Prove Det(kA)=kndetA [closed]

haidar

7 min read

·

Sep 25, 2024

29

Share

In the realm of linear algebra, determinants play a crucial role in understanding the properties of matrices. One fundamental property involves scaling a matrix by a constant factor and its impact on the determinant. This article delves into the proof of the theorem: Prove det(kA)=k^ndetA, where A is an n x n matrix and k is a scalar. We will explore the underlying concepts and provide a step-by-step demonstration of this theorem's validity.

Understanding the Theorem and its Significance

The theorem Prove det(kA)=k^ndetA states that the determinant of a matrix scaled by a constant k is equal to k raised to the power of the matrix's dimension (n) multiplied by the determinant of the original matrix. This theorem is essential in various applications, including:

• Linear transformations: Understanding how scaling a matrix affects its determinant is crucial in analyzing linear transformations.
• Eigenvalues and eigenvectors: The theorem provides insight into the relationship between eigenvalues and determinants when a matrix is scaled.
• Solving systems of linear equations: Determinants are instrumental in solving systems of linear equations, and this theorem aids in understanding how scaling coefficients affects solutions.

Proof by Induction

To prove Prove det(kA)=k^ndetA, we will employ the method of mathematical induction. This approach involves two steps:

1. Base Case: Establishing the theorem's validity for the smallest possible value of n (n=1).
2. Inductive Step: Assuming the theorem holds for a given value of n, we must show that it also holds for n+1.

Base Case (n=1)

For a 1 x 1 matrix A = [a], the determinant is simply the value of the element a. When we scale this matrix by k, we get kA = [ka]. Therefore,

det(kA) = ka = k * det(A)

This confirms that the theorem holds for n=1.

Inductive Step

Assume that the theorem holds for an n x n matrix. That is, assume that:

det(kA) = k^n det(A)

Now, we need to prove that the theorem holds for an (n+1) x (n+1) matrix. Let B be an (n+1) x (n+1) matrix. We can expand the determinant of kB using the cofactor expansion along the first row:

det(kB) = (kB)₁₁C₁₁ - (kB)₁₂C₁₂ + ... + (-1)^(n+1)(kB)_1(n+1)C_1(n+1)

where:

• (kB)_ij represents the element in the i-th row and j-th column of kB.
• C_ij represents the cofactor of the element in the i-th row and j-th column.

Since each element in kB is simply k times the corresponding element in B, we can rewrite the above equation as:

det(kB) = k(B₁₁C₁₁ - B₁₂C₁₂ + ... + (-1)^(n+1)B_1(n+1)C_1(n+1))

Notice that the expression in parentheses is precisely the determinant of B. Therefore, we can write:

det(kB) = k * det(B)

Now, let's consider the determinant of B. Since B is an (n+1) x (n+1) matrix, we can apply the inductive hypothesis:

det(B) = k^n * det(A)

where A is the n x n matrix formed by removing the first row and first column of B.

Substituting this into the previous equation, we get:

det(kB) = k * (k^n * det(A)) = k^(n+1) * det(A)

This proves that the theorem holds for n+1.

Conclusion

Through the principle of mathematical induction, we have successfully demonstrated that Prove det(kA)=k^ndetA. The theorem establishes a fundamental relationship between scaling a matrix and its determinant, providing valuable insights in various applications of linear algebra.

This theorem underlines the powerful nature of determinants in representing the properties of matrices and linear transformations. Understanding this relationship allows us to analyze and manipulate matrices more effectively in diverse mathematical and scientific contexts.