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From the answer to this question, how can one prove that $$\begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} = \begin{vmatrix} a & b \\ c & d-\lambda \end{vmatrix} + \begin{vmatrix} -\lambda & b \\ 0 & d-\lambda \end{vmatrix}$$

I mean it is not adding a multiple of a row / column to another row / column, and it is clear that the equality holds, but how to prove this ? Which proper of the determinant is used in here ?

The user who wrote the answer told me to check out this question, but I still don't get it.

Our
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    Determinant is linear as a function of each column (with the others held fixed). – quasi Jul 19 '17 at 14:59
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    You may find it useful to consider that the determinant is a multi-linear functional. – Chickenmancer Jul 19 '17 at 15:00
  • @Chickenmancer You are right, I totally missed that. – Our Jul 19 '17 at 15:03
  • Depending on the context in which you were exposed to the determinant, you may have never seen it defined that way. The determinant was a mystery to me until I learned about exterior algebras. Also, my original comment is practically useless without the caveat "in the future." – Chickenmancer Jul 19 '17 at 15:15
  • @Chickenmancer I couldn't completely understand what you mean by the caveat ? – Our Jul 19 '17 at 15:21
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    If you had thought about it that way originally, then you wouldn't have asked the question, since it would be a defining property of the determinant. Also, it seems to trivial to say "think about it in the way that makes it work." Although that is half the battle of doing anything in mathematics :P – Chickenmancer Jul 19 '17 at 15:26

2 Answers2

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You can just multiply it out using the definition of the determinant:

The left hand side is $$(a-\lambda)(d-\lambda)-bc = ad-bc-(a+d)\lambda + \lambda^{2}$$

The right hand side is

$$a(d-\lambda)-bc + (-\lambda)(d-\lambda) = ad-bc -(a+d)\lambda +\lambda^{2}$$

pwerth
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  • The OP is asking "which property of the determinant is being used." – Chickenmancer Jul 19 '17 at 15:01
  • Thanks for your answer, but I have already stated that "it is clear that the equality holds", and I also said that I'm looking for a proof. – Our Jul 19 '17 at 15:01
  • My answer is certainly a proof, it's just algebraic as opposed to using linearity of the determinant – pwerth Jul 19 '17 at 15:02
  • That is correct that your's is proof, but it doesn't give any insight about the nature of determinant, so we might called it a "trivial proof" :) – Our Jul 19 '17 at 15:05
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Let $$v_1=(a,c) $$ $$v_2=(-\lambda,0) $$ $$v_3=(b,d-\lambda) $$

we just use the fact that

$$det (v_1+v_2,v_3)=det (v_1,v_3)+det (v_2,v_3) $$

It is a bilinear form.

hamam_Abdallah
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