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# Matrix in Fp 3 watch

1. 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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2. (Original post by 123321123)

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...
3. (Original post by 123321123)

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

Posted from TSR Mobile
So you know the magnitude of is , so the unit vector is so that the magnitude becomes .

You have another vector whose magnitude is so the unit vector in this case is...?
4. (Original post by Zacken)
So you know the magnitude of is , so the unit vector is so that the magnitude becomes .

You have another vector whose magnitude is 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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5. (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?

Posted from TSR Mobile
If you have an eigenvector then any non-zero multiple of that vector is also an eigenvector. i.e: for any real non-zero is also an eigenvector.

So in your case, (2, 1, 2), (-2, -1, -2), (4, 2, 4), (-4, -2, -4), etc... are all eigenvectors.
6. (Original post by Zacken)
If you have an eigenvector then any non-zero multiple of that vector is also an eigenvector. i.e: for any real non-zero 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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7. (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.
8. (Original post by Zacken)
If you have an eigenvector then any non-zero multiple of that vector is also an eigenvector. i.e: for any real non-zero 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

Posted from TSR Mobile
9. (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
So you have an infinite number of eignenvectors, but there are only two eigenvectors with unit magnitude. You get me?
10. (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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