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Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation, given $B=((1,1,1),(1,1,0),(1,0,1))$ a base of $\mathbb{R}^3$. Suppose $(1,0,0) \in\ker T$.

$$[T]_{B} = \left(\begin{array}{ccc} 1 & 0 & 1\\ 3 & 2 & 1\\ 2 & 1 & 1 \end{array}\right).$$

Find a base for $\operatorname{Im}T$.

I know that the columns of $[T]_B$ are the coordinates vectors of the vectors which span $\operatorname{Im}T$, but I don't understand how to find a base for $\operatorname{Im}T$ using that information (and of course, using the fact that dim $\operatorname{Im}T=2$).

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  • $\begingroup$ Find the biggest familly of all linear independent vector column. Notice that $\dim \ker(T)\geq 1$ give you the information that $\dim Im(T)$ is at most 2. $\endgroup$ Commented Feb 18, 2020 at 9:50

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Since $\ker T\neq\{0\}$, $\dim\operatorname{Im}Y\leqslant2$. On the other hand, you know that $(1,1,1)+3(1,1,0)+2(1,0,1)\bigl(=(6,4,3)\bigr)$ and $2(1,1,0)+(1,0,1)\bigl(=(3,2,1)\bigr)$ belong to $\operatorname{Im}Y$. Since they are linearly independent, they form a basis of $\operatorname{Im}Y$.

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    $\begingroup$ I donwnvoted. For someone with your experience, it's a bit pity to give brute force the answer. We would expect either a comment or a Hint... $\endgroup$ Commented Feb 18, 2020 at 9:53
  • $\begingroup$ So, you are downvoting an answer that you find correct to a question that you find acceptable. $\endgroup$ Commented Feb 18, 2020 at 10:53

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