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There is a thing that I don't understand about how the determinant of a matrix could be split this way: $$ \begin{vmatrix} a & b\\ c & d \end{vmatrix}= \begin{vmatrix} a & 0\\ c & d \end{vmatrix} + \begin{vmatrix} 0 & b\\ c & d \end{vmatrix} $$

I read that this "split" is possible because of some linearity property but I still cannot figure out how the "split" is formed.

It could be even split further like: $$ \begin{vmatrix} a & 0\\ c & d \end{vmatrix} = \begin{vmatrix} a & 0\\ c & 0 \end{vmatrix} + \begin{vmatrix} a & 0\\ 0 & d \end{vmatrix} $$

How do these "splitting" go about? Thanks for any help.

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    $\begingroup$ The determinant is linear in each row and in each column. That's the property being used. $\endgroup$ Jul 27, 2011 at 19:51
  • $\begingroup$ Thanks. hmm..but I thought it has to add rows of itself? Like a+a, b+b? This one isn't adding anything so how can it split? $\endgroup$
    – xenon
    Jul 27, 2011 at 19:57
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    $\begingroup$ The first row in your first example is $(a,b) = (a,0)+(0,b)$; If you think of the determinant as a function of the rows, then you have $$D((a,b),(c,d)) = D\Bigl((a,0)+(0,b),(c,d)\Bigr) = D((a,0),(c,d))+D((0,b),(c,d)).$$In the second example, you are expressing the second rows $(c,d)$ as the sum $(c,d)=(c,0)+(0,d)$. $\endgroup$ Jul 27, 2011 at 20:08
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    $\begingroup$ Thanks Arturo! Very sharp and clear explanation! Thanks!! :) $\endgroup$
    – xenon
    Jul 27, 2011 at 20:25

2 Answers 2

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Good observation. We can indeed generalize this observation as follows. Suppose $A$ is any $n \times n$ matrix and let $A_1, A_2, \ldots, A_n$ be its rows. Now, suppose $A_1$ is written as the sum of two vectors $A_1'$ and $A_1''$. Let $A'$ be the matrix you get if you replace the row $A_1$ in $A$ by $A_1'$; similarly define the matrix $A''$. Then, we have the identity: $$ \det(A) = \det(A')+\det(A''). $$ Now, of course, the same property holds if the $i^{\rm th}$ row is considered instead of the first row; in fact, we can consider the $i^{\rm th}$ column as well.

As an example, we have: $$ \left| \begin{array}{ccc} a & b'+b'' & c \\ d & e'+e'' & f \\ g & h'+h'' & i \end{array} \right| = \left| \begin{array}{ccc} a & b' & c \\ d & e' & f \\ g & h' & i \end{array} \right| + \left| \begin{array}{ccc} a & b'' & c \\ d & e'' & f \\ g & h'' & i \end{array} \right| .$$

Can you see why this property explains your two observations?

Proof. The property I claimed has a simple proof, so I will just include that here for completeness. By the Laplace expansion of the determinant, notice that the determinant function is linear in any single row or column of the matrix, which would readily give the property.

Let me also explain the proof in more detail. Let $A_i$ be the $i^{\rm th}$ row, and suppose that we write the $i^{\rm th}$ row $A_i$ as $A_i = A'_i + A''_i$. Now, let $M_{i,j}$ be the $(i,j)^{\rm th}$ minor of $A$. Then, by expanding along the $i^{\rm th}$ row, we have

$\begin{eqnarray*} \det(A) &=& \sum_{j=1}^n (-1)^{i+j} A_{i,j} M_{i,j} = \sum_{j=1}^n (-1)^{i+j} A'_{i,j} M_{i,j} + \sum_{j=1}^n (-1)^{i+j} A''_{i,j} M_{i,j} \\ &=& \det(A')+\det(A''), \end{eqnarray*}$

which is what we wanted.

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    $\begingroup$ Thanks for showing the proof! Now, I understand. Thanks! :) $\endgroup$
    – xenon
    Jul 27, 2011 at 20:26
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Try to expand the determinants: $(ad - bc) = (ad - c\cdot 0) + (0 \cdot d - bc) = ad - bc$

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  • $\begingroup$ yea... i did this before but this is like working in reverse. it did work. but the thing is how was the matrix being split to that way. what was the process like? $\endgroup$
    – xenon
    Jul 27, 2011 at 19:52

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