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    Hi can any one tell me why the unit vector of this question can be both positive and negative? I have got the magnitude of the vector am which is 3 and magnitude is a scalar, so can't be -3. So what makes it possible to be negative. Thanks




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    (Original post by 123321123)
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    Hi can any one tell me why the unit vector of this question can be both positive and negative? I have got the magnitude of the vector am which is 3 and magnitude is a scalar, so can't be -3. So what makes it possible to be negative. Thanks
    Uh... the magnitude of both unit vectors should be 1, that's the whole point of a unit vector...
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    (Original post by 123321123)
    Name:  ImageUploadedByStudent Room1457431700.314694.jpg
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    Hi can any one tell me why the unit vector of this question can be both positive and negative? I have got the magnitude of the vector am which is 3 and magnitude is a scalar, so can't be -3. So what makes it possible to be negative. Thanks




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    So you know the magnitude of \begin{pmatrix} 2 \\ 1 \\ 2\end{pmatrix} is \sqrt{2^2 + 1^2  + 2^2 } = 3, so the unit vector is \frac{1}{3} \begin{pmatrix} 2 \\1 \\ 2 \end{pmatrix} so that the magnitude becomes 1.

    You have another vector \begin{pmatrix} -2 \\-1 \\- 2\end{pmatrix} whose magnitude is \sqrt{(-2)^2 +(-1)^2 +(-2)^2}= \sqrt{9} =3 \neq -3 so the unit vector in this case is...?
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    (Original post by Zacken)
    So you know the magnitude of \begin{pmatrix} 2 \\ 1 \\ 2\end{pmatrix} is \sqrt{2^2 + 1^2  + 2^2 } = 3, so the unit vector is \frac{1}{3} \begin{pmatrix} 2 \\1 \\ 2 \end{pmatrix} so that the magnitude becomes 1.

    You have another vector \begin{pmatrix} -2 \\-1 \\- 2\end{pmatrix} whose magnitude is \sqrt{(-2)^2 +(-1)^2 +(-2)^2}= \sqrt{9} =3 \neq -3 so the unit vector in this case is...?
    Hi thanks for ur help. I understand the how to find unit vector, but not sure why the original eigenvector can be both (2,1, 2) and (-2,-1, -2) where does the negative sign come from?


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    (Original post by 123321123)
    Hi thanks for ur help. I understand the how to find unit vector, but not sure why the original eigenvector can be both (2,1, 2) and (-2,-1, -2) where does the negative sign come from?


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    If you have an eigenvector \begin{pmatrix}x \\ y \\ z\end{pmatrix} then any non-zero multiple of that vector is also an eigenvector. i.e: \lambda \begin{pmatrix} x \\y \\ z  \end{pmatrix} for any real non-zero \lambda is also an eigenvector.

    So in your case, (2, 1, 2), (-2, -1, -2), (4, 2, 4), (-4, -2, -4), etc... are all eigenvectors.
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    (Original post by Zacken)
    If you have an eigenvector \begin{pmatrix}x \\ y \\ z\end{pmatrix} then any non-zero multiple of that vector is also an eigenvector. i.e: \lambda \begin{pmatrix} x \\y \\ z  \end{pmatrix} for any real non-zero \lambda is also an eigenvector.

    So in your case, (2, 1, 2), (-2, -1, -2), (4, 2, 4), (-4, -2, -4), etc... are all eigenvectors.
    Hi
    Do you mean that the answer could be any thing as long as the lambda is non-zero. Thanks .


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    (Original post by 123321123)
    Hi
    Do you mean that the answer could be any thing as long as the lambda is non-zero. Thanks .


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    Yes. But there are only two unit vectors.
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    (Original post by Zacken)
    If you have an eigenvector \begin{pmatrix}x \\ y \\ z\end{pmatrix} then any non-zero multiple of that vector is also an eigenvector. i.e: \lambda \begin{pmatrix} x \\y \\ z  \end{pmatrix} for any real non-zero \lambda is also an eigenvector.

    So in your case, (2, 1, 2), (-2, -1, -2), (4, 2, 4), (-4, -2, -4), etc... are all eigenvectors.
    I still confused with the value of lambda. As I calculated in part a) lambda is 4, which is a positive number, so why the answer comes out two answer and one of them is negative. Sorry for my poor understanding skill, could u explain more specific? Thanks


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    (Original post by 123321123)
    I still confused with the value of lambda. As I calculated in part a) lambda is 4, which is a positive number, so why the answer comes out two answer and one of them is negative. Sorry for my poor understanding skill, could u explain more specific? Thanks


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    So you have an infinite number of eignenvectors, but there are only two eigenvectors with unit magnitude. You get me?
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    (Original post by 123321123)
    I still confused with the value of lambda. As I calculated in part a) lambda is 4, which is a positive number, so why the answer comes out two answer and one of them is negative. Sorry for my poor understanding skill, could u explain more specific? Thanks


    Posted from TSR Mobile
    Each eigenvalue corresponds to an infinite set of eigenvectors. These form a subspace, most commonly a line. In this case it's a line, so there are only two vectors in this set that are unit vectors(have magnitude of 1).
 
 
 
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