Proof of determinants for matrices of any order

I was told that the determinant of a square matrix can be expanded along any row or column and given a proof by expanding in all possible ways, but only for square matrices of order 2 and 3.

• Is a general proof for any order even possible ?


• If so, how is this done ?



• On a similar note, how can we prove the various properties of determinants for square matrices for any order like the following:

  
  
  
  
  
  • Swap two rows/columns and all we get is a minus sign as a result.
  
  
  • $R_1 \to R_1+ aR_2$ does not change the determinant.
  
  
  • Determinant of the transpose is the same as the determinant of the original matrix.
  
  
  

Answer

Here is one possible path. We define the determinant recursively:

• if $A$ is $1\times 1$, let $\det A=A$;




• If $A$ is $(n+1)\times (n+1)$, let
  

  $$\det A=\sum_{k=1}^{n+1} (-1)^{k+1}A_{k1}\,M_{k1}^A,$$
  where $M_{k1}^A$ is the determinant of the $n\times n$ matrix obtained by removing the $k^{\rm th}$ row and the first column of $A$.

Now,

1. Show that if $B$ is obtained from $A$ by multiplying a row by $\alpha$, then

   $$\det B=\alpha\,\det A.$$ This is done by induction very easily.




2. Show that if we have $A,B,C$ with $A_{rj}=B_{rj}+C_{rj}$ for all $j$, and $A_{kj}=B_{kj}=C_{kj}$ when $k\ne r$ and for all $j$, then
   

   $$\det A=\det B+\det C.$$ Again this is done by induction. When $r=1$ the equality follows trivially from the definition of determinant (as the minors of $A,B,C$ will be all equal) and when $r\ne 1$ we use induction.





3. Show that if $B$ is obtained from $A$ by swapping two rows, then

   $$\det B=-\det A.$$ Here one first swaps rows $1$ and $r$, and then any other swapping of two rows $r$ and $s$ can be achieved by three swaps ($r$ to $1$, $s$ to $1$, $r$ to $1$). This can be used to show that one can calculate the determinant along any row (swap it with row 1).




4. It now follows that if $A$ has two equal rows, then $\det A=0$ (because $\det A=-\det A$).




5. If $B_{rj}=A_{rj}+\alpha A_{sj}$, and $B_{kj}=A_{kj}$ when $k\ne r$, then by 1. and 2.,
   

   $$\det B=\det A+\alpha\det C,$$ where $C$ is the matrix equal to $A$ but with the $s$ row in place of the $r$ row; by 4., $\det C=0$, so $$\det B=\det A$.




6. Now one considers the elementary matrices, and checks directly (using the above properties) that for any elementary matrix $E$,

   $$\det EA=\det E\,\det A.$$




7. If $B$ is invertible, then $B$ can be written as a product of elementary matrices, $B=E_1E_2\cdots E_m$, and so
   

   $$\begin{align} \det BA&=\det E_1E_2\cdots E_m A=\det E_1\det E_2\cdot\det E_m\det A\\ \ \\ &=\det (E_1\cdots E_m)\det A=\det B\det A. \end{align}$$
   
   Similarly, $\det AB=\det A\det B$.




8. If neither $A$ nor $B$ are invertible: then $AB$ is not invertible either. For a non-invertible matrix, its Reduced Row Echelon form has a row of zeroes, and so its determinant is zero; as we can move to $A$ by row operations, it follows that $\det A=0$; similarly, $\det AB=0$. So

   $$\det AB=\det A\det B$$ also when one of them is not invertible.




9. Knowing that det is multiplicative, we immediate get that, when $A$ is invertible,

   $$\det A^{-1}=\frac1{\det A}.$$




10. For an arbitrary matrix $A$, it is similar to its Jordan form: $A=PJP^{-1}$. Then
    

    $$\det A=\det (PJP^{-1})=\det P\,\det J\,\frac1{\det P}=\det J.$$
    As $J$ is triangular with the eigenvalues of $A$ (counting multiplicities) in its diagonal, we get that
    $$\det A=\lambda_1\cdots\lambda_n,$$
    where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$, counting multiplicities.




11. Since the eigenvalues of $A^T$ are the same as those from $A$, we get
    

    $$\det A^T=\det A.$$




12. Now, everything we did for rows, we can do for columns by working on the transpose. In particular, we can calculate the determinant along any column.