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I'm working my way through Franklin's Matrix Theory and am stuck on question 1.3.2.

The question:

Let $A$ have the form:

$$ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\ 0 & 0 & a_{33} & a_{34} & a_{35}\\ 0 & 0 & a_{43} & a_{44} & a_{45}\\ 0 & 0 & a_{53} & a_{54} & a_{55} \end{bmatrix} = \begin{bmatrix} A_1 & \cdots \\ \bigcirc & A_2 \end{bmatrix} $$

Show that:

$$\det{A} = \det{A_1} \cdot \det{A_2}$$

The only argument I'm confident enough in making is that $\det{A}$ will ultimately consist only of elements from $A_1$ and $A_2$.

My reasoning is this:

Each term $s(j_1,\cdots,j_n)\prod_{i=1}^{n}a_{i j_i} $ of $\det{A}$ is non-zero only if whenever $1 \leq i \leq 2$, $1 \leq j_i \leq 2$ and whenever $3 \leq i \leq 5$, $3 \leq j_i \leq 5$.

This is because if when $1\leq i \leq 2$ we chose $j_i$ such that $3 \leq j_i \leq 5$ we would unavoidably have to choose a zero when $3 \leq i \leq 5$ as we must make one choice for each row 3,4,5. We want non-zero terms so for each choice of $j_i$ we pick from columns 3,4,5. However, one of the columns that would give us an element in $A_2$ (non-zero) is no longer available due to our previous choice, forcing us to take a zero.

The case which arises when $3 \leq i \leq 5$ and we choose $1 \leq j_i \leq 2$ is obvious.

I'm not confident in asserting anything beyond that.

Any help is much appreciated!

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  • $\begingroup$ Consider polylinear , skew-symmetric function of $n$ vectors. Then $F(v_{1} \dots v_{n}) = F(e_{1}\dots e_{n})detA$, where $A$ matrix of coordinates $v_{1} \dots$ in basis $e_{1} \dots$ $\endgroup$
    – openspace
    Feb 21 '17 at 7:46
  • $\begingroup$ Thank you, unfortunately I can't see the connection though. Probably just because I'm not familiar enough with polylinear functions or skew-symmetry. $\endgroup$
    – Casey
    Feb 21 '17 at 8:21
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Hint:

Here you can see the proof for a matrix of the form $$ \begin{bmatrix}A&0\\C&D \end{bmatrix} $$

with a little effort you can modify it for your case.

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  • $\begingroup$ Ah, it's pretty obvious then in the matrix decomposition case I suppose. Following from the determinant property $\det{AB} = \det{A}\det{B}$. Now I'm going to try and learn more about a proof involving the Leibniz definition of the determinant. Thank you! $\endgroup$
    – Casey
    Feb 21 '17 at 8:18

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