Determinant of a block matrix

[@LuisBorja1981]

Pretty colourful demonstration, I LOVE it

[@JosePerez-rw6hs]

This is so good. Your cheering voice really warmed up my heart tonight after I was stuck on this as my homework problem for hours. So much gratitude.

[@paul-thestupid1761]

The enthusiasm is strong in this one & great detailed explanation....

[@SmileyHuN]

I usually use the general Laplace expansion formula, but this is so cool ^^

[@siakamnovice]

I have a question. Must the Nullmatrix be on the lower left side? . Thus this Formula applies when the null matrix is on the lower right side?

[@sebastiansiebertz1618]

Great video, as always! And here, just for fun, the way i would have done it: Let M = [A, B ] [0, C] with A, B, C, 0 being matrices of appropriate dimensions such that M is square (A, C being square). 1.) Assume |det(A)| > 0: B = [b1, ... , bk] C = [c1, ... , ck] bn, cm column vectors of column n and m. It then follows that det(M) = det( [A, b1, b2, ... , bk ] [0, c1, c2, .... , ck] ) Using the multinearity of the determinant function we get = det( [A, b1, b2, ... , bk ] + det([A, 0, b2, ... , bk ] [0, 0, c2, .... , ck] ) [0, c1, c2, ... , ck]) where det(M1) = det( [A, b1, b2, ... , bk ] = 0 [0, 0, c2, .... , ck] ) because b1 lies in the image of A, A being invertible by assumption, the resulting matrix does not have full rank which results in a determinant of 0. Repeating this process leaves us with det(M) = det( [A, 0, 0, ... , 0 ] [0, c1, c2, .... , ck] ) Which can be rewritten as det(M) = det( [A, 0] [0, C] ) Rewriting this diagonal block matrix as a product of two simpler diagonal block matrices results in [A, 0] = [A, 0] * [1, 0] = D1 * D2 [0, C] [0, 1] [0, C] The determinants can then be directly calculated by applying the Laplace formula for determinants det(D1) = det(A) and det(D2) = det(C) All of this culminates in the final formula det(M) = det(D1)*det(D2) = det(A)*det(C) 2.) Assume: det(A) = 0 This is the trivial case, because the matrix in the lower left corner of the block matrix has only zeros as entries. This means that the first m vectors of M (m being the dimension of A) are linear dependent which implies that M does not have full rank. Therefore det(M) = 0 = 0*det(C) = det(A)*det(C) This proofs the formula det(M) = det(A)*det(C) On a side note: A very important application of this formula comes from control engineering. It is used for the proof of the separation principle, which allows for a separate design of controller and observer for a given process. For those who want to know more, see: en.wikipedia.org/wiki/Separation_principle

[@davidk7212]

Awesome video, as always. Your students are very lucky!

[@guyarbel2387]

Please do a video about cauchy binet formula

[@西元幾元]

you kept it simple and clear! appriciate it!

[@jesusurieldiazmartinez656]

Thanks, I was looking for a proof of this statement for my Matrix Algebra course

[@Handelsbilanzdefizit]

Ok, next time when I have to calculate the determinant of a big matrix, I will look if I can seperate it in simpler blocks. But "det(A)det(C)" means: / A B \ det \ 0 C / = det (AC)

[@alifruslan2728]

Great way to prove the theorem

[@johelgoni9438]

Yes! More videos about block matrix pls

[@lauroteixeira8507]

Clear explanation! Thanks too much!

[@omaymaouhadi9315]

Such a great explanation thaaanks a lot!!

[@michalbotor]

dr peyam, what are some situations in which block matrices arise naturally?

[@drpeyam]

Jordan forms, Cayley Hamilton Theorem etc

[@michalbotor]

@@drpeyam thank you! :)

[@shadali9045]

Was a very helpfull video . Thank you very much

[@benjamindavid7371]

Cheers, this is very helpful!

[@masomar2318]

Thank you so much for the great explanation :)

[@TheMauror22]

This Proof is very neat!!!

[@davide467]

When a video about the inverse of a partitioned Matrix?

[@theproofessayist8441]

So Block Matrix = Partitioned Matrix?

[@davide467]

@@theproofessayist8441 yes It is

[@MrRyanroberson1]

even better: since C is a matrix that can be applied to I, then the matrix [I,0,0,C][A,B,0,I] = [A,B,0,C] and therefore since the determinant is distributive, you get that det(C)det(A) = det([A,B,0,C]), without even needing to check if C is invertable or any of that.

