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  1. MHB Craftsman

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    #1
    Hello

    I have a question in which I need to choose the wrong statement. I have 5 statements, and I managed to rule out 3 options so I am left with two.

    the options are:

    1. The dimension of the 3X3 anti-symmetric matrices subspace is 3.

    2. An nXn matrix which has different numbers on it's main diagonal, is diagonalized.

    3. A non invertible matrix has an eigenvalue of 0

    4. Every matrix has a unique canonical form matrix

    5. v1 and v2 are vectors from a vector space V. Then v1-2v2 also belongs to V.

    I managed to rule out 1, 3 and 4 (they are correct in my opinion). I don't know which one is not, is it 2 or 5 ?

  2. MHB Craftsman

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    #2 Thread Author
    I think I can rule out 5 too, but that doesn't help me understand why 2 is correct.

    am I right to say:

    let's assume that V is the space of all vectors of form (1,a,b), then:

    v1 = (1,a,b) v2 = (1,c,d)

    v1-2v2 = (-1,2-ac,b-2d) which doesn't belong to V ?

  3. MHB Seeker
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    #3
    Hi Yankel!

    If 2 vectors belong to a vector space, than any linear combination of those vectors also belongs to that vector space by definition.
    Since $v_1-2v_2$ is a linear combination, (5) is correct.

    As for (2), which values does a diagonal matrix have that are not on its main diagonal?

  4. MHB Journeyman
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    #4
    Quote Originally Posted by Yankel View Post
    2. An nXn matrix which has different numbers on it's main diagonal, is diagonalized.
    This is false. Choose for example $A=\begin{bmatrix}{1}&{-2}\\{1}&{-1}\end{bmatrix}\in \mathbb{R}^{2\times 2}$. Its eigenvalues are $\lambda=\pm i\not\in \mathbb{R}$, so $A$ is not diagonalizable on $\mathbb{R}$.

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    #5
    Quote Originally Posted by Yankel View Post
    I think I can rule out 5 too, but that doesn't help me understand why 2 is correct.

    am I right to say:

    let's assume that V is the space of all vectors of form (1,a,b), then:

    v1 = (1,a,b) v2 = (1,c,d)

    v1-2v2 = (-1,2-ac,b-2d) which doesn't belong to V ?
    As ILikeSerena told you, this is true. Your mistake is that the set $\{(1,a,b):a,b\in\mathbb{R}\}$ is not a vector space.

  6. MHB Craftsman

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    #6 Thread Author
    thank you both !

    Yes, silly example, my set wasn't a vector space since it's not close on addition

    (1+1!=1)


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