Today I heard in a lecture (some video on YouTube) that the determinant is linear as a function of each of the rows of the matrix.

I am not able to understand the above statement. I know that determinant is a special function which assign to each $x$ in $\mathbb K^{n \times n}$ a scalar. This is the intuitive idea. And this map is not linear as well. One way to see this is to consider the fact that determinant of $cA$ is $c^n\det(A)$

Can someone please explain what did the person mean by saying that the determinant is linear as a function of each of the rows of matrix?

  • $\begingroup$ I got it. It means that elementary row operations have a linear effect on determinant. Say $A=(r1,r2,...,r_n)$ is a matrix then det of $(r_1,..,cr_j +r_i,..,r_n)$ is nothing but determinant of $(r_1,..,cr_j,..,r_n)$ plus determinant of $(r_1,..,r_i,..,r_n)$.Am I right? $\endgroup$ – StammeringMathematician Apr 16 at 3:40
  • $\begingroup$ Yes. The fact that is is linear in each row separately gives rise to the combinatorial formula for the determinant. $\endgroup$ – copper.hat Apr 16 at 3:41

If $r_1, \ldots r_n$ are the rows of the matrix and $r_i = sa+tb$, where $s,t$ are scalars and $a,b$ are row vectors, then you have

$$\det\begin{pmatrix}r_1 \\ \vdots \\r_i \\ \vdots \\ r_n\end{pmatrix} = \det\begin{pmatrix}r_1 \\ \vdots \\ sa+tb \\ \vdots \\ r_n\end{pmatrix} = s\det\begin{pmatrix}r_1 \\ \vdots \\ a \\ \vdots \\ r_n\end{pmatrix} + t\det\begin{pmatrix}r_1 \\ \vdots \\ b \\ \vdots \\ r_n\end{pmatrix}$$

This holds for any row $i=1,\ldots , n$. And similarly this also applies to columns.


Let $M$ be an $n\times n$ matrix with rows $\mathbf{r}_1,\dots,\mathbf{r}_n$. Then we may think of the determinant as a function of the rows $$ \det(M)=\det(\mathbf{r}_1,\dots,\mathbf{r}_n). $$ To say that $\det$ is a linear function of the rows means that if we scale a single row by $c$, the result is scaled by $c$; that is, $$ \det(\mathbf{r}_1,\dots,\mathbf{r}_{i-1},c\mathbf{r}_i,\mathbf{r}_{i+1}\dots\mathbf{r}_n)=c\det(\mathbf{r}_1,\dots,\mathbf{r}_n). $$ Similarly if we fix all but one row (say the first), we obtain $$ \det(\mathbf{x}+\mathbf{r}_1,\mathbf{r}_2,\dots,\mathbf{r}_n)=\det(\mathbf{x},\dots,\mathbf{r}_n)+\det(\mathbf{r}_1,\dots,\mathbf{r}_n). $$ Your mistake was that you scale all the rows at once; to be linear, you can only do things "one at a time"


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