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Chapter 4: Problem 16

Prove that if \(\delta: M_{n \times n}(F) \rightarrow F\) is an alternating \(n\)-linear function, then there exists a scalar \(k\) such that $\delta(A)=k \operatorname{det}(A)\( for all \)A \in \mathrm{M}_{n \times n}(F)$.

Short Answer

Expert verified
Assume that there does not exist any scalar k such that 未(A) = k * det(A) for all A 鈭� M_{n脳n}(F). By considering the action of elementary row operations on A and their corresponding change in 未(A) and det(A), we find that they change in the same way. This implies the existence of a scalar k such that 未(A) = k * det(A), contradicting our initial assumption. Therefore, there indeed exists a scalar k such that 未(A) = k * det(A) for all A 鈭� M_{n脳n}(F).

Step by step solution

01

Set up the proof by contradiction

Assume that there does not exist any scalar k such that 未(A) = k * det(A) for all A 鈭� M_{n脳n}(F). Now, we want to find a contradiction to this assumption to prove that there indeed exists such a scalar k.
02

Apply elementary row operations on A

Consider the action of elementary row operations on A 鈭� M_{n脳n}(F) and the corresponding change in 未(A): i) Swapping adjacent rows: 未(A) changes sign, and det(A) also changes sign. ii) Multiplying a row by a scalar 伪 鈭� F: 未(A) is multiplied by 伪, and det(A) is also multiplied by 伪. iii) Adding a scalar multiple of one row to another: 未(A) remains unchanged, and det(A) also remains unchanged.
03

Observe the contradiction

From the row operations, we can see that 未(A) and det(A) change in the same way when elementary row operations are applied. This means that 未(A) and det(A) should have the same relationship for all A 鈭� M_{n脳n}(F), implying the existence of a scalar k such that 未(A) = k * det(A). This contradicts our initial assumption that such a scalar k does not exist.
04

Conclude the proof

Since we have found a contradiction to our initial assumption, we can conclude that there indeed exists a scalar k such that 未(A) = k * det(A) for all A 鈭� M_{n脳n}(F), proving the statement.

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