Proof of determinants for matrices of any order

Question

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 1

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,

0. Show by induction that $$ \det A=\sum_{\sigma\in\mathbb S_n}(\operatorname{sgn}\sigma)\,A_{1,\sigma(1)}\cdots A_{n,\sigma(n)}. $$ This implies that $\det$, when seen as a function of the rows of $A$, is a skew-symmetric multilinear map.

The next properties follow directly from the multilinearity and skew-linearity of $\det$.

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, calculate, unswap).

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 immediately 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. If $B$ is the matrix obtained by exchanging rows $1$ and $2$ in $A$, then $$ \det A=-\det B=-\sum_{j=1}^n (-1)^{1+j}b_{1j}\,M_{1j}^B =\sum_{j=1}^n (-1)^{2+j}a_{2j}\,M_{2j}^B. $$

13. 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. Exchanging with other rows we get by induction $$ \det A=\sum_{j=1}^n (-1)^{k+j}a_{kj}\,M_{kj}^A. $$

Answer 2

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 $$ \det \left\lgroup M_{ij} \right\rgroup_{\substack{i\in I\\ j\in J}} = \sum_{i=1}^{n} M_{ij}(-1)^{i+j}\det \left\lgroup M_{uv} \right\rgroup_{\substack{u\in I_i\\ v\in J_j}} $$