Reduce the skew-symmetric bilinear function to canonical form and find the matrix of the transformation.

$$ varphi (x, y) = x_1y_2-x_2y_1 + 2x_1y_3-2x_3y_1-x_1y_4 + x_4y_1 + 4x_2y_4-4x_4y_2 + x_3y_4-x_4y_3 $$

**My approach:** To let $ {e_1, e_2, e_3, e_4 } $ is the given basis for the vector space and the matrix of this form is $$ B _ { varphi} ^ {(e)} = begin {bmatrix}

0 & 1 & 2 & -1 \

-1 & 0 & 0 & 4 \

-2 & 0 & 0 & 1 \

1 & -4 & -1 & 0 \

end {bmatrix} $$

Let's find the new base in the form $ e & # 39; _1 = e_1, e & # 39; _2 = e_2 $ and $ e & # 39; _i = e_i + dfrac {b_ {2i}} {b_ {12}} e_1- dfrac {b_ {1i}} {b_ {12}} e_1 $ to the $ i geq 3 $, Where $ b_ {ij} $ are elements of the matrix above (we find the new basis so that $ varphi (e & # 39; _1, e & # 39; _i) = varphi (e & # 39; _2, e & # 39; _i) = 0 $ to the $ i geq 3 $).

Then you can show that $ e & # 39; _3 = e_3-2e_2 $ and $ e & # 39; _4 = e_4 + 4e_1 + e_2 $,

The basic calculation also shows this $$ varphi (e & # 39; _3, e & # 39; _4) = varphi (e_3-2e_2, e_4 + 4e_1 + e_2) = b_ {34} -2b_ {24} + 4b_ {31} -2b_ {21 } + b_ {32} = – 13. $$ Let's take the new base $ (e & # 39; & # 39;) = {e & # 39; _1, e & # 39; _2, -e & # 39; _3 / 13, e & # 39; _4 } $ and then it follows that on this basis the matrix has a canonical form, i.e. $$ B _ { varphi} ^ {(e & # 39; & # 39;)} = begin {bmatrix}

0 & 1 & 0 & 0 \

-1 & 0 & 0 & 0 \

0 & 0 & 0 & 1 \

0 & 0 & -1 & 0 \

end {bmatrix} $$ and the function has the following canonical form $$ (u_1v_2-u_2v_1) + (u_3v_4-u_4v_3) $$ and the transformation matrix is this

$$ begin {bmatrix}

1 & 0 & 0 & 0 \

0 & 1 & 0 & 0 \

0 & 2/13 & -1/13 & 0 \

4 & 1 & 0 & 1 \

end {bmatrix} $$

Can someone tell me that my reasoning and the answers are correct, please?

Would be very grateful!