AM.
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  2. #2
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    Quote Originally Posted by sofwhere View Post
    Hi I have a problem in Diagonalizability concept.

    A= [7 -4 0
    5 -5 0
    6 -6 0]

    A is 3x3 matrix.

    I got eigen values as 3, [1+ 2 sqrroot(15)], 1- 2 sqrroot(15)]
    Well, you got wrong then! Without doing any calculations, your matrix has a column of 0's so its determinant is 0 which means that it must have 0 as an eigenvalue.
    The eigenvalue equation is \left|\begin{array}{ccc}7-\lambda & -4 & 0 \\ 5 & -5-\lambda & 0 \\ 6 & -6 & -\lambda\end{array}\right|= 0. Expanding on the last column, that is -\lambda \left|\begin{array}{cc} 7- \lambda & -4 \\ 5 & -5-\lambda\end{array}\right| = (-\lambda)(\lambda^2- 2\lambda- 35- 20) =(-\lambda)(\lambda^2- 2\lambda- 55)= 0. Thus, the eigenvalues are 0, 1+ \sqrt{56}= 1+ 2\sqrt{14}, and 1- \sqrt{56}= 1- 2\sqrt{14}.

    Need to find eigen vectors and matrix D, where [D= Q^(-1)AQ]
    D being the diagonal matrix.

    If there is any easy methodology plz suggest

    Need help from eigen vectors onwards.

    Thanks in advance.
    Last edited by HallsofIvy; August 26th 2009 at 05:47 PM.
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  3. #3
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    Let A = \left[<br />
\begin{array}{ c c c }<br />
7 & -4 & 0 \\<br />
5 & -5 & 0 \\<br />
6 & -6 & 0 \\<br />
\end{array} \right]

    We can see that rank(A) = 2, thus 0 is an eigenvalue. If you solve det(A - \lambda I)=0, you will get that -3, 5 are the remaining eigenvalues.

    Now, the diagonal matrix D which satisfies Q^{-1}AQ = D simply has A's eigenvalues as its diagonal entries. Do you understand why?

    To find the eigenvectors for 0, -3, 5 -- you need to solve (A - \lambda I)x = 0 for \lambda = 0, -3, 5.

    @HallsofIvy: = (-\lambda)(\lambda^2- 2\lambda- 35- 20) should be = (-\lambda)(\lambda^2- 2\lambda - 35 + 20); you're also missing an = sign!
    Last edited by Defunkt; August 26th 2009 at 01:25 PM.
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  4. #4
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    Question Is the matrix Diagonalizable

    Sorry to say the element in the matrix is not zero. Its 3.
    I also updated in my first post.

    Also 2nd row first element is 8.
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    Question

    A= [7 -4 0
    8 -5 0
    6 -6 3]
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  6. #6
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    Well, then the new eigenvalues are 3,3,-1, so D will have them as its diagonal entries. To find the eigenvectors now, solve (A-\lambda I)x = 0 for \lambda = 3,1

    (this is for A = \left[<br />
\begin{array}{ c c c }<br />
7 & -4 & 0 \\<br />
8 & -5 & 0 \\<br />
6 & -6 & 3 \\<br />
\end{array} \right])
    Last edited by Defunkt; August 27th 2009 at 06:32 AM.
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  7. #7
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    lambda(denote it as t) = 3,-1 ,3
    Then i went for finding eigen vectors

    B1= [A-(t1)I]

    Then for t2=-1
    B2= [ 8 -4 0
    8 -4 0
    6 -6 4 ]
    the coefficients X= [x1, x2, x3]

    [A-(t2)I]X= 0

    need to find values of x1,x2,x3
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  8. #8
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    Quote Originally Posted by sofwhere View Post
    lambda(denote it as t) = 3,-1 ,3
    Then i went for finding eigen vectors

    B1= [A-(t1)I]

    Then for t2=-1
    B2= [ 8 -4 0
    8 -4 0
    6 -6 4 ]
    the coefficients X= [x1, x2, x3]

    [A-(t2)I]X= 0

    need to find values of x1,x2,x3
    So, for eigenvalue -1, you want to solve 8x- 4y= 0, 8x- 4y= 0, and 6x- 6y+ 4z= 0. Those first two equations are obviously the same so they just say that y= 2x. Putting that into the last equation, 6x- 6(2x)+ 4z= 0 or -6x+ 4z= 0 or z= (6/4)x= (3/2)x. An eigenvector corresponding to eigenvalue -1 is x<1, 2, 3/2> and, if we take x= 2, that is <2, 4, 3>.

