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Let

$$ A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\ -1 & 1 & 3 & -1 & 0\\ -2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix} $$

Find a $ 5 \times 5$ matrix $M$ with rank 2 such that $AM =0_{4\times5}$

My logic was to row reduce $A$ into a matrix $B$ such that:

$$ A = E_n \dots E_2 \cdot E_1 \cdot B $$

therefore $AM = 0$ becomes

$$ AM = (E_n \dots E_2 \cdot E_1 \cdot B) \cdot M = E_n \dots E_2 \cdot E_1 \cdot (B \cdot M) = 0$$

This allows me to find a matrix M that, multiplied left by B, will give me the zero matrix. I got that

$$ B = \begin{bmatrix} 1 & 0 & -1 & 0 &0\\ 0 & 1 & 2 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} $$

However, the only column I can think of that would give me a zero is in the form: $$ x = \begin{bmatrix} c\\ -2c\\ c\\ 0\\ 0\\ \end{bmatrix} $$

Setting that as one of the columns of M and filling the other entries with zeroes would only give me a matrix of rank 1.

Am I missing something or is my logic totally flawed?

Bryden C
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    I think you had some problem with the derivation of the RREF. – user Nov 06 '18 at 22:58
  • Actually I think that might be the problem. I did it by hand and put in into a calculator, not sure how that happened. Is everything else logically sound? – Bryden C Nov 06 '18 at 23:01
  • By hand I've found $rank=2$ and then we can construct the required matrix M using any two linearly independent vectors of $N(A)$. – user Nov 06 '18 at 23:03

1 Answers1

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We have that

$$A = \begin{bmatrix} 1 & 0 & -1 & 2 &1\\ -1 & 1 & 3 & -1 & 0\\ -2 & 1 & 4 & -1 & 3\\ 3 & -1 & -5 & 1 & -6\\ \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -1 & 2 &1\\ 0 & 1 & 2 & 1 & 1\\ 0 & 1 & 2 & 3 & 5\\ 0 & -1 & -2 & -5 & -9\\ \end{bmatrix}\to$$

$$\begin{bmatrix} 1 & 0 & -1 & 2 &1\\ 0 & 1 & 2 & 1 & 1\\ 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & -2 & -4\\ \end{bmatrix}\to \begin{bmatrix} 1 & 0 & -1 & 2 &1\\ 0 & 1 & 2 & 1 & 1\\ 0 & 0 & 0 & 2 & 4\\ 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$

therefore since the rank of the matrix is equal to $r=3$ the null space has dimension $\dim(N(A))= n-r=5-3=2\,$ and we can find $x_1$ and $x_2$ linearly independent such that

$$Ax_1=Ax_2=0$$

user
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