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Math Help - Diagonalizing 2x2 matrices

  1. #1
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    Diagonalizing 2x2 matrices

    Diagonalize the following matrix:

    <br />
\begin{bmatrix}<br />
  2 & 3      \\<br />
  4 &1      \\ <br />
\end{bmatrix}<br />

    I don't know if I made an algebraic mistake or if the matrix is just not diagonalizable. o.O

    _______________________________________
    My work:

    det(A-\lambda I) = det\begin{bmatrix}<br />
  2-\lambda & 3      \\<br />
  4 &1-\lambda      \\ <br />
\end{bmatrix} = (2-\lambda)(1-\lambda)-(3)(4) =(\lambda-5)(\lambda+2)

    Therefore: \lambda = 5, \lambda = -2
    ____________________

    For \lambda = 5:

    \begin{bmatrix}<br />
  2 & 3      \\<br />
  4 & 1      \\ <br />
\end{bmatrix} - \begin{bmatrix}<br />
5 & 0      \\<br />
   0 & 5      \\ <br />
 \end{bmatrix} = \begin{bmatrix}<br />
-3 & 3      \\<br />
   4 & -4      \\ <br />
 \end{bmatrix} ~ \begin{bmatrix}<br />
 1 & -1      \\<br />
    0 & 0      \\ <br />
  \end{bmatrix}

    (A - 5\lambda)X = 0: x_1 -x_2 = 0 and x_2 is free. So x_1 = x_2, but x_2 =1 , which means x_1 = 1 too.

    v_1=  \begin{bmatrix}1 \\<br />
1 \\<br />
\end{bmatrix}
    ____________________

    Using the same method for \lambda = -2, I get:

    (A+2I)X= \begin{bmatrix}<br />
  4 & 3      \\<br />
  4 & 3      \\ <br />
\end{bmatrix} ~ \begin{bmatrix}<br />
   4 & 3      \\<br />
   0 & 0      \\ <br />
 \end{bmatrix}

    so

    v_2=  \begin{bmatrix}3 \\<br />
4 \\<br />
\end{bmatrix}
    __________________________

    Now:

    P= [v_1 v_2] =  \begin{bmatrix}1 & 3 \\<br />
1 & 4 \\<br />
\end{bmatrix}

    ^ Is that wrong?
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  2. #2
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    Doesn't look too diagonal to me (:
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  3. #3
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    Quote Originally Posted by Cursed View Post
    Diagonalize the following matrix:

    <br />
\begin{bmatrix}<br />
  2 & 3      \\<br />
  4 &1      \\ <br />
\end{bmatrix}<br />

    I don't know if I made an algebraic mistake or if the matrix is just not diagonalizable. o.O

    _______________________________________
    My work:

    det(A-\lambda I) = det\begin{bmatrix}<br />
  2-\lambda & 3      \\<br />
  4 &1-\lambda      \\ <br />
\end{bmatrix} = (2-\lambda)(1-\lambda)-(3)(4) =(\lambda-5)(\lambda+2)

    Therefore: \lambda = 5, \lambda = -2
    ____________________

    For \lambda = 5:

    \begin{bmatrix}<br />
  2 & 3      \\<br />
  4 & 1      \\ <br />
\end{bmatrix} - \begin{bmatrix}<br />
5 & 0      \\<br />
   0 & 5      \\ <br />
 \end{bmatrix} = \begin{bmatrix}<br />
-3 & 3      \\<br />
   4 & -4      \\ <br />
 \end{bmatrix} ~ \begin{bmatrix}<br />
 1 & -1      \\<br />
    0 & 0      \\ <br />
  \end{bmatrix}

    (A - 5\lambda)X = 0: x_1 -x_2 = 0 and x_2 is free. So x_1 = x_2, but x_2 =1 , which means x_1 = 1 too.

    v_1=  \begin{bmatrix}1 \\<br />
1 \\<br />
\end{bmatrix}
    ____________________

    Using the same method for \lambda = -2, I get:

    (A+2I)X= \begin{bmatrix}<br />
  4 & 3      \\<br />
  4 & 3      \\ <br />
\end{bmatrix} ~ \begin{bmatrix}<br />
   4 & 3      \\<br />
   0 & 0      \\ <br />
 \end{bmatrix}

    so

    v_2=  \begin{bmatrix}3 \\<br />
4 \\<br />
\end{bmatrix}
    __________________________

    Now:

    P= [v_1 v_2] =  \begin{bmatrix}1 & 3 \\<br />
1 & 4 \\<br />
\end{bmatrix}

    ^ Is that wrong?
    .

    Yes, it is wrong: \begin{pmatrix}2&3\\4&1\end{pmatrix}\begin{pmatrix  }3\\4\end{pmatrix}\neq(-2)\begin{pmatrix}3\\4\end{pmatrix} , so that isn't an eigenvector corresponding to the eigenvalue -2...

    Tonio
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  4. #4
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    Characteristic polynomials of 2x2 matrices can be done without computing det(A-\lambda I)
    P_{\lambda}(A)=\lambda^2-\lambda tr(A)+det(A)

    Thus, tr(A)=3 \ \mbox{and} \ det(A)=-10\rightarrow \ \lambda^2-3\lambda-10=(\lambda-5)(\lambda+2)
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  5. #5
    A Plied Mathematician
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    I agree with Tonio that the v_{2} eigenvector is incorrect. Your steps are good up until the moment you say that v_{2}=\dots. Don't skip steps there!

    As for constructing P, you may need to run the Gram-Schmidt procedure on your eigenvectors. P has to be orthogonal, which means that the eigenvectors have to be orthonormal. Also, don't forget that the final result has to be your diagonal matrix. Remember how to get the diagonal matrix?
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  6. #6
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    Perhaps I'm being confused, but why would P need to be orthogonal?
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  7. #7
    A Plied Mathematician
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    Quote Originally Posted by Defunkt View Post
    Perhaps I'm being confused, but why would P need to be orthogonal?
    Maybe I'm thinking of orthogonally diagonalizing a matrix. The advantage there is that inverting the P is as simple as transposition. You are correct (implying I was wrong before) in saying that P does not have to be orthogonal. It does have to be invertible.
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