Question

Can we represent a matrix in a matrix?

Answer 1

For instance, if A is a 2x2 matrix,\[\LARGE AI = \left[\begin{matrix}A & 0 \\ 0 & A\end{matrix}\right]\]

Answer 2

hmmmmm, not sure. havent made it that far XD

Answer 3

Suppose \[A=\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\] Then we have: \[\left[\begin{matrix}a & b\\ c & d\end{matrix}\right] I = \left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right]\] Which is really equivalent to saying we can represent all nxn matrices as (kn)x(kn) matrices as bands down the diagonal... which seems wrong since then it seems to imply that the determinant is squared meaning it's different... But that's not really enough for me to say that this is completely wrong yet.

Answer 4

We can have a block matrix:
http://en.wikipedia.org/wiki/Block_matrix

Answer 5

Let \(A=\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\), then
\[\left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right]=\left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right] = AI\]

Answer 6

Sure, but this seems to violate the rule since AI=A they should have the same determinant.

\[\Large \det(AI)=\det(A) \\ \Large \det(A)^2 \ne \det(A)\]

You see what I mean, when we take the determinant of the block matrix? However I am not opposed to the idea of det(A) being the "principal" determinant or something.

Answer 7

Here is another interesting "weird" use I found using the Cayley-Hamilton Theorem where it doesn't quite line up. A is a 2x2 matrix and has determinant d and trace t.

\[\left[\begin{matrix}t & -d\\ 1 & 0\end{matrix}\right] \left(\begin{matrix}A \\ I\end{matrix}\right) = \left(\begin{matrix}A^2 \\ A\end{matrix}\right)\]

You see if we expand the "vectors" as 2x2 matrices we are multiplying a 2x2 matrix with a 4x2 matrix which is technically not allowed. But it works out since these matrices are indeed linear combinations of each other.

Answer 8

This makes me wonder that perhaps the regular matrix multiplication is really in fact just a special case of a more general matrix multiplication where you simply resize what you are multiplying to match.

Answer 9

\[\Large A_{ij}B_{jk}=C_{ik}\] what I'm saying is change this to do something like we had in the last example I showed with matrices in the vectors I could write it with just their indicies perhaps:\[\Large A_{22}B_{42}=A_{22}[B_{21}(b_{22})]=[C_{21}(c_{22})]=C_{42}\] By exchanging common factors and factoring maybe.

Answer 10

I don't understand your claim:

"Sure, but this seems to violate the rule since AI=A they should have the same determinant.
\(\det(AI)=\det(A)\), \(\det(A)^2≠\det(A)\) "

Answer 11

To make it clear,
\[\left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right]=\left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right] = \left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right]\times \left[\begin{matrix}I_4 & 0\\ 0 & I_4\end{matrix}\right] = AI_8\]

Answer 12

Ahh ok, so what I'm saying is I am pulling the matrix A inside the other matrix like you would commonly pull in a scalar: \[ AI =\left[\begin{matrix}a & b\\ c & d\end{matrix}\right] \left[\begin{matrix}1& 0\\ 0 & 1\end{matrix}\right] = \left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right]\]

Answer 13

This may be better or worse...
Let \(A=\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\),\(A' = \left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right] =\left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right] \)

\[\det(AI) = \det( \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\times \left[\begin{matrix}1 & 0\\ 0 & 1\end{matrix}\right]) =\det ( \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]) = \det(A)\]

\[\det(A'I) = \det( \left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right] \times \left[\begin{matrix}I_4 & 0\\ 0 & I_4\end{matrix}\right]) =\det ( \left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right]) = \det(A^2) = (\det(A))^2\]

Answer 14

I think no, because if the dimension of the matrix at left side is n*n, then dimension of the Matrix at right side is (2n)*(2n)

Answer 15

for example, from physics, if your thesis is true, then we ca say that Pauli matrices and Dirac matrices are the same matrices, which is the case n=2

Answer 16

@Michele_Laino What are you referring to when you say "I think no, ..."?

