Show that the maps \(\mathbf{A} \rightarrow \omega_{A}^{1}\) and \(\mathrm{A} \rightarrow \omega_{A}^{2}\) establish isomorphisms of the lineat spacx \(\mathbb{R}^{3}\) of vectors \(A\) with the linear spaces of 1 -forms on \(\mathbb{R}^{3}\) and 2 -forms on \(\mathbb{R}^{3}\). If we choose an orthonormal oriented coordinate system \(\left(x_{1}, x_{2}, x_{3}\right)\) on \(\mathbb{R}^{3}\). then $$ \omega_{A}^{1}=A_{1} x_{1}+A_{2} x_{2}+A_{3} x_{3} $$

Recall the given map for 1-forms: $$ \omega_A^1 = A_1 x_1 + A_2 x_2 + A_3 x_3 $$

To establish isomorphism, we need to show that the 1-form components are linearly independent. Consider the following: $$ c_1\omega_{A_1}^1 + c_2\omega_{A_2}^1 + c_3\omega_{A_3}^1 = 0 $$ where \(c_1, c_2, c_3\) are constants. By substituting the expressions for each \(\omega_A^1\), we get: $$ c_1(A_{11}x_1 + A_{12}x_2 + A_{13}x_3) + c_2(A_{21}x_1 + A_{22}x_2 + A_{23}x_3) + c_3(A_{31}x_1 + A_{32}x_2 + A_{33}x_3) = 0 $$ The only way for this equation to hold true for all \(x_1, x_2, x_3\) is if the coefficients of \(x_1, x_2, x_3\) are all zero: \(c_1A_{11} + c_2A_{21} + c_3A_{31} = 0\) \(c_1A_{12} + c_2A_{22} + c_3A_{32} = 0\) \(c_1A_{13} + c_2A_{23} + c_3A_{33} = 0\) Since the matrix A is non-singular, we can conclude that \(c_1 = c_2 = c_3 = 0\). Thus, the 1-form components are linearly independent.

To further establish isomorphism, we must show that the linear spaces of vectors A and 1-forms have the same dimension. Since there are three independent 1-form components and ℝ³ has dimension 3, the linear spaces have the same dimension.

Recall the given map for 2-forms: $$ \omega_A^2 = A_1x_2x_3 + A_2x_3x_1 + A_3x_1x_2 $$

As before, to establish isomorphism, we need to show that the 2-form components are linearly independent. Consider the following: $$ c_1\omega_{A_1}^2 + c_2\omega_{A_2}^2 + c_3\omega_{A_3}^2 = 0 $$ where \(c_1, c_2, c_3\) are constants. By substituting the expressions for each \(\omega_A^2\): $$ c_1(A_{11}x_2x_3 + A_{12}x_3x_1 + A_{13}x_1x_2) + c_2(A_{21}x_2x_3 + A_{22}x_3x_1 + A_{23}x_1x_2) + c_3(A_{31}x_2x_3 + A_{32}x_3x_1 + A_{33}x_1x_2) = 0 $$ The only way for this equation to hold true for all \(x_1, x_2, x_3\) is if the coefficients of \(x_2x_3, x_3x_1, x_1x_2\) are all zero: \(c_1A_{11} + c_2A_{21} + c_3A_{31} = 0\) \(c_1A_{12} + c_2A_{22} + c_3A_{32} = 0\) \(c_1A_{13} + c_2A_{23} + c_3A_{33} = 0\) Since the matrix A is non-singular, we can conclude that \(c_1 = c_2 = c_3 = 0\). Thus, the 2-form components are linearly independent.

As before, we must show that the dimensions of the spaces of vectors A and 2-forms are the same. Since there are three independent 2-form components and ℝ³ has dimension 3, the linear spaces have the same dimension.

The maps \(\mathbf{A} \rightarrow \omega_A^1\) and \(\mathbf{A} \rightarrow \omega_A^2\) establish isomorphisms for both the 1-forms and 2-forms on ℝ³, as the components are linearly independent and the dimensions of the linear spaces are the same.