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Given the linear transformation:

$$T_3 : \mathbb{C}^3 \to \mathbb{C}^3 , (x_1, x_2, x_3) \mapsto (x_1+x_2, x_1+x_3, x_1 - x_2+2x_3)$$

I need to find the image and kernel, and basis for both of them. I know that the kernel is the set of vectors for which the linear transformation will map to zero, and this can be found by setting up a system of equations like such:

$$x_1 + x_2 = x_1 + x_3 = x_1 - x_2 +2x_3 = 0$$

But once here, how would I format my answer for $ker(T_3)$, and how would I then find a basis for that? Secondly I understand the image of a linear transformation to be the span of the target vectors from the linear transformation, so would the $im(T_3)$ just be $\mathbb{C}^3$ ?

Thank you!

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The only solution of that system is $(0,0,0)$. So, $\ker(T_3)=\{(0,0,0)\}$, the only basis of which is $\emptyset$. And, by the rank-nullity theorem $\dim\operatorname{Im}(T_3)=3$ and therefore $\operatorname{Im}(T_3)=\mathbb C^3$.

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