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Math Help - Determinants - Proving that |A|*|B|=|A*B|

  1. #1
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    Determinants - Proving that |A|*|B|=|A*B|

    How can I prove this:



    ?

    Thank you in advance.

    BTW - If you have sites dedicated for explanations of mathematical proofs, then I'll be glad to have a link to them. Thank you very much!
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  2. #2
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    the usual proof is long and cumbersome. the sophisticated one uses things you do not know. for reading good articles on the web plz see the blog "the unapologetic mathematician". i wish i knew this blog when i was starting maths.
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  3. #3
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    Well, I did found one proof, and it is long and sophisticated, and since I have my first math-test this Monday I just wished there would be an 'easier way', which is something that is many-times possible in mathematical proofs.

    I need to know the proofs of the lemmas and theorems in Linear Algebra 1, and the rest of them I was able to find and understand in books and google.

    BTW - Any tips for a first test? what to do if I don't seem to remember a proof by heart, and too stressful to figure it out at the moment?

    Thank you very much!
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  4. #4
    Super Member Matt Westwood's Avatar
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    Given it's your first maths test it is VERY doubtful whether you'll be asked to prove this.

    I tried to post up a proof of it on ProofWiki:

    Determinant of Matrix Product - ProofWiki

    but I had to abandon it because of lack of finding a decent way of getting the sign right without a lot of extra complication.

    Check out ProofWiki btw, it's growing all the time and may well be what you're looking for.
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    Well... Thank you very much, but where I study is a special program for teenagers in high-school, and they expect us to work very hard... so I prefer knowing as much as I can

    BTW - I also need the name of these theorems (or a link to a proof):

    If V,W are subspaces:
    dim(V+W)=dim(V)+dim(W)-dim(V∩W)

    If T is a linear transformation, then:
    dim(Im(T))+dim(Ker(T))=dim(V)

    (I don't study in English, so I need to figure out their names in order to be able to look up for them)

    Thank you
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  6. #6
    Super Member Matt Westwood's Avatar
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    Got this one:

    Dimension of Sum and Intersection of Vector Spaces - ProofWiki

    I don't think we've got the other one up on that site yet - our coverage of linear transformations is a bit limited yet. I think you're probably way ahead of me already anyway! Best of luck.
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  7. #7
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    Quote Originally Posted by adam63 View Post
    I also need the name of these theorems (or a link to a proof):

    If V,W are subspaces:
    dim(V+W)=dim(V)+dim(W)-dim(V∩W)

    If T is a linear transformation, then:
    dim(Im(T))+dim(Ker(T))=dim(V)
    I don't know of a name for the first of those theorems. The second one is often known as the "rank plus nullity theorem".
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  8. #8
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    Quote Originally Posted by adam63 View Post
    Well... Thank you very much, but where I study is a special program for teenagers in high-school, and they expect us to work very hard... so I prefer knowing as much as I can

    BTW - I also need the name of these theorems (or a link to a proof):

    If V,W are subspaces:
    dim(V+W)=dim(V)+dim(W)-dim(V∩W)

    If T is a linear transformation, then:
    dim(Im(T))+dim(Ker(T))=dim(V)

    (I don't study in English, so I need to figure out their names in order to be able to look up for them)

    Thank you
    Let V , W be vector spaces over a field F and T:V \to W a linear transformation. Then, dim(V) = dim (Im (T)) + dim (Ker (T))

    Proof:

    Let k = dim(ker (T)) , n = dim(V) and let (v_1, v_2, ... , v_k) be a base of Ker (T). We will complete it to (v_1, v_2, ... , v_k, v_{k+1}, ... , v_n) - a base of V. (we can do this since v_1,...,v_k are linearly independent)

    We want to show that dim(Im (T)) = n-k. If we prove that (v_{k+1},...,v_n) is a base of Im (T), we will be done. First, we will show that (v_{k+1},...,v_n) span Im (T), and then we will show that they are linearly independent, and thus a base to Im (T).

    (I) Let v \in V, then v = a_1v_1 + a_2v_2+...+a_kv_k+a_{k+1}v_{k+1}+...+a_nv_n where a_1,...,a_n are scalars from F. Then:
    T(v) = T(a_1v_1 + ... + a_kv_k + a_{k+1}v_{k+1} + ... + a_nv_n) and thus T(v) = a_1T(v_1) + a_2T(v_2) + ... + a_kT(v_k) + a_{k+1}T(v_{k+1}) + ... + a_nT(v_n), but v_1,v_2,...,v_k \in Ker(T) \Rightarrow T(v) = a_{k+1}T(v_{k+1}) + ... + a_nT(v_n) and this gives us that a_{k+1},...,a_n spans Im(T)

    (II) We will now show that a_{k+1},...,a_n are linearly independent. Assume \beta_{k+1}v_{k+1} + ... + \beta_nv_n = 0. We want to show that \beta_{k+1} = \beta_{k+2} = ... = \beta_n = 0

    The RHS of this equation is also equal to T(v_1 + v_2 + ... +v_k) = \gamma_1v_1 + \gamma_2v_2 + ... + \gamma_kv_k since v_1,...,v_k \in Ker(T). Substitute these expressions and we get:

    \beta_{k+1}v_{k+1} + ... + \beta_nv_n = \gamma_1v_1 + ... + \gamma_kv_k This gives us:

    -\gamma_1v_1 - ... - \gamma_kv_k + \beta_{k+1}v_{k+1} + ... + \beta_nv_n = 0 but we know (v_1,...,v_n) is a base to V and thus linearly independent, so we get:

    \gamma_1 = \gamma_2 = ... = \gamma_k = \beta_{k+1} = \beta_{k+2} = ... = \beta_n = 0 and so (v_{k+1},...,v_n) is a base of Im(T) and thus dim (Im(T)) = dim(V) - dim(Ker(T)) as required.
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  9. #9
    Super Member Matt Westwood's Avatar
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    Ah - looks like we do have this in there after all:

    Sum of Nullity and Rank of Linear Transformation - ProofWiki
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  10. #10
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    Thank you all, very much!! I really appreciate your good will to help!
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