I'd like to prove the following properties of the determinant map.

  1. $\det I = 1$
  2. $\det$ is linear in the rows of the input matrix

The determinant map is defined on $n\times n$ matrices $A$ by: $$\det \begin{bmatrix}a\end{bmatrix} = a$$$$\det A = a_{11}\det A_{11}-a_{21}\det A_{21}\pm \dots\pm a_{n1}\det A_{n1}$$

Where $A_{xy}$ is the matrix obtained from $A$ by removing the $xth$ row and the $yth$ column and $a_{xy}$ are the entries of the matrix $A$.


  1. $\det I_n = 1\times \det I_{n-1}+0=\det I_{n-2}=\dots=det I_1 = 1$
  2. The proof is by induction on the size of the matrix. For $2\times 2$ matrices:

$\det \begin{bmatrix}a & b\\ c+x & d+y\end{bmatrix}=a(d+y)-b(c+x)=(ad-bc)+(ay-bx)=\det \begin{bmatrix}a & b\\ c & d\end{bmatrix}+\det \begin{bmatrix}a & b\\ x & y\end{bmatrix}$

Let $ C=\begin{bmatrix} \dots\\ R+S\\ \dots \end{bmatrix}$ be a $n\times n$ matrix for $n\gt 2$, we compute $\det C$ in terms of $\det R$ and $\det S$ where $R = \begin{bmatrix}\dots\\R\\\dots\end{bmatrix}$ and $S = \begin{bmatrix}\dots\\S\\\dots\end{bmatrix}$. Let $i$ denote the index of the rows $R, S, R+S$. $$\det C = \sum_{j\neq i}\pm c_{j1} \det C_{j1} \pm c_{i1} \det C_{i1}$$. By the induction hypothesis, we can write: $$\det C = \sum_{j\neq i}\pm c_{j1}(\det R_{j1}+\det S_{j1}) \pm c_{i1}\det C_{i1}$$

We know that $\det C_{i1}=\det R_{i1}=\det S_{i1}$ and $c_{i1}=r_{i1}+s_{i1}$ and $c_{j1}=r_{j1}=s_{j1}$ for $j\neq i$. So $$\det C = \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm (r_{i1} + s_{i1})\det C_{i1}= \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm ( r_{i1}\det C_{i1} + s_{i1}\det C_{i1})= \sum_{j\neq i}\pm r_{j1}\det R_{j1} + \sum_{j\neq i}\pm s_{j1}\det S_{j1} \pm (r_{i1}\det R_{i1} + s_{i1}\det S_{i1})= \sum_{j}\pm r_{j1}\det R_{j1} + \sum_{j}\pm s_{j1}\det S_{j1}=\det R + \det S$$. A simular induction argument shows that $$\det \begin{bmatrix}\dots \\ c\times R \\ \dots\end{bmatrix}= c\times \det \begin{bmatrix}\dots \\ R \\ \dots\end{bmatrix}$$

Is my proof correct?

  • 2
    $\begingroup$ The determinant is multi-linear in its rows, not linear. I believe you also left out some $\rm{det}$s in your definition of the determinant - this looks like it should be recursive to me. $\endgroup$ – Anthony Carapetis Aug 6 '13 at 15:56
  • 2
    $\begingroup$ The usual $\det$ is defined in a different way (by Laplace): $$\det A = a_{11} \det A_{11} - a_{21} \det A_{21} + a_{31} \det A_{31} - \ldots$$ See en.wikipedia.org/wiki/… $\endgroup$ – AlexR Aug 6 '13 at 15:56
  • $\begingroup$ It might go down easier if you first show that the determinant can be evaluated along any line/row, and then use the 'special' line for calculations. $\endgroup$ – Jonathan Y. Aug 6 '13 at 16:01
  • 1
    $\begingroup$ @AnthonyCarapetis Artin, in his book, Algebra says it's linear. $\endgroup$ – saadtaame Aug 6 '13 at 16:06
  • 1
    $\begingroup$ For 2., it is easier to start the induction at $n=1$. $\endgroup$ – Tony Huynh Aug 6 '13 at 16:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.