I've been confused with the following matrix. I'm trying to find the basis of the image of A:
$$A= \left( \begin{array}{ccc} 6&4&10\\ 4&-1&3\\ -2&-5&-7\\ -10&-3&-13 \end{array}\right)$$
Normally for these types of problems I'm used to row reducing and not having any issues, however for this problem since it is a 4x3 matrix I'm honestly not sure what to do. I got my matrix to be reduced to the following:
$$A= \left( \begin{array}{ccc} 1&0&1\\ 0&1&1\\ 0&0&0\\ 0&0&0 \end{array}\right)$$
I thought the answer was going to be the following however it is incorrect:
$$\left( \begin{array}{ccc} -1\\ -1\\ 1\\ 0\end{array}\right)$$
What am I doing wrong? I figure its something simple to fix.
$\endgroup$2 Answers
$\begingroup$Let denote by $c_i$ the $i$-th column of $A$. Notice that $c_3=c_1+c_2$ and that $c_1$ and $c_2$ are not colinear so they are linearly independent so $(c_1,c_2)$ is a basis for the image of $A$.
$\endgroup$ $\begingroup$The vector you gave is a basis for the kernel (or nullspace) of $A$, not the image (or column space) of $A$.
Hints: The image of $A$ (or more properly the image of the linear transformation $x \mapsto Ax$) is the column space of $A$.
In the Gaussian elimination algorithm, the columns of the original matrix $A$ that have pivots in them are a basis for the column space of $A$.
Also, you should not write that $A =$ the reduced row echelon form. The matrix $A$ is the one given in the problem; the reduced row echelon form matrix is a different matrix (many people call it $R$) that is obtained from the Gaussian elimination algorithm.
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