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Math Help - diagonalising a 3x3 matrix

  1. #1
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    diagonalising a 3x3 matrix

    im really struggling with this problem, i keep getting so far and then hitting a brick wall,
    the question asks: diagonalise the matrix B=
    [0 1 1]
    [0 2 2]
    [0 0 3]
    in other words find an invertible matrix P and a diagonal matrix D so P'BP=D where P'= P inverse

    can anybody please help, as i am really struggling!! thank you
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by linearalgebradunce View Post
    im really struggling with this problem, i keep getting so far and then hitting a brick wall,
    the question asks: diagonalise the matrix B=
    [0 1 1]
    [0 2 2]
    [0 0 3]
    in other words find an invertible matrix P and a diagonal matrix D so P'BP=D where P'= P inverse
    To diagonalize B you first need to find its eigenvalues : the roots of the polynomial P(\lambda)=\det(B-\lambda I_3). Once you know the three eigenvalues \lambda_1,\lambda_2 and \lambda_3, you can find the eigenvectors v_i by solving Bv_i=\lambda_iv_i for i \in\{1,2,3\}. The matrix P you're looking for will simply be a matrix which columns are the three eigenvectors.
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  3. #3
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    thanks for your help, however the problem there is so far ive found 2 different sets of eigenvalues, and i cannot find corresponding eigenvectors for either set. this is really important i get this right, my next year at university rests on ensuring answers to problems like these are correct.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    however the problem there is so far ive found 2 different sets of eigenvalues, and i cannot find corresponding eigenvectors for either set.
    What did you find for P(\lambda) ?
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  5. #5
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    do you mean the polynomial I found, - lambda cubed + 5lambda squared - 6lambda. (sorry i couldnt find the symbols)
    however i have also read that for a triangular square matrix the eigenvalues are the diagonal values?
    i then cannot find the eigenvectors?! ive been attempting this question for a number of days now, please help.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    do you mean the polynomial I found, - lambda cubed + 5lambda squared - 6lambda. (sorry i couldnt find the symbols)
    That's it. Factoring the polynomial, we get
    <br />
-\lambda^3+5\lambda^2-6\lambda=-\lambda(\lambda^2-5\lambda+6)=-\lambda(\lambda-2)(\lambda-3)
    so the eigenvalues are 0, 2 and 3.
    however i have also read that for a triangular square matrix the eigenvalues are the diagonal values?
    I didn't know that but that's right.
    i then cannot find the eigenvectors?! ive been attempting this question for a number of days now, please help.
    Let's find the eigenvector v=\begin{bmatrix}x\\y\\z\end{bmatrix} whose corresponding eigenvalue is 2.

    We know that Bv=2v <br />
\Longleftrightarrow \begin{bmatrix} 0& 1& 1\\<br />
0& 2& 2\\<br />
0& 0& 3 <br />
\end{bmatrix}<br />
\cdot <br />
\begin{bmatrix}<br />
x\\<br />
y\\<br />
z<br />
\end{bmatrix}<br />
=2\begin{bmatrix}x \\ y \\ z \end{bmatrix}<br />
\Longleftrightarrow \begin{bmatrix} y +z\\  2y+2z\\ 3z \end{bmatrix}<br />
=\begin{bmatrix}2x \\ 2y \\ 2z \end{bmatrix} thus we have to solve  \begin{cases}<br />
y+z=2x&(1)\\<br />
2y+2z=2y&(2)\\<br />
3z=2z&(3)\\<br />
\end{cases}<br />

    From (3) we get that z=0, (2) is useless (0=0) and (1) tells us that y=2x so v=\begin{bmatrix}x\\2x\\0 \end{bmatrix} for x\neq 0. We can, for example, choose x=1 to get v=\begin{bmatrix}1\\2\\0 \end{bmatrix}.
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  7. #7
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    thank you so much for your help, however for eigenvalue 0 I am still struggling to find a non-zero eigenvector. I have the eigenvector for 3. When I then place the 3 eigenvectors into matrix form does it matter which column I place each vector in?!
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    thank you so much for your help, however for eigenvalue 0 I am still struggling to find a non-zero eigenvector.
    This time we have to solve
     \begin{cases}<br />
y+z=0&(1)\\<br />
2y+2z=0&(2)\\<br />
3z=0&(3)\\<br />
\end{cases}<br />
    From (3) we get that z=0, from (2) and (1) that y=-z=0. As there is no condition on x, the solutions are the vectors \begin{bmatrix}x\\0\\0\end{bmatrix} for x\neq 0. If you choose, for example, x=1, you get a non-zero vector.

    I have the eigenvector for 3. When I then place the 3 eigenvectors into matrix form does it matter which column I place each vector in?!
    No, it does not matter.
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  9. #9
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    thank you, however, now i have all 3 eigenvectors, no matter which way I place them into the matrix P I cannot find the diagonal vector D. i am always one number out.
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  10. #10
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by linearalgebradunce View Post
    thank you, however, now i have all 3 eigenvectors, no matter which way I place them into the matrix P I cannot find the diagonal vector D. i am always one number out.
    I don't understand how you can be "one number out", the diagonal entries of D are the three eigenvalues of B.
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  11. #11
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    because BP does not equal PD

    for BP the resultant matrix is
    [0 2 3]
    [0 4 6]
    [0 0 0]

    and for PD the resultant matrix is
    [0 2 3]
    [0 4 9]
    [0 0 0]
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  12. #12
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    Quote Originally Posted by linearalgebradunce View Post
    because BP does not equal PD

    for BP the resultant matrix is
    [0 2 3]
    [0 4 6]
    [0 0 0]

    and for PD the resultant matrix is
    [0 2 3]
    [0 4 9]
    [0 0 0]
    Check your eigenvector for the eigenvalue 3. If you get it right, you should find that both BP and PD are \begin{bmatrix}0&2&3\\0&4&6\\0&0&3\end{bmatrix}.
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  13. #13
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    i found my mistake, thank you very much to both of you for all your help!!
    the people on this forum are brilliant!! thanks again!
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