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Math Help - eigenvalues and eigen vectors!

  1. #1
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    eigenvalues and eigen vectors!

    helloo!

    find the eigenvalues and eigen vectors of

    A= |1 2|
    |5 4|

    i dont think its long to work out, but ive just started this topic and cannot work it out
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  2. #2
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    Quote Originally Posted by charlotte_usa View Post
    helloo!

    find the eigenvalues and eigen vectors of

    A= |1 2|
    |5 4|

    i dont think its long to work out, but ive just started this topic and cannot work it out
    The eigen values are the roots of the equation (quadratic equation in this case):

    \det(A-\lambda I)=0

    Why is that?

    RonL
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  3. #3
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    The Eigenvalues are solutions to:
    \left| \begin{array}{cc}k-1&-2\\-5&k-4 \end{array} \right| = 0
    Thus,
    (k-1)(k-4) - 10 = 0
    Now solve.
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  4. #4
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    A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.

    Ax=kx

    Where k is the eigenvalue.

    They're tied together. Lambda is mostly used, but I will use k.

    You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.

    \begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end  {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}

    \begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr  ix}-5&2\\5&-2\end{bmatrix}

    rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm  atrix}1&1\\0&0\end{bmatrix}

    rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}

    In the first matrix, we have x=-y

    Let y=t, then we have x=-t and y=t

    In the second one, we have x=\frac{2}{5}y

    Let y=t, then x=\frac{2}{5}t

    Therefore, the eigenvectors are

    \begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}

    and

    \begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}
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  5. #5
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    Quote Originally Posted by galactus View Post
    A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.

    Ax=kx

    Where k is the eigenvalue.

    They're tied together. Lambda is mostly used, but I will use k.

    You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.

    \begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end  {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}

    \begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr  ix}-5&2\\5&-2\end{bmatrix}

    rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm  atrix}1&1\\0&0\end{bmatrix}

    rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}

    In the first matrix, we have x=-y

    Let y=t, then we have x=-t and y=t

    In the second one, we have x=\frac{2}{5}y

    Let y=t, then x=\frac{2}{5}t

    Therefore, the eigenvectors are

    \begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}

    and

    \begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}




    i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by college_clare View Post




    i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...
    Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation Ax = \lambda x):
    \left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]

    So we have the simultaneous equations
    a + 2b = -a
    and
    5a + 4b = -b

    Solving this for a and b gives a = a and b = -a. So the eigenvector is
    \left [ \begin{array}{c} a \\ -a \end{array} \right ]  = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]
    (This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)

    We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
    \left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]

    I'll leave it to you to work out the rest.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation Ax = \lambda x):
    \left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]

    So we have the simultaneous equations
    a + 2b = -a
    and
    5a + 4b = -b

    Solving this for a and b gives a = a and b = -a. So the eigenvector is
    \left [ \begin{array}{c} a \\ -a \end{array} \right ]  = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]
    (This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)

    We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
    \left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]

    I'll leave it to you to work out the rest.

    -Dan
    i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)

    but earlier it was suggested that the answer was 2/5 so have i gone wrong?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by charlotte_usa View Post
    i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)

    but earlier it was suggested that the answer was 2/5 so have i gone wrong?
    You are getting an eigenvector of
    a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ]

    Again, this is simply a modification of galactus' answer. Let a = (2/5)t. Then
    a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = \frac{2}{5} t \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = t \left [ \begin{array}{c} \frac{2}{5} \\ 1 \end{array} \right ]
    which is the same as galactus' eigenvector. So the two are essentially the same.

    -Dan
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