Prove that $\det(kA) = k^n\det(A)$ for all $A \in M_{n\times n}(F)$ So I was just looking over an old homework problem, and my proof doesn't seem right. I got full credit for it, but it seems to be circular reasoning. Please tell me if this is actually a valid proof:
Suppose $A \in M_{n\times n}(F)$. Then $$\det(kA) = \det(kI_nA) = \det(kI_n)\det(A) = k^n\det(I_n)\det(A) = k^n(1)\det(A) = k^n\det(A).$$
I am wondering if claiming that $\det(kI_n) = k^n\det(I_n)$ was a valid move, since that is basically what the problem asks me to prove. Is it just a given for the identity matrix? I don't remember what I was thinking when originally doing this problem.
(Side note, how do I do a vertically aligned string of equalities in LaTex? )
 A: This all really depends what your definition of the determinant is, and what lemmas were previously proven.  For example, if your definition is the one with signed permutations,
$$\det(A) = \sum_\pi (-1)^\pi a_{1\pi(1)} \cdots a_{n\pi(n)},$$
then it follows trivially that $\det(kI_n) = k^n$, since there is only one permutation for which the corresponding product of terms does not contain a zero, and that product is of the $n$ diagonal copies of $k$.  (Of course, with this definition it's also obvious straight from the definition that multiplying each entry of $A$ by $k$ multiplies each term of that sum by $k^n$.)
If your definition is by Laplace expansion,
$$\det(A) = \sum_{i = 1}^n (-1)^i a_{i1} \det(A_{i1}),$$
then you require some kind of inductive argument whether or not you factor out $kI_n$ as you did.
If your definition is axiomatic, defining $\det$ to be the unique scalar-valued function on square matrices such that


*

*$\det$ is linear in each row of its argument,

*$\det(A) = 0$ if two rows of $A$ are equal, and

*$\det(I) = 1$,


then the combination of 1 and 3 implies $\det(kI_n) = k^n$.  On the other hand, just 1 alone implies $\det(kA) = k^n \det(A)$ also.
In other words, your argument may or may not be circular, but I don't think it is actually a simplification.
A: I think the claim that $det(kI_n)=k^ndet(I_n)$ is valid. $kI_n$ looks like the identity matrix but with $k$ in place of each 1 on the main diagonal. Because the determinant of any triangular matrix is the product of the entries on the diagonal, $det(kI_n)$ equals $k$ multiplied by itself $n$ times, i.e. $k^n$. So $det(kI_n)=k^n=k^ndet(I_n)$. 
