Definition of linearity: Let $A_i$ denote the $i$th row of matrix $A$. Let $A, B, D$ be matrices, all of whose entries are equal except for those in row $k$. Suppose furthermore that $D_k = cA_k + c'B_k$ for scalars $c, c'$. Then $\det D = c\det A + c'\det B $

Artin gives a proof by induction. He assumes that this property is true for all $n-1 \times n-1$ matrices and then tries to show that $d_{v1} \det(D_{v1}) = ca_{v1}\det(A_{v1}) + c'b_{v1}\det(B_{v1})$ for every index $v$. He splits the problem into two cases, $v=k$ (which I understand) and $v\not=k$ (which I don't).

Here is his explanation of the second case: If we let $A'_k, B'_k, D'_k$ denote the vectors obtained from rows $A_k, B_k, D_k$ respectively by dropping the first entry, then $A'_k$ is the row of the minor $A_{v1}$, etc. I understand this concept. However, I don't understand the rest of the proof: Here $D'_k = cA'_k + c'B'_k$ and by induction on n, $\det D'_{v1} = c\det(A'_{v1}) + c'\det B_{v1}$. On the other hand, since $v\not=k$, $a_{v1}, b_{v1}, c_{v1}$ are equal. So $d_{v1} \det D_{v1} = ca_{v1}\det A_v1 + c'b_{v1}\det B_{v1}$

Specifically, I don't understand the induction step, and how to get from $\det D'_{v1} = c\det A'_{v1} + c'\det B_{v1}$ to $d_{v1} \det D_{v1} = ca_{v1}\det A_{v1} + c'b_{v1}\det(B_{v1})$.

  • $\begingroup$ Not an answer, but linearity follows directly from the expansion $\det A = \sum_\sigma \operatorname{sgn} \sigma A_{1,\sigma_1}\cdots A_{n,\sigma_n}$. $\endgroup$
    – copper.hat
    Jan 21, 2014 at 7:28
  • $\begingroup$ @copper.hat Why? $\endgroup$
    – user122719
    Jan 21, 2014 at 7:31

2 Answers 2


Here is an answer (as opposed to my comment response above).

It is important to keep in mind that for $v \neq k$ that $D_{vj} = A_{vj} = B_{vj}$ and similarly $d_{v1} = a_{v1} = b_{v1}$. Furthermore, for $v=k$, we have $d_{vj} = ca_{vj} + c' b_{vj}$.

The cofactor expansion gives $\det D = \sum_{v=1}^n (-1)^{v+1} d_{v1}\det D_{v1}$, so showing that $d_{v1}\det D_{v1} = c a_{v1}\det A_{v1} + c' b_{v1}\det B_{v1}$ is certainly sufficient to show that $\det D = c \det A + c' \det B$.

Case $v=k$: \begin{eqnarray} d_{v1}\det D_{v1} &=& (ca_{v1} + c' b_{v1}) \det D_{v1}\\ &=& ca_{v1} \det D_{v1}+ c' b_{v1}\det D_{v1} \\ &=& ca_{v1} \det A_{v1}+ c' b_{v1}\det B_{v1} \end{eqnarray}

Case $v \ne k$: We have $d_{v1} = a_{v1} = b_{v1}$, so focus on $\det D_{v1}$. Remember that the $k$ row has elements $d_{kj} = ca_{kj} + c' b_{kj}$ and all other elements of $D_{v1}$ equal the corresponding elements of both $A_{v1}$ and $B_{v1}$.

In particular, the $(n-1) \times (n-1)$ matrix $D_{v1}$ has the form of being equal everywhere except for Row $k$ which is the sum of elements as above. Hence the induction step tells us that $\det D_{v1} = c \det A_{v1} + c' \det B_{v1}$ and so \begin{eqnarray} d_{v1}\det D_{v1} &=& d_{v1} (c \det A_{v1}+ c' \det B_{v1}) \\ &=& d_{v1} c \det A_{v1}+ d_{v1} c' \det B_{v1} \\ &=& a_{v1} c \det A_{v1}+ b_{v1} c' \det B_{v1} \end{eqnarray} as required.

