# [10 - 11] Cross Product

Lightman·2023년 2월 15일
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## MATH 𝒳𝒴𝒵

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Why is this function f in
"$f(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}) = det\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix} \right)$" is linear based on properties of the determinant?

The function f defined by $f(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}) = det\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix} \right)$ is linear because the determinant satisfies the following properties:

1. Multiplying a row of the matrix by a scalar multiplies the determinant by the same scalar. ( In Geogrphical view, you just stretch the length of bottom line of parallelogram )
• For example, if we multiply the first row of the matrix by a scalar k, the new matrix is:

$\begin{bmatrix} kx & kv_1 & kw_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix}$

and the determinant of the new matrix is k times the determinant of the original matrix:

$k*det\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix} \right)$

1. Adding a scalar multiple of one row to another row does not change the determinant. ( In Geogrphical view, you just moving the left top point of parallelogram along the parallel line to bottom line )
그런데, 이 부분을 유심히 생각해보면 한 벡터에 다른 벡터를 더할 때 그 벡터가 나머지 두 벡터의 선형조합인 벡터라면 paralellopiped의 부피는 변하지 않는다. 만약 선형조합으로 구할 수 없는 벡터라면 paralellopiped의 부피는 x2 y2 z2가 만들어내는 높이만큼 더해진다. 따라서 det가 분리된다는 사실은 시각적으로 명확하다!
• For example, if we add k times the first row to the second row, the new matrix is:

$\begin{bmatrix} x & v_1 & w_1 \\ y+kx & v_2+kv_1 & w_2+kw_1 \\ z & v_3 & w_3 \\ \end{bmatrix}$

and the determinant of the new matrix is the same as the determinant of the original matrix.

These two properties of the determinant imply that the function f is linear, because:

$f\left(\begin{bmatrix} x_1 \\ y_1 \\ z_1 \\ \end{bmatrix} + \begin{bmatrix} x_2 \\ y_2 \\ z_2 \\ \end{bmatrix}\right) \\= f\left(\begin{bmatrix} x_1 + x_2 \\ y_1 + y_2 \\ z_1 + z_2 \\ \end{bmatrix}\right) \\\\= det\left(\begin{bmatrix} x_1 + x_2 & v_1 & w_1 \\ y_1 + y_2 & v_2 & w_2 \\ z_1 + z_2 & v_3 & w_3 \\ \end{bmatrix} \right) \\\\= det\left(\begin{bmatrix} x_1 & v_1 & w_1 \\ y_1 & v_2 & w_2 \\ z_1 & v_3 & w_3 \\ \end{bmatrix} \right) + det\left(\begin{bmatrix} x_2 & v_1 & w_1 \\ y_2 & v_2 & w_2 \\ z_2 & v_3 & w_3 \\ \end{bmatrix} \right) \cdots ① \\\\= f\left(\begin{bmatrix} x_1 \\ y_1 \\ z_1 \\ \end{bmatrix}\right) + f\left(\begin{bmatrix} x_2 \\ y_2 \\ z_2 \\ \end{bmatrix}\right)$

And for any scalar k:

$f(k\begin{bmatrix} x\\ y \\ z \\ \end{bmatrix}) \\ = f\left(\begin{bmatrix} kx \\ ky \\ kz \\ \end{bmatrix}\right) \\= kdet\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix} \right) \\= kf\left(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}\right)$

Therefore, f satisfies the linearity property of a linear transformation, and hence is a linear function.

①에 대한 증명

This equation holds because of the distributive property of the determinant. The determinant of a 3x3 matrix can be expanded using the Laplace expansion along the first row as follows:

$det\left(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} \right) \\ = a\det\left(\begin{bmatrix} e & f \\ h & i \\ \end{bmatrix} \right) - b\det\left(\begin{bmatrix} d & f \\ g & i \\ \end{bmatrix} \right) + c\det\left(\begin{bmatrix} d & e \\ g & h \\ \end{bmatrix} \right)$

In the case of the function f defined by

$f(\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}) = det\left(\begin{bmatrix} x & v_1 & w_1 \\ y & v_2 & w_2 \\ z & v_3 & w_3 \\ \end{bmatrix} \right)$

the Laplace expansion along the first row gives:

$f(\begin{bmatrix} x \\ y \\ z \ \end{bmatrix}) \\= x\det\left(\begin{bmatrix} v_2 & w_2 \\ v_3 & w_3 \\ \end{bmatrix} \right) - v_1\det\left(\begin{bmatrix} y & w_2 \\ z & w_3 \\ \end{bmatrix} \right) + w_1\det\left(\begin{bmatrix} y & v_2 \\ z & v_3 \\ \end{bmatrix} \right)$

Using this expansion, we can see that the determinant of the matrix

$\begin{bmatrix} x_1 + x_2 & v_1 & w_1 \\ y_1 + y_2 & v_2 & w_2 \\ z_1 + z_2 & v_3 & w_3 \\ \end{bmatrix}$

is equal to the sum of the determinants of the matrices

$\begin{bmatrix} x_1 & v_1 & w_1 \\ y_1 & v_2 & w_2 \\ z_1 & v_3 & w_3 \\ \end{bmatrix}$

and

$\begin{bmatrix} x_2 & v_1 & w_1 \\ y_2 & v_2 & w_2 \\ z_2 & v_3 & w_3 \\ \end{bmatrix}$

where we expand each of these determinants using the Laplace expansion along the first row.

This shows that the equation

$det\left(\begin{bmatrix} x_1 + x_2 & v_1 & w_1 \\ y_1 + y_2 & v_2 & w_2 \\ z_1 + z_2 & v_3 & w_3 \\ \end{bmatrix} \right) = det\left(\begin{bmatrix} x_1 & v_1 & w_1 \\ y_1 & v_2 & w_2 \\ z_1 & v_3 & w_3 \\ \end{bmatrix} \right) + det\left(\begin{bmatrix} x_2 & v_1 & w_1 \\ y_2 & v_2 & w_2 \\ z_2 & v_3 & w_3 \\ \end{bmatrix} \right)$

holds for any values of x1, y1, z1, x2, y2, and z2, as long as the matrices have entries in some field, such as the real numbers or the complex numbers.

# CONCLUSION

• The matrix represents a scale factor that changes with a linear transformation.

$det(AB) = det(A)det(B)$

• The scale change of two linear transformations is equal to the product of the scale changes of each linear transformation.

$det\left(\begin{bmatrix} k\vec x & \vec y & \vec z \end{bmatrix} \right) = k*det\left(\begin{bmatrix} \vec x & \vec y & \vec z \end{bmatrix} \right)$

• If you multiply the vectors involved in the linear transformation by a factor of k, the linear transformation is also multiplied by a factor of k.

$det\left(\begin{bmatrix} \vec x_1+\vec x_2 & \vec y & \vec z \end{bmatrix} \right) = det\left(\begin{bmatrix} \vec x_1 & \vec y & \vec z \end{bmatrix} \right) + det\left(\begin{bmatrix} \vec x_2 & \vec y & \vec z \end{bmatrix} \right)$

• Moving a vector involved in a linear transformation increases the scale of the linear transformation by the number of vectors moved. If you move it parallel to the floor, it will increase by 0. If you move it non-parallel to the floor, it will have a non-zero value.
현직 데이터 분석가 / 데이터 과학의 정도를 따라 🚲 / About DEV DA ML