# Finding the basis of an image of linear transformation

Let $$\vec{a}, \vec{b} \in \mathbb{R^3}$$. Let $$A : \mathbb{R^3} \rightarrow \mathbb{R^3}$$ be a linear transformation and $$A\vec{x} = \langle \vec{x}, \vec{a} \rangle \vec{b} + 2 \langle \vec{x},\vec{b} \rangle \vec{a}$$.

When $$\vec{a} = (1,1,0), \vec{b} = (0,1,1)$$, find a basis of an image of a linear transformation $$A$$.

My attempted solution:

Since $$\vec{a}$$ and $$\vec{b}$$ are linearly independent, a set $$\{\vec{a}, \vec{b}, \vec{a}\times \vec{b}\}$$ should be a basis for $$\mathbb{R^3}$$.

I've then tried calculating linear transformations of these vectors, and have gotten:

$$A\vec{a} = (2,4,2), A\vec{b} = (4,5,1), A(\vec{a} \times \vec{b}) = 0$$

Since $$A\vec{a}$$ and $$A\vec{b}$$ are linearly independent, i suppose the basis of an image should be $$\{A\vec{a}, A\vec{b}\}$$, because an image of a linear transformation is a a linear span of images of basis vectors.

I'm not sure whether this is currect because a given soluton for a basis is $$\{(0,1,1),(2,3,1)\}$$.

The given solution cannot possible be correct, because the range of $$A$$ consists of linear combinations of $$\vec a$$ and $$\vec b$$, and $$(2,3,1)$$ is not one of them.
Your answer is correct, but note that, since $$\vec a\times\vec b$$ is orthogonal to both $$\vec a$$ and $$\vec b$$, it follows from the definition of $$A$$ that $$A\left(\vec a\times\vec b\right)$$ is automatically equal to $$0$$; there was no need to actually compute it.
Finally, note that every real vector space other than $$\{0\}$$ has infinitely many bases. So, it is not amazing that your answer is different from the given one.