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.


 A: Just use a suitable notation to prove the various properties of the determinant.
For $I=\{1,\ldots, i,\ldots,n\}$ and $J=\{1,\ldots, j,\ldots,n\}$ fix the index set notation: $I_i=I-\{i\}$ for $i=1,2,\ldots,n$   and $J_j=J-\{j\}$ for $j=1,2,\ldots,n$. Now fix the index notation for matrices:
$$
M= \left\lgroup M_{ij} \right\rgroup_{\substack{ i\in I\\j\in J}}=
\begin{pmatrix}
M_{11} &\ldots & M_{1j-1}\;M_{1j}\;M_{1j+1}&\ldots & M_{1n}\\
\vdots &       &\vdots &       &\vdots \\
M_{i-11} &\ldots & M_{i-1j-1}\; M_{i-1j}\; M_{i-1j+1}&\ldots & M_{i-1n}\\
M_{i1} &\ldots & M_{ij-1}\; M_{ij}\; M_{ij+1}&\ldots & M_{in}\\
M_{i+11} &\ldots & M_{i+1j-1}\; M_{i+1j}\; M_{i+1j+1}&\ldots & M_{i+1n}\\
\vdots &       &\vdots &       &\vdots \\
M_{n1} &\ldots & M_{nj-1}\;M_{nj}\;M_{nj+1}&\ldots & M_{nn}\\
\end{pmatrix}_{n\times n}
$$ 
$$
\left\lgroup M_{uv} \right\rgroup_{\substack{ u\in I_i\\v\in J_j}}
=
\begin{pmatrix}
M_{11} &\ldots & M_{1j-1}\;M_{1j+1}&\ldots & M_{1n}\\
\vdots &       &\vdots &       &\vdots \\
M_{i-11} &\ldots & M_{i-1j-1}\; M_{i-1j+1}&\ldots & M_{i-1n}\\
M_{i+11} &\ldots & M_{i+1j-1}\; M_{i+1j+1}&\ldots & M_{i+1n}\\
\vdots &       &\vdots &       &\vdots \\
M_{n1} &\ldots & M_{nj-1}\;M_{nj+1}&\ldots & M_{nn}\\
\end{pmatrix}_{(n-1)\times (n-1)}
\\
\left\lgroup M_{uv} \right\rgroup_{\substack{ u\in I_i\\v\in J_j}}
=
\begin{pmatrix}
M_{11} &\ldots & M_{1v}&\ldots & M_{1 n-1}\\
\vdots &       &\vdots &       &\vdots \\
M_{u1} &\ldots & M_{uv}&\ldots & M_{u n-1}\\
\vdots &       &\vdots &       &\vdots \\
M_{n-1 1} &\ldots & M_{n-1v}&\ldots & M_{n-1n-1}\\
\end{pmatrix}_{(n-1)\times (n-1)}
$$ 
Then the expansion on line $i$ is 
$$
\det \left\lgroup
M_{ij}
\right\rgroup_{\substack{i\in I\\ j\in J}}
=
\sum_{j=1}^{n} M_{ij}(-1)^{i+j}\det 
\left\lgroup
M_{uv}
\right\rgroup_{\substack{u\in I_i\\ v\in J_j}}
$$
and the expansion on column $j$ is 
$$
\det \left\lgroup
M_{ij}
\right\rgroup_{\substack{i\in I\\ j\in J}}
=
\sum_{i=1}^{n} M_{ij}(-1)^{i+j}\det 
\left\lgroup
M_{uv}
\right\rgroup_{\substack{u\in I_i\\ v\in J_j}}
$$
A: 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
$$
\det A=\sum_{k=1}^{n+1} (-1)^{k+1}A_{1k}\,M_{1k}^A,
$$
where $M_{st}^A$ is the determinant of the $n\times n$ matrix obtained by removing the $s^{\rm th}$ row and the $t^{\rm th}$  column  of $A$.
Now,

*

*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.


*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.


*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).


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


*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 $\det B=\det A$.


*Now one considers the elementary matrices, and checks directly (using the above properties) that for any elementary matrix $E$, $$\det EA=\det E\,\det A.$$


*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, $\det AB=\det A\det B$.


*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.


*Knowing that det is multiplicative, we immediate get that, when $A$ is invertible, $$\det A^{-1}=\frac1{\det A}.$$


*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.


*Since the eigenvalues of $A^T$ are the same as those from $A$, we get
$$
\det A^T=\det A. 
$$


*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.