[@MrRyanroberson1]

one further step of abstraction: [I,0,0,C][I,B,0,I][A,0,0,I] = [A,B,0,C] -> det(C)det(I)det(A) = det([A,B,0,C]) and since det(I) = 1, you get the result without needing to prove anything more than the determinants of *those* three matrices.

[@drpeyam]

You need to check if C is invertible

[@saththiyambharathiyan8175]

Why for 3×3 matrix determinant is calculated in the way it is defined ...?

[@drpeyam]

Watch my video on Area of a Parallelogram it’s a similar idea in 3 dimensions

[@michalbotor]

sarrus' rule is just a trick: when you're calculating a determinant of a 3 by 3 matrix M = [a b c], you're really answering the question: are these 3 3 dimensional column vectors a, b and c coplanar (detM = 0) or not. if they are, then the transformation x --> Mx = y, where x and y are 3 dimensional vectors, is going to be information destructive in a sense, that there will be no way to find matrix M^(-1), such that y --> M^(-1)y = x. We will say then, that M is not invertible or singular (behaves badly).

[@saththiyambharathiyan8175]

@@drpeyam i have already watched.... but I dont get satisfied with the explanation you gave..... for 2x2 matrix determinat calculated by the rule of elimination of one variable of 2 variable linear equations... Like ax+by=k1,cx+dy=k2 if we want to eliminate the y then adx-bcx,(ad-bc)x=k1d-k2b, since ad-bc determines solution of equations it is called determinant.... Like wise determinant of 3x3 matrix also should come from elimination of variables of 3 variable linear equation considering all elements matrix are coefficients of variables of linear equations.... Your geometrical interpretation of determinant such as area of parallelogram or volume of parallelopiped are later conceived interpretation using geometry.... But I don't get why 3×3 matrix determinat is calculated in the way it is defined.... When it try to find the determinant through eliminating variable .... I don't get the right answer.... i searched all internet and text books .... All the places i found the rule for calculation of determinant of 3×3 matrix .... But not concept behind that rule......

[@drpeyam]

It’s the same thing like for the parallelogram, but for a parallelipiped: You take a box and subtract pyramids from it until you get a parallelipiped

[@saththiyambharathiyan8175]

@@drpeyam it should be until getting parallelogram.... I tried eliminating variable .... a 1x+b1y+c1z=k1,. a2x+b2y+c2z=k2, a2x+b2y+c2z=k3 to make this as parallelogram we need to eliminate third row and third column..... How to do this....?

[@rigorless6330]

I’m.... a little confused as to what counts as a block matrix. Isn’t every matrix a block matrix? Does it have to have all 0’s in the lower left corner?

[@drpeyam]

Yeah, all 0s in the lower left corner, and the upper left and lower right blocks have to be square here

[@madhuragrawal5685]

@@drpeyam this isn't true in every context though, right? I.e. that won't always be a constraint when talking of block matrices in general?

[@meqdamqudah2331]

amazing 🔥thanks

[@nrgaimingnrgaiming6730]

this is the first time i learned some thing from a gay and to be honest he save us tomorrow in the algebra exam

[@raokondaMathematics]

What will happen when the matrix is of type [0 A; B 0]?

[@drpeyam]

Maybe -det(A)det(B)

[@raokondaMathematics]

@@drpeyam I don't think so....

[@raokondaMathematics]

@@drpeyam Actually I need the eigenvalues of the matrix of type [ 0 A; B 0] . That is first row of the matrix is [0 A], and second row of the matrix is [B, 0], where A,B are square matrices of same order and 0 is a null matrix. Here order of Null matrix=order of A=Order of B

[@biswabismitabag9017]

Thank you sir 🙏🙂

[@TheNachoesuncapo]

Hi peyam!i'm your fan number π

[@ssdd9911]

i like ur shirt and what does a matrix with a T at the top right mean?

[@drpeyam]

Transpose :) There’s a video on that

[@rocioobispo8401]

OMG :0 THANK YOU!

[@sabyasachi36]

Lovely bro

[@junhanliao1444]

nice shirt and nice vid :)

[@mohammedal-haddad2652]

Very clever.

[@BabaFroga]

lol, lucky...today we were doing determinantes on University classes xD

[@morriganfayehaney1372]

So happy

[@MegaICS]

you could have showed the original thesis by noting that [A, B; 0, C] = [A, B/2; 0, I] * [I, B/2; 0, C] and using det(M1*M2) = det(M1) * det(M2) but maybe det(M1*M2) = det(M1)*det(M2) uses this very proof so you would have a circular argument...

[@shrutikagawade1155]

I don't understand what your teaching 😔😔

[@drpeyam]

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