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PROBLEM: Use as many columns of $A$ as possible to construct a matrix $B$ with the property that the equation $B x = 0$ has only the trivial solution.

Matrix A

Thoughts: Since I am doing this problem using Mathematica, I can pick and choose certain columns that will give me a matrix that reduces to Reduced Echelon Form. I chose columns $X_1$, $X_2$, $X_4$, and $X_5$. This new matrix $B$ row reduces completely, such that there is a pivot in each column.

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Do I have the right idea? Is this what the question asked me to do?

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  • $\begingroup$ Yes, you are correct. That is exactly what they want. $\endgroup$
    – jgon
    Jan 22 '18 at 5:48
  • $\begingroup$ What are the sizes of the matrix B and x?If you form the matrix B with four columns of A such that B is invertible then the system Bx=0 will always have trivial solution. $\endgroup$
    – Dastan
    Jan 22 '18 at 6:28
  • $\begingroup$ Based on what you did you've already shown how to do it with 4 columns. What remains to be argued is why it cannot work with 5 columns. Though that is almost trivial. $\endgroup$
    – Florian
    Jan 22 '18 at 13:54
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What you did is correct. You are searching for a matrix with kernel/null space equal to $\{0\}$. Here is the idea behind your answer.

As this cannot hold for five columns, because this means there will be a linearly dependent column, you must remove a linearly dependent column.

Now you found a matrix that has reduced echelon form $I$, which means it is invertible, thus bijective, so only $x=0$ is the solution to $Bx=0$.

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