Originally Posted by

**arbolis** This is an exercise I had on the Linear Algebra final exam I failed. I could only do a part and I'm still stuck for most of it. So I'll write it entirely and I'll ask help only for part 1) until we finish it, then ask for further help.

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Let $\displaystyle \bold B = \{ \alpha_1, \alpha_2, \alpha_3 \}$ be the ordered basis of $\displaystyle \mathbb{R}^3$ where $\displaystyle \alpha_1=(1,0,1)$, $\displaystyle \alpha_2=(0,1,1)$, $\displaystyle \alpha_3=(0,-1,1)$.

Let $\displaystyle T : \mathbb{R}^3 \to \mathbb{R}^4$ be defined by $\displaystyle T(\alpha_1)=(1,-2,1,0)$, $\displaystyle T(\alpha_2)=(2,1,0,-1)$, $\displaystyle T(\alpha_3)=(1,0,1,0)$.

1)Give an explicit description of $\displaystyle \ker T$, calculate its dimension and give a basis.

2)Give an explicit description of $\displaystyle \Im T$, calculate its dimension and give a basis.

3)Calculate $\displaystyle [T]_{\bold B}^{\bold B'}$, where $\displaystyle \bold B' =\{ (1,1,0,0), (0,0,1,1), (1,0,0,4), (0,0,0,1) \}$.

4)If $\displaystyle \alpha=(-3,-1,3)$, find the coordinates of $\displaystyle T(\alpha)$ with respect to $\displaystyle \bold B'$.

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1) My attempt : only thoughts. I must find all the vectors $\displaystyle v \in \mathbb{R}^3$ such that $\displaystyle T(v)=0$. I'm at a loss. I don't even know if the transformation is linear. I've absolutely no idea about how to find the kernel of it.