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$T$ is a linear transformation such that: $$T:R_3[x] \to R_3[x]$$ $$T(ax^2+bx+c) = (a+b+c)x^2+(2a-(k^2-6)b +2c)x+a+b-c$$

For which values of $k$, $T$ is invertible?

My attempt:

I thought to look at the representing matrix, call it $A$, and show that it is invertible, so the linear transformation is invertible (if $A$ is invertible, then the $\operatorname{ker}(T) = \{0\}$, so its injective, the rank of the matrix is full, so its surjective, namely, $T$ is bijective, so it exists $T^{-1}$ and $[T^{-1}] = A^{-1}$. ).

So I looked at the standard basis: $$ e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1) $$

\begin{align} T(e_3) &= T(1) = cx^2 + 2cx - c = x^2 + 2x - 1 \\ T(e_2) &= T(x) = bx^2 + (k^2 - 6)bx + b = x^2 + (k^2 - 6)x + 1 \\ T(e_1) &= T(x^2) = ax^2 + 2ax + a = x^2 + 2x + 1 \\ &\Downarrow \end{align}

$$[T]_E = \begin{bmatrix}1&1&1 \\ 2&(k^2-6)&2 \\ 1&1&-1\end{bmatrix} \xrightarrow{RREF} \begin{bmatrix}1&1&1 \\ 0&(k^2-8)&2 \\ 0&0&-2\end{bmatrix} $$ Therefore the answer is for all $k \in R, k \neq \pm 8.$

Is this correct?

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    $\begingroup$ Looks like there's a typo in your definition of T. And T(e2) seems wrong. $\endgroup$ Commented Jan 2, 2020 at 13:54
  • $\begingroup$ can you tell me why? $\endgroup$ Commented Jan 2, 2020 at 14:14
  • $\begingroup$ It looks like an x is missing in (2a-(k^2-6)b+2c), and if that is the case T(e2) should be x^2+(6-k^2)x+1. $\endgroup$ Commented Jan 2, 2020 at 14:16
  • $\begingroup$ I dont understand you $\endgroup$ Commented Jan 2, 2020 at 14:19

1 Answer 1

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There are some calculation mistakes. First note that $T(x) = x^2 + (6 - k^2) x + 1$, therefore, the corresponding matrix is

$$\begin{pmatrix} 1 & 1 & 1 \\ 2 & 6 - k^2 & 2 \\ 1 & 1 & -1 \end{pmatrix} \quad \sim\quad\begin{pmatrix} 1 & 1 & 1 \\ 0 & 4 - k^2 & 0 \\ 0 & 0 & -2 \end{pmatrix}.$$

This matrix is invertible if only if $k \neq \pm 2$. Therefore, $T$ is invertible for all $k \in \mathbb{R} \setminus \{ \pm 2 \}$.

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  • $\begingroup$ Ok now i see the mistake, thank you. $\endgroup$ Commented Jan 3, 2020 at 9:26

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