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Finding out if non-invertible matrices are subspaces of the vector space Mn(R)

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    I have Mn(R) is a vector space of all nxn matrices with n>=2.

    Now I need to determine if the set A in Mn(R) is a subspace of Mn(R) where A is the set of all non-invertible matrices.

    Now, I know that in vector fields, to prove that a set is a subspace of a vector field we need to show that the 3 axioms hold:

    (S1) A is not equal to the empty set.

    (S2) A is closed under addition.

    (S3) A is closed under multiplication.

    I also know that a square matrix is non-invertible if its determinant = 0.

    I'm just not really sure how to go about finding out if this set A is a subspace of Mn(R).

    Any hints?
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    As always, the first step is to decide whether or not it's true (then you have to either prove it or find a counterexample).

    If you're not sure, one thing to think about: Why might it be true? If you can't think of any reason for it to be true, maybe it isn't. If you can find a reason for it to be true, that will give you a hint towards how you might prove it.
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    Ok, I've decided that the non-invertible matrices A are NOT a subspace of Mn(R) because it is not closed under addition:

    If you take a non-invertible matrix and add to it another non-invertible matrix, the result is an invertible matrix.

    What's the best way to show this?
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    Would an explicit example suffice? And if so, how do I go about finding two non-invertible matrices?
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    Yes, all you need is one counter-example to show it's not closed under addition.

    To find a non-invertible matrix, think of what other properties it has, as compared to an invertible matrix.
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    Hello again ghostwalker. Properties non-invertible matrices have... their determinants are equal to 0?
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    (Original post by Sun Tower)
    Hello again ghostwalker. Properties non-invertible matrices have... their determinants are equal to 0?
    Yep - should be easy to find a couple and experiment.
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    Ok, so after trying a few it appears that adding two non-invertible matrices DOES give you a non-invertible matrix. Is this always true?
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    (Original post by Sun Tower)
    Ok, so after trying a few it appears that adding two non-invertible matrices DOES give you a non-invertible matrix. Is this always true?
    That's what you have to find out.

    Choose the most basic invertible matrix you can imagine, now try and split it up into the sum of two non-invertible matrices.

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Updated: March 9, 2009
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