Okay, so I am in a linear algebra course and we are going though the derivation of the determinant function for matrices. I am struggling with some of the properties of determinant functions i.e. functions which are n-linear, alternating, and give a 1 for the identity matrix. The question I am currently stuck on is as follows.

Let $K$ be a commutative ring with identity and $D$ an $n$-linear function on $n\times n$ matrices over $K$. Show that $D(B)=D(A)$, if $B$ is obtained from $A$ by adding a scalar multiple of one row of $A$ to another.

What I have so far is as follows.

Let $A$ be an $n\times n$ matrix. Let $A=(\alpha_1, \cdots,\alpha_i,\cdots,\alpha_j,\cdots, \alpha_n)$ where $a_k$, $k=1,...,n$ denote the rows of $A$. Let $B=(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_n)$ for $a\in K$. That is let $B$ be the matrix formed from $A$ where the $i^{th}$ row of $A$ is replaced by $a$ times the $i^{th}$ of $A$ row plus the $j^{th}$ row of $A$. Then

$D(B)=D(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_n)=aD(\alpha_1,\cdots,\alpha_i,\cdots,\alpha_j,\cdots,\alpha_n)+D(\alpha_1,\cdots,\alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)\text{($n$-linearity) }=aD(A)+0=aD(A).$

Therefore $D(B)=aD(A)$

I am clearly missing something obvious because the $a$ should not be there according to the question but for the life of me I can't see what is going on. I am assuming it is something obvious and easy but I have been stuck for far too long. Thanks.

  • $\begingroup$ In your first expression for $D(B)$, it looks like you are adding $a$ times the $i$th row to the $j$th row, but then you write an expression where the *$j$th* row has been repeated. In fact, it is the first term that has a repeated row $\alpha_i$ (and multiplication by $a$), and the second term is equal to $D(A)$. $\endgroup$ – BaronVT Oct 26 '18 at 17:17
  • $\begingroup$ I clarified how I constructed $B$. $\endgroup$ – Walt Oct 26 '18 at 17:30

Your proof seems right to me. But, you must define $B$ as

$$B=(\alpha_1,\cdots,\alpha_i+a\cdot \alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)$$

because $B$ is obtanined by adding $a$ times the row $j$ of $A$ to the row $i$ of $A$. In this case you will get $D(A)=D(B)$ since:

$$D(B)=D(\alpha_1,\cdots,\alpha_i+a\cdot \alpha_j,\cdots,\alpha_j,\cdots,\alpha_n) =D(\alpha_1,\cdots,\alpha_i,\cdots,\alpha_j,\cdots,\alpha_n)+a\cdot D(\alpha_1,\cdots,\alpha_j,\cdots,\alpha_j,\cdots,\alpha_n)=D(B)+a\cdot 0=D(B).$$

  • $\begingroup$ Why must $B$ be defined this way. I see that it works this way but what am I missing with the way I defined $B$? $\endgroup$ – Walt Oct 26 '18 at 17:26
  • $\begingroup$ Because, B is obtained from A by adding a scalar multiple of one row of A and not multiplying a row of A and then adding after other row of A(this is what you wrote before) $\endgroup$ – Hector Blandin Oct 26 '18 at 17:27
  • $\begingroup$ I'm sorry I'm still not connecting the dots. The only difference between the definitions is the order of the addition. The way I am reading the question is that I am just picking two rows, multiplying one of them by a scalar, adding them, and then replacing one of the chosen rows with the result. The order of the addition shouldn't matter right? $\endgroup$ – Walt Oct 26 '18 at 17:34
  • $\begingroup$ Ok. Suppose you pick a row $r_1$ and then you compute $a\cdot r_1+r_2$ and you replace $r_1$ with $a\cdot r_1+r_2$ (This is what you wrote initially). This is not the same thing as taking a row $r_1$ and adding a multiply of another row $r_2$, this will be $r_1+a\cdot r_2$. The last operation is the actual condition you have to prove. $\endgroup$ – Hector Blandin Oct 26 '18 at 17:37
  • $\begingroup$ I think I see it. I think my issue is what position I'm consider the new row to be in. If we write $a\alpha_i+\alpha_j$ then this row is the $j^{th}$ row of $B$. Otherwise if we write $\alpha_i+a\alpha_j$ then this is the $i^{th}$ row of $B$. $\endgroup$ – Walt Oct 26 '18 at 17:41

You started heading for trouble when you re-wrote $$D(\alpha_1,\cdots,\alpha_i, \cdots, \alpha_j,\cdots,\alpha_n)$$ (which calls out the $i$th and $j$th positions) as

$$D(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_n)$$ which leaves it unclear as to which position the new expression sits in. You can't tell if it's supposed to represent $$D(\alpha_1,\cdots,\alpha_i, \cdots, a\alpha_i+\alpha_j,\cdots,\alpha_n)$$ or $$D(\alpha_1,\cdots,a\alpha_i+\alpha_j,\cdots,\alpha_j, \cdots, \alpha_n)$$

HectorBlandin's answer made it explicit and so didn't lead to the same confusion.


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