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Math Help - Re: Linear Algebra-Transition Matrices

  1. #1
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    Re: Linear Algebra-Transition Matrices

    how to post the problem? Please, guide me
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  2. #2
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    Re: Linear Algebra-Transition Matrices

    Quote Originally Posted by Loser66 View Post
    how to post the problem? Please, guide me
    Just the same way as you posted this. If you are trying to post a picture or an equation we can help you there. Just say the word.

    -Dan
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  3. #3
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    Re: Linear Algebra-Transition Matrices

    Let A in L(U,V), B in L(V,W)
    a) Prove that rank (A) <= min{dim(U) , dim(V)}
    b) Prove that rank (BA) <= min{rank (A), rank(B)}
    Please, help
    It should be in the new post, but I don't know how to do.
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  4. #4
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    Re: Linear Algebra-Transition Matrices

    Quote Originally Posted by Loser66 View Post
    Let A in L(U,V), B in L(V,W)
    a) Prove that rank (A) <= min{dim(U) , dim(V)}
    Let $(e_1,\dots,e_n)$ be a basis of $U$. Then $\text{Im}(A)=\text{span}(Ae_1,\dots,Ae_n)$ (why?), so $\text{rank}(A)=\dim(\text{Im}(A))\le n$. I'll leave it to you to show that $\text{rank}(A)\le\dim(V)$.

    Quote Originally Posted by Loser66 View Post
    b) Prove that rank (BA) <= min{rank (A), rank(B)}
    I assume $A$ is applied first in $BA$. Obviously, $\text{Im}(BA)\subseteq\text{Im}(B)$, so
    \[
    \text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(B))=\text{rank}(B).
    \]
    In addition, $\text{Im}(BA)=B(\text{Im}(A))$, so using statement a),
    \[
    \text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(A))=\text{rank}(A).
    \]
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  5. #5
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    Re: Linear Algebra-Transition Matrices

    Thanks for reply. My work is Problem 1.pdf
    at problem1, if you have time, please, check the next one and guide me more if I am not right. Thanks in advance. This is my practice review for final. I desperate need help.
    Last edited by Loser66; May 5th 2014 at 03:06 AM.
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  6. #6
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    Re: Linear Algebra-Transition Matrices

    I think problem 1 (a) and (b) are largely correct. If I could make a general advice, it would be to not hesitate to use more English text in complete sentences. It would make the role of the grader, or any reader, much easier.

    I assume Col(A) means the column space of A. But in the beginning of problem 1, A and B are defined as linear operators, not matrices. You need to say something about a basis.

    Let x ∈ R^p $\to$ x ∈ Col BA, then BA(x) = B(Ax) ∈ Col B
    I believe you mean, "If $x\in\text{Col}(BA)$, then $x=BAy$ for some $y$, so $x=B(Ay)$, which means $y\in\text{Col}(B)$".

    Note the the solution in post #4 does not use the rank-nullity theorem.

    Problem 6.
    Let a =(a1,a2,a3) and v = (v1,v2,v3) then, a x v = a2v3 a3v2 a1v3+a3v1 +a1v2-a2v1
    Is $a$ a fixed vector? This should be said explicitly. The right-hand side of the last equality is a number while it should be a vector, i.e., three numbers. The matrix of A is not Av, and it should not depend on v1, v2, v3 (the first matrix). I am not sure how you determine the matrix of $A$. In my book, its columns are images of the basis vectors. For example, (1, 0, 0) is mapped to $(0, a_3,-a_2)$, so that would be the first column. Overall, the matrix of $A$ is
    \[
    \begin{pmatrix}
    0&-a_3&a_2\\
    a_3&0&-a_1\\
    -a_2&a_1&0
    \end{pmatrix}
    \]
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