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Math Help - matrix and vector problem

  1. #1
    Senior Member Sambit's Avatar
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    Question matrix and vector problem

    Let A be a nx n matrix such that |A|=0. Identify whether there exists any non-null vector \tilde{x} such that A\tilde{x}=\tilde{0}

    All I can make out is if we write A=(\tilde{a_1},\tilde{a_2},.....,\tilde{a_n}), we have x_1\tilde{a_1}+x_2\tilde{a_2}+...+x_n\tilde{a_n}=0. Then I find nowhere to proceed. How to use the fact that |A|=0?
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  2. #2
    A Plied Mathematician
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    How about using the following type of argument: a matrix with zero determinant can have elementary row operations performed upon it such that the solution space is unchanged, and the matrix is diagonal with one's on the main diagonal except that the last row (at least) is all zeros. That is, the matrix A is similar to this:

    \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
0 &0 &0 &\dots &0<br />
\end{bmatrix}

    Does this suggest anything to you?
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  3. #3
    Senior Member Sambit's Avatar
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    This implies \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
0 &0 &0 &\dots &0<br />
\end{bmatrix}\begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
\x_n\\<br />
\vdots\\<br />
\end{bmatrix}=\tilde{0}
    or, \begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_{n-p}\\<br />
0\\<br />
0\\<br />
\vdots\\<br />
0\\<br />
\end{bmatrix}=\tilde{0}which means \tilde{x} is a null vector.
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  4. #4
    Senior Member Sambit's Avatar
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    This implies \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
\end{bmatrix}\begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
\vdots\\<br />
x_n\\<br />
\end{bmatrix}=\tilde{0}
    or, \begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_{n-p}\\<br />
0\\<br />
0\\<br />
\vdots\\<br />
0\\<br />
\end{bmatrix}=\tilde{0}which means \tilde{x} is a null vector.
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  5. #5
    Senior Member Sambit's Avatar
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    This implies \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
\end{bmatrix}\begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_n\\<br />
\end{bmatrix}=\tilde{0}
    or, \begin{bmatrix}<br /> <br /> <br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_{n-p}\\<br />
0\\<br />
0\\<br />
\vdots\\<br />
0\\<br />
\end{bmatrix}=\tilde{0}which means \tilde{x} is a null vector.


    SORRY, I posted it thrice. Problem with my P.C. Hope I won't be penalized
    Last edited by Sambit; January 1st 2011 at 08:25 AM. Reason: to apologize
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  6. #6
    A Plied Mathematician
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    Well, I wouldn't put the zeros in the last part of the vector, because when you do the matrix multiplication, you don't get zeros everywhere. How about zeros in the first part of the vector? Then the zero rows of the matrix will kill the nonzero last part of the vector.
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  7. #7
    Senior Member Sambit's Avatar
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    what I mean to say is, according to you, the last few columns of A are zeros. So when A is multiplied by \begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_n\\<br />
\end{bmatrix}, a matrix is obtained whose last few elements are zeros. What is the mistake here?
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  8. #8
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    Quote Originally Posted by Sambit View Post
    This implies \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
\end{bmatrix}\begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_n\\<br />
\end{bmatrix}=\tilde{0}
    NO, it doesn't. That is what you want to prove

    or, \begin{bmatrix}<br /> <br /> <br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_{n-p}\\<br />
0\\<br />
0\\<br />
\vdots\\<br />
0\\<br />
\end{bmatrix}=\tilde{0}which means \tilde{x} is a null vector.
    Again, no. On the left you have the result of your multiplication, on the right what you want it to equal. Obviously you must have x_1= x_2= \cdot\cdot\cdot= x_{n- p}= 0 but what about all of the x_{n-p+1}, x_{n-p+2}, \cdot\cdot\cdot, x_n?


    SORRY, I posted it thrice. Problem with my P.C. Hope I won't be penalized
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  9. #9
    Senior Member Sambit's Avatar
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    From the equation A\tilde{x}=\tilde{0} we need to solve for \tilde{x} and check whether it can be non-null or not. So we consider the equation, so why is the RHS not supposed to be present?

    We don't want to prove  \begin{bmatrix}<br />
1 &0 &0 &\dots &0\\<br />
0 &1 &0 &\dots &0\\<br />
0 &0 &1 &\dots &0\\<br />
\vdots &\vdots &\vdots &\ddots &\vdots\\<br />
\end{bmatrix}\begin{bmatrix}<br />
x_1\\<br />
x_2\\<br />
\vdots\\<br />
x_n\\<br />
\end{bmatrix}=\tilde{0}. Rather, given the equation, we want to check whether \tilde{x} can be non-null. Isn't it?
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