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Consider the block $2n \times 2n$ matrix

$$\begin{bmatrix} A&B\\ 0&D \end{bmatrix}$$

where $A,B,D$ are $n \times n$ blocks. Show that

$$\det\begin{bmatrix} A&B\\ 0&D \end{bmatrix}=\det(A)\det(D)$$

So, I'm thinking maybe induction on the size of the matrices $A,B,D$ is the best way to approach this problem.

For $n=1, A=[a], B=[b], D=[d]$ $$\det\begin{bmatrix} a&b\\ 0&d \end{bmatrix}=ad-b\cdot 0=\det(A)\det(D)$$ Now let us suppose that the claim is true for the $n$ by $n$ matrices $A,B,D$.

We have to prove that this is true for the matrices $A,B,D$ of size $n+1$ by $n+1$.

$$\det\begin{bmatrix} A_{n+1\text{x}n+1}&B_{n+1\text{x}n+)}\\ 0&D_{n+1\text{x}n+1} \end{bmatrix}$$

I feel like I just ran into a circular argument. How can this proof be done with induction, if at all?

Martin Sleziak
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Mary
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  • This could be done using the definition of determinant that involves signs and patterns, noting that any pattern which uses an element of $B$ must also use an element of $0$ and will contribute nothing to the overall sum. Further, any pattern of $A$ can be combined with any pattern of $D$ to get a pattern from the larger matrix. After some tricky reorganizing of terms and factoring, you arrive at the intended result. – JMoravitz Jun 07 '16 at 23:10
  • Possible duplicate of Determinant of a block lower triangular matrix – JMoravitz Jun 07 '16 at 23:12
  • Split into two cases, if $A$ is singular, then the matrix is singular hence the result is true. If $A$ is invertible, then you can multiply by $\operatorname{diag} (A^{-1}, I)$ (whose $\det$ is ${1 \over \det A}$) to get the result. – copper.hat Jun 07 '16 at 23:15
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    @JMoravitz: That other question isn't about a proof by induction. Arturo mentions induction in a comment, but neither the question nor the answers do. – joriki Jun 07 '16 at 23:33
  • There is another question which asks for proof by induction: http://math.stackexchange.com/questions/1184825/upper-triangular-block-matrix-determinant-by-induction – Martin Sleziak Jun 08 '16 at 04:31
  • I think that you should be able to prove $\det\begin{pmatrix}I&U\0&V\end{pmatrix}=\det(V)$ and $\det\begin{pmatrix}X&Y\0&I\end{pmatrix}=\det(X)$ by induction (on the size of the block with the identity matrix). Then you could try to get the given matrix as a product of matrices of these types. – Martin Sleziak Jun 08 '16 at 04:35
  • But if you want to try this by induction for the matrix of the form $\begin{pmatrix}A&B\0&D\end{pmatrix}$, then induction on the size of $A$ seems reasonable. Inductive step could be Laplace expansion w.r.t. the first column. (However, you will have to prove a stronger claim - sizes of $A$ and $D$ can be different.) – Martin Sleziak Jun 08 '16 at 04:38

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