Let A be a square matrix that can be partitioned as
$$M =\begin{pmatrix}A&B\\0&C\end{pmatrix}$$ A and C are square matrices. Such a matrix is said to be in block (Lower) triangular form. Prove that det = (det A)(det C).
Hint: Try a proof of induction on the number of rows of A
Okay, I was hoping someone could give me a more helpful hint of where to start. I know that induction is where you solve for a base case. Then you assume it is true for n and try to prove that it is true for n +1. If it is true for n +1 then it is true for all numbers from 1 to n. I understand that concept. I just have no idea where to begin.
Best Answer
If $P$ is 1 by 1, then, using the cofactor expansion for the determinant along the first row $$ \det A = \det \pmatrix{P & 0 \\ Q & S} = \det \pmatrix{p_{11} & 0 \\ Q & S} = p_{11}\cdot\det S = \det P \cdot \det S $$ since the determinant of a 1 by 1 matrix is just the value of its entry.
Then, assume this holds for matrices $P$ that are $n-1$ by $n-1$. Let us denote $P^{(i,j)}$ to be the matrix $P$ with the $i$th row and $j$th column removed and $Q^{(j)}$ be the matrix $Q$ with the $j$th column removed. Then, if $P$ is $n$ by $n$, we can again expand along the first row \begin{align*} \det A &= \det \pmatrix{P & 0 \\ Q & S} = (-1)^0\cdot p_{11}\cdot \det \pmatrix{P^{(1,1)} & 0 \\ Q^{(1)} & S} + \cdots +(-1)^n\cdot p_{1n}\cdot \det \pmatrix{P^{(1,n)} & 0 \\ Q^{(n)} & S} \\ &= (-1)^0 \cdot p_{11}\cdot (\det P^{(1,1)}\cdot\det S) + \cdots +(-1)^{n-1}\cdot p_{1n}\cdot (\det P^{(1,n)}\cdot\det S) \\&= \det S\cdot((-1)^0\cdot p_{11}\cdot \det P^{(1,1)} + \cdots +(-1)^n\cdot p_{1n}\cdot \det P^{(1,n)})\\&=\det S\cdot \det P \end{align*} Going from the second to the third line, we make use of the fact that $P^{(1,i)}$ are necessarily $n-1$ by $n-1$, so the induction hypothesis applies. Going from the third to the fourth line is just the cofactor expansion of the determinant along the first row.
Hope this helps! I am relatively new to linear algebra, so if something's off please tell me.