Construct an example to show that the following statement is not true for all square matrices A and B of the same size: \[det( A + B ) = det(A) + det(B)\]
A simple counterexample to the sum \(C\) of \(A\) and \(B\) below:
Let the matrix \[C = \left[ \begin{array} {rr} 1 & 1 \\ 1 & 1 \end{array} \right] = \left[ \begin{array} {rr} 1 & 1 \\ 0 & 1 \end{array} \right] + \left[ \begin{array} {rr} 0 & 0 \\ 1 & 0 \end{array} \right] = A + B \] Clearly \[ det(C) = (1)(1) - (1)(1) = 0\]
However, the determinants of the right hand side do not add up to zero.
\[ det(A) = det\left(\left[ \begin{array} {rr} 1 & 1 \\ 0 & 1 \end{array} \right]\right) = (1)(1)- (0)(1) = 1 \]
\[ det(B) = det\left(\left[ \begin{array} {rr} 0 & 0 \\ 1 & 0 \end{array} \right]\right) = (0)(0)- (0)(1) = 0 \]
This provides the required counterexample because:
\[ 0 = det(C) \neq det(A) + det(B) = 1 + 0 = 1\]
We use Theorem DRCM (Determinant for Row or Column Multiples) to prove the result.
First, we define notation for the column vectors of a square \(n \times n\) matrix \(A\) as \(a_1, a_2, ..., a_n\). Thus, we would write: \[ A = \left[ a_1, a_2, ..., a_n \right]\] Next, we consider the series of matrices \(B_r\) for \(r = 0,1,2, ... ,n\) where a scalar \(\alpha\) is multipled by all entries of \(A\) whose column \(j \leq r\). We may denote \(B_r\) as: \[B_r = \left[ \alpha a_1, \alpha a_2, .. \alpha a_r, a_{r+1}, ... , a_n \right]\]
It is now clear that \(A = B_0\). Likewise, if we multiple every column vector of \(A\) by \(alpha\) this is equivalent to \(B_n\). This proves \(\alpha A = B_n\).
By theorem DRCM, we know that \[ det(B_1) = \alpha det(B_0)\]. This proves \[det(B_1) = \alpha \cdot det( A ) \]
This proves \[det(B_r) = \alpha^r \cdot det(A)\] for \(r = 1\). Assume the result holds for \(r-1\).
Then, by DRCM, \[ det( B_r ) = \alpha \cdot det( B_{r-1})\]
By the inductive hypothesis, \[ det(B_r) = \alpha \cdot \left( det( B_{r-1} ) \right) = \alpha \cdot \left( \alpha^{r-1} det(A) \right) = \alpha^{r} det(A)\] which proves the general case for \(r\).
Finally, since \(B_n = \alpha \cdot A\), this implies \[ det(\alpha \cdot A) = det(B_n) = \alpha^{n} \cdot det(A)\] This completes the proof.