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Given a square matrix $A,D\in \Bbb{R}^{n \times n}$

a. $\det(\bigl(\begin{smallmatrix} A & 0 \\ 0 & I \end{smallmatrix} \bigr)\bigr)$ = det(A) (Already done)

b. The determinant of a triangular matrix equals to the product of its diagonal entries. (Already done)

c. Any real square matrix $A$ may be decomposed as $A = QR$ where $Q$ is an orthogonal matrix (its columns are orthogonal unit vectors meaning $Q^TQ = QQ^T = I$) and $R$ is an upper triangular matrix.

If $A$ is invertible, then the factorization is unique. This is called $QR$ decomposition.

Prove that det $(\bigl( \begin{smallmatrix} A & B \\ 0 & D \end{smallmatrix} \bigr)) = \det(A)\cdot \det(D)$

Hint: First prove that $(\bigl( \begin{smallmatrix} A & 0 \\ 0 & D \end{smallmatrix} \bigr)) = \det(A) \cdot \det(D)$ using the fact that $\det(AB) = \det(A)\cdot \det(B)$ (for every two matrices $A$, $B$) and part a and then use the $QR-$decomposition.

knownothing
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  • I figuired out previous 2 parts of this question, but have no idea how to deal with this one. I can tell det(A B, 0 D) = det(A)det(D) because it is one of the special cases of Laplace Expansion but i don't know how to theoretically prove it. – knownothing Sep 13 '20 at 13:22
  • You should add it to your question. It's better. Otherwise the post will be deleted or closed. – user577215664 Sep 13 '20 at 13:23
  • Part a, as I mentioned above, asked for a proof of det(A 0, 0 D) = det(A) and I just used Laplace expansion so that it should be equal to det(A)det(I) and the determinant of identity is 1, so that turns out to be det(A), For part b, it just asked us to prove that the determinant of a triangular matrix equals to the product of its diagonal entries, Basically, that's it. – knownothing Sep 13 '20 at 13:27
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    Wiil do, thanks for the reminder. – knownothing Sep 13 '20 at 13:27
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    Not exactly the same, I think it would be easier if it is not asking for a QR-decomposition-involved solution. Thanks, anyway. – knownothing Sep 13 '20 at 13:33

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