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Define T : $P_3(\mathbb{R}) → M_2(\mathbb{R})$ by $T(f)$ = $\begin{bmatrix}f'(0) & 2f(1)\\0 & f''(2)\end{bmatrix}$

a) Show that T is a linear transformation.

b) Find bases for Ker(T) and range(T)

I'm so confused with the matrix. What should be my approach?

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    $\begingroup$ Your approach would be the same as always, don't "mind" the matrix. Verify, for part (a), whether $T(\alpha f + \beta g) = \alpha \, T( f )+ \beta \, T( g)$. Can you start? $\endgroup$
    – StackTD
    Jun 6 '17 at 21:03
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If T is a linear transformation then for every pair polynomials $p,q \in\mathbb P_3 ; T(p + q) = T(p) + T(q)$

b) $Ker(T):$ what is the set of $p$ such that $T(p) = \mathbb 0$

$p(x) = ax^3 +bx^2 + cx + d$

Find $a,b,c,d$ such that

$p(1) = 0\\ p'(0) = 0\\ p''(2) = 0$

The range, are there any matrices that cannot be formed? you must have a $0$ in the bottom left hand corner. Any other restrictions?

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  • $\begingroup$ so that would be $T(f+g) = \begin{bmatrix}f'(0) & 2f(1)\\0 & f''(2)\end{bmatrix} + \begin{bmatrix}g'(0) & 2g(1)\\0 & g''(2)\end{bmatrix} = \begin{bmatrix}f'(0)+g'(0) & 2f(1) + 2g(1)\\0 & f''(2)+g''(2)\end{bmatrix} = T(f)+T(g)$ and $T(rf) = \begin{bmatrix}rf'(0) & 2rf(1)\\0 & rf''(2)\end{bmatrix} = r(T(f))$ but how about the range and kernel? $\endgroup$
    – Antt
    Jun 6 '17 at 21:09

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