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Math Help - [SOLVED] cannot find 3rd eigen vector of this matrix for diagonalization

  1. #1
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    [SOLVED] cannot find 3rd eigen vector of this matrix for diagonalization

    matrix A

    2 0 0
    0 2 0
    1 1 3

    eigen value #1 = 2
    eigen vector #1 is

    -1
    1
    0


    eigen value #2 = 2
    eigen vector #2 is

    -1
    0
    1

    eigen value #3 = 3
    eigen vector #3 is

    I cannot find this because I get this matrix after substituting llamda into the original matrix along the diagonal

    -1 0 0
    0 -1 0
    1 1 0

    I am stuck here. Please help.
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  2. #2
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    Quote Originally Posted by thekrown View Post
    matrix A

    2 0 0
    0 2 0
    1 1 3

    eigen value #1 = 2
    eigen vector #1 is

    -1
    1
    0


    eigen value #2 = 2
    eigen vector #2 is

    -1
    0
    1

    eigen value #3 = 3
    eigen vector #3 is

    I cannot find this because I get this matrix after substituting llamda into the original matrix along the diagonal

    -1 0 0
    0 -1 0
    1 1 0

    I am stuck here. Please help.

    \begin{bmatrix}<br />
1 & 0 & 0\\ <br />
0 & 1 & 0\\ <br />
0 & 0 & 0<br />
\end{bmatrix}\Rightarrow x_3\begin{bmatrix}<br />
0 \\ <br />
0 \\ <br />
1 <br />
\end{bmatrix}
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  3. #3
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    I don't understand how you end up with that. What must I do?
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  4. #4
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    \begin{bmatrix}<br />
-1 & 0 & 0\\ <br />
0 & -1 & 0\\ <br />
1 & 1 & 0<br />
\end{bmatrix} use elementary row operations to obtain the rref.

    Multiple rows 1 and 2 by -1

    \begin{bmatrix}<br />
1 & 0 & 0\\ <br />
0 & 1 & 0\\ <br />
1 & 1 & 0<br />
\end{bmatrix}

    Subtract row 1 and 3 and row 2 and 3

    \begin{bmatrix}<br />
1 & 0 & 0\\ <br />
0 & 1 & 0\\ <br />
0 & 0 & 0<br />
\end{bmatrix}

    x_1=0, x_2=0, and x_3 is a free variable.

    \begin{bmatrix}<br />
 0\\ <br />
 0\\ <br />
x_3<br />
 \end{bmatrix}\rightarrow x_3\begin{bmatrix}<br />
  0\\ <br />
  0\\ <br />
 1<br />
  \end{bmatrix}
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  5. #5
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    Okay if you don't mind I will say what I understood in my own words.

    First I find the eigen values, subtract it from the diagonal. This gave us our matrix

    1 0 0
    0 1 0
    1 1 0

    from this point I can manupulate this matrix to obtain a reduced echelon form matrix.

    Once I have my leading 1's, I can then find the value for x1, x2, x3.

    In our case, it was simple once the reduction was made and we have the matrix

    1 0 0
    0 1 0
    0 0 0

    the third row can be removed, leaving us with a free variable for x3 and x1=0, x2=0.
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  6. #6
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    Quote Originally Posted by thekrown View Post
    Okay if you don't mind I will say what I understood in my own words.

    First I find the eigen values, subtract it from the diagonal. This gave us our matrix

    1 0 0
    0 1 0
    1 1 0

    from this point I can manupulate this matrix to obtain a reduced echelon form matrix.

    Once I have my leading 1's, I can then find the value for x1, x2, x3.

    In our case, it was simple once the reduction was made and we have the matrix

    1 0 0
    0 1 0
    0 0 0

    the third row can be removed, leaving us with a free variable for x3 and x1=0, x2=0.
    I wouldn't say the 3rd row can be removed. The third element can be any value.
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  7. #7
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    Okay so I will leave it there, my teacher would remove it, maybe he was doing that to save chalk. Who knows.
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