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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 $\{(0,1,1),(2,3,1)\}$.

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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.

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