Answer 17

to the this equality:
\[A=\left[\begin{matrix}A & 0 \\ 0 & A\end{matrix}\right]\]

Answer 18

oops.. to this equality:....

Answer 19

Essentially what I'm saying is perhaps we should really consider matrices to be able to be represented periodically and that
\[A=\left[\begin{matrix}A & 0 \\ 0 & A\end{matrix}\right]\]

Is in fact a correct representation.

Answer 20

No, it is not. That why I need to define A'

Answer 21

You say it is not, but there is no reason why it can't be. I can choose to define it to be this way if I choose.

Answer 22

\[A=\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\]

\[A' = \left[\begin{matrix} \left[\begin{matrix}a & b\\ c & d\end{matrix}\right]& \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] \\ \left[\begin{matrix}0 & 0\\ 0 & 0\end{matrix}\right] &\left[\begin{matrix}a & b\\ c & d\end{matrix}\right]\end{matrix}\right] =\left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right]\]

Two matrices can only be identical when
1) they are of the same size
2) the corresponding entries are identical.
This is the definition.

Answer 23

Sorry, @Kainui if your equality, is true, then Puli matrices are able to act into a space of four-spinors, and of course that is false.

Answer 24

When you define in that way, you are just abusing the notation and make it unclear.

Answer 25

oops.. Pauli matrices...

Answer 26

Now to account for it, all I have to say is det(A) is the "principal determinant" and the version of this matrix with the direct sum of a matrix n times has the same determinant to the nth power.

Arguing with this definition as you currently are is like saying there is only a single square root of 1. You can say the square root of 1 is defined to be 1 but I am saying it can be both 1 and -1 depending on what you need.

Answer 27

So, what is your question?

Answer 28

The question in the question box "Can we represent a matrix in a matrix?"
--> Yes. Example: Block matrix

Question about determinant: "\(\det(AI)=\det(A)\), \(\det(A)^2\ne \det(A)\)"
--> I don't know since I don't know if the A are 2x2 matrix are a 4x4 matrix. I have shown you both cases in 6 posts above. An amendment in that post, the last line should be
\(\det(A'I)\)
\(= \det( \left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right] \times \left(\begin{matrix}I_2 & 0\\ 0 & I_2\end{matrix}\right]) \)
\(=\det ( \left[\begin{matrix}A & 0\\ 0 & A\end{matrix}\right]) \)
\(= \det(A^2) = (\det(A))^2\)

Just to make my presentation clearer, I chose to define A' instead of using A since A' and A are not the same.

Answer 29

I am not disagreeing with you, I just think it might be time to depart from the standard definition, and in doing so, does this work? If so, how, or are there holes in it that cause it to be irreconcilable?

Answer 30

One more thing, when two matrices are the same, no matter how you name them, they are still the same.
Example:
If \(A=\left[\begin{matrix}1 & 2\\ 3 & 4\end{matrix}\right]\), \(B=\left[\begin{matrix}1 & 2\\ 3 & 4\end{matrix}\right]\), then A=B
BUT if
\(A=\left[\begin{matrix}1 & 2\\ 3 & 4\end{matrix}\right]\), \(B=\left[\begin{matrix}1 & 2\\ 4 & 3\end{matrix}\right]\), then A≠B
Also, if
\(A=\left[\begin{matrix}1 & 2\\ 3 & 4\end{matrix}\right]\), \(A=\left[\begin{matrix}1 & 2 & 5\\ 3 & 4 & 6\end{matrix}\right]\), then the first A is not the same as the second A

Answer 31

Yeah, I am not really worried about basic things like this. From the traditional standpoint these are obvious statements of fact.

Answer 32

So far, I know we can compute ordinary multiplication using blocks as entries, given that the blocks are compatible.

Answer 33

If the blocks are not compatible, you'd better re-partition it.

Answer 34

search up Kronecker Product

Answer 35

WOW! This seems to fill some of the missing part in my book about block matrix! Thanks a lot!!!

Answer 36

@watchmath Ahh this is interesting, thanks.