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Math Help - Problem regarding rank

  1. #1
    Senior Member Pinkk's Avatar
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    Problem regarding rank

    Are there values for r and s for which:

    \left(\begin{array}{ccc}1&0&0\\0&r-2&2\\0&s-1&r+2\\0&0&3\end{array}\right)

    has rank 1 or 2? And if so, find those values.

    Well, it's obvious that if r=2 and s=1. It will have rank 2, and I can see that it can never have rank 1, but I am not sure how to exactly show this. I know in order to have rank 1, each row (or column) must be a scalar of every other column. So, would it be fair to say that aC_{1}+bC_{2}+cC_{3}+dC_{4}=(0,0,0,0) for some non-trivial solution of a,b,c,din order for that to be true. I think using the rows won't work in this situation since the rows can be thought of as vectors in R^{3} and a set of 4 vectors in R^{3} must be linearly dependent. I'm guessing if I can show that the a linear combination of the three column vectors can only equal a column of 0's with only the trivial solution, than that eventually leads to the fact that it is never possible to have any given row (or column) represented as a scalar multiple of all of the other rows or columns.

    Any help would be appreciated.
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  2. #2
    Senior Member Pinkk's Avatar
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    I just thought of an easier way, but again, I don't know if it suffices. Since 1 \ne a0 for any a \in R, none of the rows or columns can all be scalar multiples of one another.
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  3. #3
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    You've got the right idea. Here is a fleshed out version of what you're getting at.

    Let r and s be any real numbers. For the matrix to have rank 1, the three columns vectors must span a 1-dimensional subspace. In this case, any non-zero vector in that subspace would serve as a spanning vector, namely (1,0,0,0) (the first column vector). But obviously the third column vector, no matter what r and s you choose, cannot be a scalar multiple of the first, by what you just said. And that's it.
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  4. #4
    Senior Member Pinkk's Avatar
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    Thanks for clarifying.
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