    For eigenvalue 3, We have A- 3I= \begin{bmatrix}7-3 & -4 & 0 \\ 8 & -5- 3 & 0 \\ 6 & -6 & 3-3\end{bmatrix}= \begin{bmatrix}4 & -4 & 0 \\ 8 & -8 & 0 \\ 6 & - 6 & 0\end{bmatrix} so you want to solve 4x- 4y= 0, 8x- 8y= 0, and 6x- 6y= 0. All of those, of course, reduce to y= x so any vector of the form <x, x, 0>= x<1, 1, 0> is an eigenvector corresponding to eigen value 3. But there was no "z" in those equations so z can be anything. <0, 0, z>= z< 0, 0, 1> is also an eigenvector corresponding to eigenvalue 3.

    The fact that this matrix has 3 independent eigenvectors is what tells us it can be diagonalized. Let B be the matrix having those eigenvectors as columns and you should get B^{-1}AB= \begin{bmatrix}-1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix} (the order of numbers on the diagonal will depend on the order of the eigenvectors as columns).
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  9. #9
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    Unhappy

    I took D = Q^(-1)AQ

    Where Q is matrix of eigen vectors (Its the B u referred)
    I took

    Q= [ 1 1 0
    2 1 0
    3/2 0 1]

    and so;ved the equation by finding Q^(-1) and then Q^(-1)AQ

    I got D= [-1 0 0
    0 3 0
    9/4 0 3]
    Is this correct?



    Or should I take the eigen vector for lambda= -1 as ( 2,4,3) as u mentioned and redo.
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  10. #10
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    Quote Originally Posted by sofwhere View Post
    I took D = Q^(-1)AQ

    Where Q is matrix of eigen vectors (Its the B u referred)
    I took

    Q= [ 1 1 0
    2 1 0
    3/2 0 1]

    and so;ved the equation by finding Q^(-1) and then Q^(-1)AQ

    I got D= [-1 0 0
    0 3 0
    9/4 0 3]
    Is this correct?



    Or should I take the eigen vector for lambda= -1 as ( 2,4,3) as u mentioned and redo.
    What you should have gotten is this: D = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}. The eigenvectors you take are irrelevant to your solution, as they all should produce the same one.

    You can also find D without directly calculating Q^{-1}AQ -- it is a direct result of the definition of eigenvectors and eigenvalues.
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  11. #11
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    Quote Originally Posted by sofwhere View Post
    lambda(denote it as t) = 3,-1 ,3
    Then i went for finding eigen vectors

    B1= [A-(t1)I]

    Then for t2=-1
    B2= [ 8 -4 0
    8 -4 0
    6 -6 4 ]
    the coefficients X= [x1, x2, x3]

    [A-(t2)I]X= 0

    need to find values of x1,x2,x3
    sofwhere, isn't the last (or the first) eigenvalue -3 (instead of +3)?
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  12. #12
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    Finally det(T-tI)= (3-t)(t+1)(t-3) = O

    therefore t1=3, t2=-1, t3=3
    These r the three eigen values and eigen vectors are

    v1= t( 1 1 0)

    v2= t( 1 2 3/2 )

    v3 = t( 0 0 1)
    where tbelongs to R

    This is how i started.
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  13. #13
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    Thumbs up

    Quote Originally Posted by courteous View Post
    sofwhere, isn't the last (or the first) eigenvalue -3 (instead of +3)?

    Thanks guyz finally got it.
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