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    ~What happens when, for one of the eigenvalues, lambda, of a matrix A, the matrix (A - lambdaI), required for working out the corresponding eigenvector, is a an identity matix.

    i.e. to workout the eigenvector you must solve x1+ x2 +x3 = 0.
    how do you work this out?

    Thanks
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    So you want a non-zero vector \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} satisfying x_1+x_2+x_3=0. Remember that the elements of the vector are allowed to be negative so you could put one of the elements to be x_1=-1, say, then choose suitable values for x_2, x_3 to satisfy the equation.
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    (Original post by ttoby)
    So you want a non-zero vector \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} satisfying x_1+x_2+x_3=0. Remember that the elements of the vector are allowed to be negative so you could put one of the elements to be x_1=-1, say, then choose suitable values for x_2, x_3 to satisfy the equation.
    what if you have two equal eigenvalues. what are the corresponding eigenvectors. how do you work them out
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    (Original post by sonic23)
    what if you have two equal eigenvalues. what are the corresponding eigenvectors. how do you work them out
    It depends on whether the question is asking for two linearly independent eigenvectors or whether you need some sort of general form of the eigenvectors. For the eigenvectors, it's always true that x_3=-x_1-x_2 so any vector of the form \begin{pmatrix}a \\ b \\ -a-b\end{pmatrix} where a and b are scalars (not both zero) would be an eigenvector.

    If you want two linearly independent eigenvectors then just choose suitable values of a and b to achieve this. An easy choice would be a=1,b=0 then a=0,b=1 to give \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} and \begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}.
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    (Original post by ttoby)
    It depends on whether the question is asking for two linearly independent eigenvectors or whether you need some sort of general form of the eigenvectors. For the eigenvectors, it's always true that x_3=-x_1-x_2 so any vector of the form \begin{pmatrix}a \\ b \\ -a-b\end{pmatrix} where a and b are scalars (not both zero) would be an eigenvector.

    If you want two linearly independent eigenvectors then just choose suitable values of a and b to achieve this. An easy choice would be a=1,b=0 then a=0,b=1 to give \begin{pmatrix}1 \\ 0 \\ -1\end{pmatrix} and \begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}.
    but the dot product of the eigenvectors you found is 1, not 0; i.e. they are not orthognal, and the must be since the initial matrix is symmetric.

    ?????
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    (Original post by sonic23)
    but the dot product of the eigenvectors you found is 1, not 0; i.e. they are not orthognal, and the must be since the initial matrix is symmetric.

    ?????
    You didn't say you wanted orthogonal eigenvectors. There is a way to get orthogonal eigenvectors but this way will only work for up to three dimensions. If you want something that works more generally, I would suggest asking someone else doing/teaching your course, unless someone else here has a better method. Anyone?

    But to get orthogonal eigenvectors, you could try picturing the 3-dimensional graph x_3=-x_1-x_2. As x_1 and x_2 increase, the graph slopes downwards but if x_1 increases and x_2 decreases then x_3 remains constant. The eigenvectors would have to be parallel to this plane. You could get perpendicular eigenvectors by choosing x_1=x_2=1 for one eigenvector and x_1=1, x_2=-1 for the other eigenvector. This gives you eigenvectors of \begin{pmatrix}1 \\ 1 \\ -2\end{pmatrix} and \begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} and if you dot product these, you'll find that they're perpendicular.

    I don't think this is the best way of doing the question though so if someone else has a better way please do post it!
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    (Original post by ttoby)
    You didn't say you wanted orthogonal eigenvectors. There is a way to get orthogonal eigenvectors but this way will only work for up to three dimensions. If you want something that works more generally, I would suggest asking someone else doing/teaching your course, unless someone else here has a better method. Anyone?

    But to get orthogonal eigenvectors, you could try picturing the 3-dimensional graph x_3=-x_1-x_2. As x_1 and x_2 increase, the graph slopes downwards but if x_1 increases and x_2 decreases then x_3 remains constant. The eigenvectors would have to be parallel to this plane. You could get perpendicular eigenvectors by choosing x_1=x_2=1 for one eigenvector and x_1=1, x_2=-1 for the other eigenvector. This gives you eigenvectors of \begin{pmatrix}1 \\ 1 \\ -2\end{pmatrix} and \begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix} and if you dot product these, you'll find that they're perpendicular.

    I don't think this is the best way of doing the question though so if someone else has a better way please do post it!
    I realised something. ''If A is an n x n symmetric matrix, then any two eigenvectors that come from distinct eigenvalues are orthogonal''

    This means that only eigenvectors corresponding to different eigenvalues are orthogonal. If the eigenvalues are the same (both 1), then they shouldnt be orthogonal. Correct?
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    (Original post by sonic23)
    I realised something. ''If A is an n x n symmetric matrix, then any two eigenvectors that come from distinct eigenvalues are orthogonal''

    This means that only eigenvectors corresponding to different eigenvalues are orthogonal. If the eigenvalues are the same (both 1), then they shouldnt be orthogonal. Correct?
    Yes, that's saying that if the eigenvalues are different then the eigenvectors have to be orthogonal, but if you have two eigenvectors for the same eigenvalue then they're not necessarily orthogonal (i.e. they may or may not be orthogonal and my above posts give examples of both situations).
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    (Original post by ttoby)
    Yes, that's saying that if the eigenvalues are different then the eigenvectors have to be orthogonal, but if you have two eigenvectors for the same eigenvalue then they're not necessarily orthogonal (i.e. they may or may not be orthogonal and my above posts give examples of both situations).
    so how do you know which are the correct pair of eigenvectors?
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    (Original post by sonic23)
    so how do you know which are the correct pair of eigenvectors?
    It depends what the question asks for. If it says give two orthogonal eigenvectors then do that. Otherwise just give any eigenvector(s) that fulfil the requirements in the question. Since there are an infinite number of eigenvectors for any eigenvalue, the nature of the question will mean that there are many possible answers you could give.
 
 
 
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