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Let $T: R_3[x] \to R_3[x]$ linear transformation which is giving by $$[T]_B = \left(\begin{matrix} 1 & 2 & 3 \\ 1 & 0 & -1 \\ 0 & 1 & 2 \end{matrix} \right)$$ by the base $B = (1,1+x,1+x+x^2)$.

Q: Find the base and dimension of $\text{ker}T$ and $\text{Im}T$.

I don't really understand what is the correct way to find the base and dimension I was guided that I need to find the kernel by $[T]_B \underline x = \underline 0$ and to return by the base $B$, but I wasn't sure how to use it in a matrix to find the solutions.

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  • $\begingroup$ You can identify the polynomials in your basis $B$ with vectors in $\mathbb{R}^3$ via the mappings $x \mapsto (1,0,0), 1+x \mapsto (1,1,0)$ and $1+x+x^2 \mapsto (1,1,1)$. The columns of your matrix are now nothing else than the images of those vectors under your linear mapping, which can be seen by noteing that $[T]_B \cdot \vec{x} = (\vec{b}_1,\vec{b}_2,\vec{b}_3)\cdot \vec{x} = \vec{b}_1 x_1 + \vec{b}_2 x_2+\vec{b}_3 x_3$ for a general vector $\vec{x}$. All you are left with is determining a basis from the three column vectors of your matrix in order to determine the dimension of the image. $\endgroup$
    – gst
    Jun 12, 2015 at 22:48
  • $\begingroup$ For $\mathrm{ker} T$ you you determine the general solution for $[T]_B\cdot \vec{x} = \vec{0}$. The amount of "free parameters" of the solution gives you the dimension of your kernel, i.e. it determines how many basis vectors you need to span your kernel. $\endgroup$
    – gst
    Jun 12, 2015 at 22:53

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