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Let det $A = \det(\begin{bmatrix}B& 0\\ 0& I_mI\end{bmatrix})$; $B$ and $D$ are square matrices. $I_m$ is an identity matrix of size $m$.

I keep reading that it is obvious that we can view the RHS as a function of the $n$ rows of $A$ and that this function satisfies all three axioms for a determinant.

However, I fail to see that this is true.

  1. $\det(\begin{bmatrix}B& 0\\ 0 &I_m\end{bmatrix}) = \det(\begin{bmatrix}B& 0\\ 0& I_m\end{bmatrix}$)?
  2. $F = tA$, then $\det F = \det(\begin{bmatrix}tB &0\\ 0& tI_m\end{bmatrix} = t\det(A)$?
  3. determinant of the identity matrix = 1. ?

The question marks mean that these facts are apparently so obvious to everyone else that no one proves them but just states them.

(Big picture. I'm hoping getting clear on this will help me on my bigger question: how to prove $\det A = \det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$. I have found this question elsewhere (Determinant of a block lower triangular matrix) but it seems way too handwavy. Other answers seem way too convoluted with too little explanation.)

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    $\begingroup$ The answer in the question you linked is not handwavy at all, in fact it is quite constructive. $\endgroup$ – Emily Apr 23 '14 at 20:43
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@larry:

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

seems wrong, did you mean

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

2) if $F = tA$ then $det(F) = t^m det(A)$ this is easy to see if $A = I_m$, using the remark (2a) below

2a) the determinant of a diagonal matrix is just a product of diagonal elements, say $\det(\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{bmatrix}) = 2*3*4 = 24$ and yes, $det(I_m) = 1$

3) I don't know a better way to prove $\det A = \det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$ rather than using the definition of the determinant. This is tedious, and will take a while to set up here.

But you can get the general idea from the following example $\det A = \det(\begin{bmatrix}b & 0\\ C &D\end{bmatrix}) = b \det(D)$, that is, matrix B is now 1-by-1.

This should follow if you decompose the determinant using the first row. (Look up "cofactor decomposition", a 3-3 example is here: http://en.wikipedia.org/wiki/Determinant#3.C2.A0.C3.97.C2.A03_matrices).

3b) We can keep nibbling at the general formula by considering $\det A = \det(\begin{bmatrix}I_m & 0\\ C &D\end{bmatrix}) = \det(D)$, that is, matrix B is now $I_m$. Now, when you try to decompose, you should get $\det A = 1\cdot 1 \cdot \dots 1 \cdot \det(D)$

4) Come to think of it, we can finish the proof, if you are familiar with Gaussian elimination. I will be a bit sketchy with details, though.

Step 1: Do row operations on matrix D to make it low triangular, say $L = TD$. The result of applying the same row operations on the whole matrix is $\begin{bmatrix}B & 0\\ TC & L\end{bmatrix}$

BTW, row operations (take $Row_i + b*Row_j$, put into $Row_i$) do not change the determinant.

Step 2: Divide each of the rows of L (and also TC) by the element on the diagonal of L (this is like doing a second step of Gaussian elimination). Since det(L) is the product of its diagonal elements (you can see this by repeatedly using cofactor expansion), this is equivalent to dividing the whole thing by det(L) = det(D).

So far, we have $det(A) = det(D) * det(\begin{bmatrix}B & 0\\ FTC & FL\end{bmatrix})$ where FL has ones on the diagonal.

Step 3: Apply more row operations (which are equivalent by multiplying by some other matrix, G) to turn FL into $I_m$. Thus, you get

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

Now you can go back to the note 3b from above.

QED!

By the way: $det(\begin{bmatrix}B & C\\ E & D \end{bmatrix})$ generally does not equal $det(B)det(D) - det(C) det(E)$. Go figure!

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To understand that $$\det A = \det(\begin{bmatrix}B& 0\\ C &D\end{bmatrix}) = \det(B)(\det(D))$$ try to use determinant definition $$\det(A) = \sum_{for\ all\ (i_1,i_2,\dots,i_n)}(-1)^{inversionsCount((i_1,i_2,\dots,i_n))}a_{1i_1}a_{2i_2}\cdots a_{ni_n}$$ where $(i_1,i_2,\dots,i_n)$ transposition of $(1,2,\dots,n)$.

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  • $\begingroup$ No, you shouldn't use the def of a determinant to answer this problem. $\endgroup$ – larry Apr 1 '14 at 23:54
  • $\begingroup$ That would be difficult, as any theorems about determinant identities are simply extensions of the definition.... $\endgroup$ – Emily Apr 23 '14 at 20:13

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