I am currently studying for a final exam and am confused about the proof of $\det(AB)=\det(A)\det(B)$ given in Hoffman/Kunze. I'll type out the entire thing so that my question will be in the correct context.

Theorem: Let $K$ be a commutative ring with identity, and let $A$ and $B$ be $n \times n$ matrices over $K$. Then, $\det(AB)=\det(A)\det(B)$. Proof:

Let $B$ be a fixed $n \times n$ matrix over $K$, and for each $n \times n$ matrix $A$ define $D(A)=\det(AB)$. If we denote the rows of $A$ by $\alpha_{1},...,\alpha_{n}$, then $D(\alpha_{1},...,\alpha_{n})=\det(\alpha_{1}B,...,\alpha_{n}B)$. Here $\alpha_{j}B$ denotes the $1 \times n$ matrix which is the product of the $1 \times n$ matrix $\alpha_{j}$ and the $n \times n$ matrix $B$. Since, $(c \alpha_{i} + \alpha_{i}')B = c \alpha_{i}B + \alpha_{i}' B$ and $\det$ is $n$-linear, it is easy to see that $D$ is $n$-linear. If $\alpha_{i}=\alpha_{j}$, then $\alpha_{i}B=\alpha_{j}B$, and since $\det$ is alternating, $D(\alpha_{1},...,\alpha_{n})=0$. Hence $D$ is alternating. Now, $D$ is an alternating $n$-linear function, and so $D(a)=\det(A)D(I)$, but $D(I)=\det(IB)=\det(B)$ and thus, $\det(AB) = D(A) = \det(A)\det(B)$.

I think I understand the entire proof, how you show it and why each step is taken, but I am just confused about how the part right before "it is easy to see that $D$ is $n$-linear" show that $D$ is $n$-linear. If someone could just fill in the details or give an explanation of why that is sufficient would be helpful.

Thank you.


1 Answer 1


We have $$\begin{align*} D(\alpha_1,\ldots,c\alpha_i+\alpha'_i,\ldots,\alpha_n) &= \det(\alpha_1B,\ldots,(c\alpha_i+\alpha'_i)B,\ldots,\alpha_nB)\\ &= \det(\alpha_1B,\ldots,c(\alpha_iB) + \alpha'_iB,\ldots,\alpha_nB)\\ &= c\det(\alpha_1B,\ldots,\alpha_iB,\ldots,\alpha_nB) + \det(\alpha_1B,\ldots,\alpha'_iB,\ldots,\alpha_nB)\\ &= cD(\alpha_1,\ldots,\alpha_i,\ldots,\alpha_n) + D(\alpha_1,\ldots,\alpha'_i,\ldots,\alpha_n), \end{align*}$$ which shows that $D$ is linear in the $i$th coordinate. Of course, the same computation works in each component, so $D$ is $n$-linear.

The first equality is by definition of $D$. The second is by the first observation (that $(c\alpha_i + \alpha'_i)B = c\alpha_iB + \alpha'_iB$); the third because $\det$ is $n$-linear, and the final equality by the definition of $D$.

  • $\begingroup$ Thanks again Arturo, your answers are always very helpful! The step I missed was the third one. $\endgroup$ Apr 15, 2012 at 20:55

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