  • $\begingroup$ I guess I'm not going to get the bounty. Was really looking forward to the MSE jumpsuit. $\endgroup$
    – copper.hat
    Feb 17, 2022 at 7:15
  • $\begingroup$ What is MSE jumpsuit? $\endgroup$
    – Labbsserts
    Feb 17, 2022 at 18:54
  • 1
    $\begingroup$ @Labbsserts It is a long running joke with someone on MSE chat, there is no such thing :-). (My comment above was also a joke, along similar lines.) In any event, I hope the answer helped. $\endgroup$
    – copper.hat
    Feb 17, 2022 at 20:22
  • 1
    $\begingroup$ Yes, thank you for taking the time. Also I love inside jokes, hope to be part of one someday. $\endgroup$
    – Labbsserts
    Feb 17, 2022 at 20:39
  • 1
    $\begingroup$ Just show up, its informal and marginally abusive :-). $\endgroup$
    – copper.hat
    Feb 17, 2022 at 21:16

I understand that he is calculating a determinant by "expanding" the first column:

$$\left|\begin{matrix}d_{11} & d_{12} & \dots & d_{1n}\\ d_{21} & d_{22} & \dots & d_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ d_{n1} & d_{n2} & \dots & d_{nn}\end{matrix}\right| = d_{11}\left|\begin{matrix} d_{22} & \dots & d_{2n}\\ \vdots & \ddots & \vdots\\ d_{n2} & \dots & d_{nn}\end{matrix}\right| - d_{21}\left|\begin{matrix} d_{12} &d_{13} & \dots & d_{1n}\\ d_{32} & d_{33} & \dots & d_{3n} \\ \vdots & \vdots & \ddots & \vdots\\ d_{n2} & d_{n3} & \dots & d_{nn}\end{matrix}\right| + \dots + (-1)^{n+1}d_{n1}\left|\begin{matrix} d_{12} &d_{13} & \dots & d_{1n}\\ d_{22} & d_{23} & \dots & d_{2n} \\ \vdots & \vdots & \ddots & \vdots\\ d_{(n-1)2} & d_{(n-1)3} & \dots & d_{(n-1)n}\end{matrix}\right|$$

The left hand side is $\det(D)$, and in the right hand side, the $s$-th summand is $\pm d_{s1}$ multiplied by the determinant of the matrix $D$ "without the first column and $s$-th row".

You could write the same expression for the matrices $A$ and $B$. The $k$-th summands of these three expansions are multiplied by the same "minor" (the same $(n-1)\times(n-1)$ determinant, and that is because that determinant is obtained by erasing the only row which is different between $A, B, D$), and we have $d_{k1} = ca_{k1} + c'b_{k1}$, so you can see that the $k$-th summand for $D$ is $c$ times the $k$-th summand for $A$ plus $c'$ times the $k$-th summand for $B$. This is the part you understand.

Now, for $s\neq k$, let's compare the $s$-th summand in the expansion for $D$ with the correspondings summands for $A$ and $B$.

By hypothesis, $d_{s1} = a_{s1} = b_{s1}$, because $s$-th row is equal for the three matrices. So, we must see that the $s$-th minor of $D$ is the required linear combination of the $s$-th minors for $A$ and $B$. Let's denote by $m(A)_s, m(B)_s, m(D)_s$ these matrices, obtained from $A$, $B$ and $D$ by eliminating the first column and $s$-th row.

These three matrices are equal except for one row $j$. The position $j$ of this row is not too important. If $s>k$, then we have taken out a row below the $k$-th one, so $j = k$ retains the same position. If $s<k$, then we have deleted a row above the $k$-th one, so $j = k-1$.

Anyway, these matrices are $(n-1)\times(n-1)$ matrices, equal except for the row $j$, and we have that the $j$-th row of $m(D)_s$ is equal to $c$ times the $j$-th row of $m(A)_s$ plus $c'$ times the $j$-th row of $m(B)_s$. By induction hypothesis, the theorem is true for these matrices, that is:

$$\det(m(D)_s) = c\det(m(A)_s) + c'\det(m(B)_s)$$

This proves the assertion, because, as we saw, in each expansion these minors are multiplied by the same $a_{s1} = b_{s1} = d_{s1}$.

  • 1
    $\begingroup$ That said, I think this is one of the worst ways to convince oneself of the linearity of determinants... This has been said 1000 times, but the determinant should be defined as the factor by which all hypervolumes are multiplied under transformation by $A$ (seen as a linear transformation of space). That, or the (signed) volume of the parallelepiped generated by its rows/column (using the standard basis as unit). Anyway, something involving volume. That way, linearity on columns/rows (which geometrically represent edges) is obvious. $\endgroup$
    – Compacto
    Feb 16, 2022 at 15:53

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