Proving CA⁻¹B+D=0 to Demonstrate Rank(A)=Rank([A B])=n

[jwqwerty]

if A B C D are nxn matrices such that rank(A)=rank([A B])=n ([A B] : 2nx2n matrix)
[C D] [C D]

we want to show that D=CA‑¹B (A‑¹: inverse matrix of A)

this is what i have tried:
[In 0] [A B] = [A B ] (here, In refers to identity nxn matrix)
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]

det( [In 0] [A B] ) = det ( [A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]

det( [In 0] ) det( [A B] ) = det ([A B ] )
[-CA‑¹ In] [C D] [0 -CA‑¹B+D]

by property of determinant of block matrix,

det(In)det(In)det( [A B] ) = det(A)det(CA‑¹B+D)
[C D]

since A is invertible and [A B] is not invertible, their determinents are not zero and zero respectively
[C D]

Thus, det(CA‑¹B+D)=0

If i can show that CA‑¹B+D=0, then my proof ends.
But i can't show this.

can anyone show that CA‑¹B+D=0?

 

[Fredrik]

I don't understand the question. What do you mean by [A B] for example? You said that it's 2n×2n, but not how it's related to A and B. And if it's 2n×2n, then how are you multiplying it by an n×n matrix? ("In" is a very confusing notation by the way).

 

[FactChecker]

I don't understand the question. What do you mean by [A B] for example? You said that it's 2n×2n, but not how it's related to A and B. And if it's 2n×2n, then how are you multiplying it by an n×n matrix? ("In" is a very confusing notation by the way).

More questions. Maybe I am missing something. I assume that [A B] is just the nx2n matrix adjoining columns of A and B. But what does "[C D] [C D]" in line 2 say about C and D? And why are there other miscellaneous lines of "[C D]" scattered around? And I don't what "ln" means.

 

[Fredrik]

And I don't what "ln" means.

He explained that one. It's an I, not an l, and In is his notation for the n×n identity matrix.

 

[jwqwerty]

1

 

[jwqwerty]

Ok i have rewritten my question since it looks very confusing

 

[jwqwerty]

More questions. Maybe I am missing something. I assume that [A B] is just the nx2n matrix adjoining columns of A and B. But what does "[C D] [C D]" in line 2 say about C and D? And why are there other miscellaneous lines of "[C D]" scattered around? And I don't what "ln" means.

I have rewritten my question

 

[jwqwerty]

He explained that one. It's an I, not an l, and In is his notation for the n×n identity matrix.

I have rewritten my question Fredrik!

 

[Fredrik]

The calculation looks correct, but I don't understand why you're doing it. You still haven't stated the problem in a way that makes sense. In particular, you haven't said anything to indicate that C and D are anything but arbitrary n×n matrices.

 

[jwqwerty]

The calculation looks correct, but I don't understand why you're doing it. You still haven't stated the problem in a way that makes sense. In particular, you haven't said anything to indicate that C and D are anything but arbitrary n×n matrices.

I want to show that det(D-CA‑¹B )=0 implies D-CA‑¹B=0
But i don't know how. Or are there any other ways?

 

[Fredrik]

But you haven't given any conditions on C and D. What if for example $A=I$ and $B=0$? Then $D-CA^{-1}B=D$. So you're trying to prove that the arbitrary(?) matrix D is zero.

 

[FactChecker]

you haven't said anything to indicate that C and D are anything but arbitrary n×n matrices.

The statement that the rank of the full matrix \bigl(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\bigr)is the same as the rank of A says something about C and D

P.S. Seems like it takes forever now for the LaTex to be interpreted. It did previews immediately, but it looks like the post will never be converted.

 

[Fredrik]

The statement that the rank of the full matrix \bigl(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\bigr)is the same as the rank of A says something about C and D

The statement in post #1 just says that A and [A B] both have rank n. I don't see a way to interpret the post as saying anything about C or D.

 

[FactChecker]

The statement in post #1 just says that A and [A B] both have rank n. I don't see a way to interpret the post as saying anything about C or D.

Yes. You have to look at the PDF attachment in a later post to see the correctly formatted problem statement. In the original post, the [C D] was left justified and should have been the lower half of the [A B]. It should have been
\bigl(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\bigr)

 

[Fredrik]

Yes. You have to look at the PDF attachment in a later post to see the correctly formatted problem statement. In the original post, the [C D] was left justified and should have been the lower half of the [A B]. It should have been
\bigl(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\bigr)

You mean the JPEG that was attached to three different posts? There's no problem statement in it. There's just a definition of [A B] and some calculations that may or may not be relevant to the problem. It's impossible to tell without a problem statement.

OK, I see now, by quoting post #1, that the OP intended [C D] to be the lower half and [A B] the upper half. So now the only problem is that I've done so much work trying to decode what he meant that I don't want to look at this problem anymore.

 

[jwqwerty]

The statement that the rank of the full matrix \bigl(\begin{smallmatrix} A&B\\ C&D \end{smallmatrix}\bigr)is the same as the rank of A says something about C and D

you've got any idea on solving the problem?

 

[FactChecker]

you've got any idea on solving the problem?

No. My gut feeling is that it is wrong. Saying that the determinant is 0 usually just means that a non-trivial set is mapped to zero. I don't see how to prove that everything is mapped to zero. I have my doubts.

 

[FactChecker]

1) I think the concentrating on the determinants is a mistake. It looses too much information. Saying that a determinant is zero is not a strong enough statement to draw the conclusion. I think that the initial conditions imply that C, D, and B are transformations of A, where all the transformations are invertible. I think you should use that in your proof instead of determinants.

2) I also think that there is a sign error. If we define B, C and D as B = C = D = A, doesn't that satisfy the initial conditions? But then we would be trying to prove that CA‑¹B+D = 2A =0. That can't be right. Oh. I see that the original problem was to show that D = CA‑¹B. So that is CA‑¹B - D =0.