Find an explicit isomorphism from [latex]P_{2} [/latex] onto the vector space of 2 × 2 symmetric matrices [latex]S_{2×2}[/latex].

To use the method given in the proof of Theorem 11, first let [latex] B_{1} = \left\{1, x, x²\right\}[/latex] and [latex]B_{2} = \left\{\begin{bmatrix}0 &0 \\ 0&1 \end{bmatrix}, \begin{bmatrix}0 &1 \\ 1&0 \end{bmatrix}, \begin{bmatrix}1 &0 \\ 0&0 \end{bmatrix} \right\} [/latex] be ordered bases for [latex] P_{2} and S_{2×2}[/latex], respectively. Let [latex] T_{1} and T_{2}[/latex] be the respective coordinate maps from [latex] P_{2} and S_{2×2}[/latex] into R³. Then [latex] T_{1}(ax² + bx + c) = \begin{bmatrix} c \\ b \\ a \end{bmatrix} and T_{2} \left(\begin{bmatrix} a &b\\ b&c\end{bmatrix} \right) = \begin{bmatrix} c \\ b \\ a \end{bmatrix} [/latex] Observe that [latex] T^{-1}_{2} : R³→ S_{2×2} [/latex] maps the vector [latex] \begin{bmatrix} c \\ b \\ a \end{bmatrix}[/latex]to the symmetric matrix [latex]\left(\begin{bmatrix} a &b\\ b&c\end{bmatrix} \right)[/latex] Thus, the desired isomorphism is given by [latex] \left(T^{-1}_{2} \circ T_{1} \right) : P_{2}→S_{2×2}[/latex] with [latex] \left(T^{-1}_{2} \circ T_{1} \right)(ax² + bx + c) = \begin{bmatrix} a &b\\ b&c\end{bmatrix}[/latex] For example, [latex] \left(T^{-1}_{2} \circ T_{1} \right) (x² − x + 2) = T^{-1}_{2} (T_{1} (x² − x + 2))= T^{-1}_{2}\left(\begin{bmatrix} 2\\ -1\\ 1 \end{bmatrix} \right) = \begin{bmatrix} 1& -1\\ -1&2 \end{bmatrix}[/latex]

Question 4.3.4: Find an explicit isomorphism from P2 onto the vector space o...

Introduction to Linear Algebra with Applications

Find an explicit isomorphism from $P_{2}$ onto the vector space of 2 × 2 symmetric matrices $S_{2×2}$ .

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To use the method given in the proof of Theorem 11, first let

$B_{1} = \left\{1, x, x²\right\}$ and $B_{2} = \left\{\begin{bmatrix}0 &0 \\ 0&1 \end{bmatrix}, \begin{bmatrix}0 &1 \\ 1&0 \end{bmatrix}, \begin{bmatrix}1 &0 \\ 0&0 \end{bmatrix} \right\}$

be ordered bases for $P_{2} and S_{2×2}$ , respectively. Let $T_{1} and T_{2}$ be the respective coordinate maps from $P_{2} and S_{2×2}$ into R³. Then

$T_{1}(ax² + bx + c) = \begin{bmatrix} c \\ b \\ a \end{bmatrix} and T_{2} \left(\begin{bmatrix} a &b\\ b&c\end{bmatrix} \right) = \begin{bmatrix} c \\ b \\ a \end{bmatrix}$

Observe that $T^{-1}_{2} : R³→ S_{2×2}$ maps the vector

$\begin{bmatrix} c \\ b \\ a \end{bmatrix}$ to the symmetric matrix $\left(\begin{bmatrix} a &b\\ b&c\end{bmatrix} \right)$

Thus, the desired isomorphism is given by $\left(T^{-1}_{2} \circ T_{1} \right) : P_{2}→S_{2×2}$ with

$\left(T^{-1}_{2} \circ T_{1} \right)(ax² + bx + c) = \begin{bmatrix} a &b\\ b&c\end{bmatrix}$

For example,

$\left(T^{-1}_{2} \circ T_{1} \right) (x² − x + 2) = T^{-1}_{2} (T_{1} (x² − x + 2))= T^{-1}_{2}\left(\begin{bmatrix} 2\\ -1\\ 1 \end{bmatrix} \right) = \begin{bmatrix} 1& -1\\ -1&2 \end{bmatrix}$