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# 矩陣積的分配性質 (英): 矩陣的積展線分配性質

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- Let's say we have three matrices, A, B, and C.
- And let's say that B and C are both m by n matrices, and that
- A is a, let's call it a k by m matrix.
- And what I want to do is figure out whether matrix
- products exhibit the distributive property.
- So let's test out A times B plus C.
- And of course these are all matrices.
- So B, just to make things clear, the matrix B could be
- represented as just a bunch of column vectors, B1, B2, all
- the way to Bn.
- And the matrix C can also be represented as just a bunch of
- column vectors.
- So could the matrix A, but I don't have to
- draw that just yet.
- So the matrix C could just be represented as a bunch of
- column vectors, C1, C2, all the way to Cn.
- Maybe I should've drawn this taller.
- These are column vectors, so they actually have some
- verticality to them.
- I think you've seen that multiple times.
- So what is A times B plus C?
- Well, let's figure out what B plus C is.
- This is equal to A times B plus C.
- When you add B plus C, the definition of matrix addition
- is, you just add the corresponding columns.
- Which essentially boils down to adding the
- corresponding entries.
- So the first column is going to be equal to B1 plus C1.
- The second column is going to be B2 plus C2.
- And you're going to go all the way to the nth column.
- It's going to be Bn plus Cn.
- Now by our definition of matrix-matrix products, this
- product right here is going to be equal to the matrix, where
- we take the matrix A and multiply it by each of the
- column vectors of this matrix here, of B plus C.
- Which as you can imagine, these are both m by n.
- In fact they both have to have the same dimensions for this
- addition to be well defined.
- So this is going to be an m by n matrix.
- I already you told you that A is a k by m.
- And we know this is well defined because A has the same
- number of columns as B plus C has of rows.
- So this is well defined.
- And this is going to be equal to-- let me switch colors
- again-- A times the column vector B1 plus C1.
- The second column is going to be A times the column vector
- B2 plus C2.
- I'm running out of space.
- This is going to be all the way to A times the column
- vector Bn plus Cn.
- This is our definition of matrix-matrix products.
- You just take the first matrix and you multiply it times each
- of the column vectors of the second matrix.
- And we can say that because we've already defined
- matrix-vector products.
- So what is this thing on the right equal to?
- I'll keep switching colors.
- We know that matrix-vector products exhibit the
- distributive property.
- I don't even remember when I did that video.
- But we've assumed it for a while.
- It's a very trivial thing to prove.
- So each of these columns is going to be equal to, let me
- write this way.
- This guy right here can be re-written.
- The first column is going to be A times column vector B1,
- plus A times the column vector C1.
- This term right there is the same thing as
- that term right there.
- The next one is going to be AB2 plus matrix A
- times the vector C2.
- And then the nth column is going to be the matrix-- keep
- going-- A times the column vector Bn, plus matrix A times
- the column vector C.
- Just like that.
- Now we can write this matrix as the sum of
- two different matrices.
- So what is this going to be equal to?
- This is equal to-- let me see, I'll just write it right
- here-- AB1 as the first column, AB2 as the second
- column, all the way to ABn as the third column.
- So that's these terms right there.
- And then if I were to add to that the matrix A times vector
- C1, A times the column vector C2-- these are just the
- different columns of this matrix-- and we just then have
- the matrix A times the column vector Cn.
- These represent these terms.
- So clearly if I add these two matrices, I just add the
- corresponding column vectors and I'll get
- this matrix up here.
- But what is this equal to?
- This right here, by definition, this is the matrix
- A times the matrix B.
- The definition of matrix products is you take the first
- matrix and multiply times the column vectors
- of the second matrix.
- And by the same argument, I guess you could say, this is
- equivalent to A times C.
- And all of this-- remember we just had a bunch of equal
- signs-- is equal to A times B plus C.
- So now we can say definitively that as long as the products
- are well defined and the sums are well difined, so they all
- have to have the correct dimensions, that A times B
- plus C is equal to AB plus AC.
- So matrix products do exhibit the distributive property, at
- least in this direction.
- And I say that because remember matrix products are
- not commutative.
- So we don't know necessarily that B plus C times A is
- equivalent to that.
- In fact, most of the time these two things are not
- equivalent.
- So we don't know quite yet that if we reversed this,
- whether it's still going to exhibit the
- distributive property.
- So let's try to do that.
- And I'll do it a little bit quicker, because I think you
- know the general argument here.
- So let's take B plus C times A.
- And I'll just write A as its column vectors.
- A1, A2, all the way to-- A has m columns if I remember
- correctly, right A has m columns-- so
- all the way to Am.
- And by the definition of matrix products, this is going
- to be equal to the matrix-- B plus C is
- just a matrix, right?
- We can represent it as the sum of two matrix, but
- it is just a matrix.
- So it's B plus C times each of the column vectors of A.
- So it's going to be equal to B plus C times a1, B plus C
- times a2, all the way to B plus C times an.
- And, once again, it was many videos ago that I think we
- showed that matrix-vector products are distributive, so
- we can just distribute this vector
- along these two matrices.
- And if I haven't proven it yet, it's actually a very
- straightforward proof to do.
- So we could say that this is equal to Ba1 plus Ca2.
- That's the first column.
- The second column is B times a2, plus C times a2, all the
- way to B times an, plus C times an.
- And then what is this equal to?
- I'll write it out.
- This is equivalent to B times-- this is a1, no this is
- an a1 right here-- a1, and then B times a2, all the way
- to Ban plus the vector C times a1, C times a2, all the way to
- C times an, right?
- This guy represents these terms right there, and this
- guy represents the first terms in each of
- these column vectors.
- And this, by the definition of matrix products, is just
- equivalent to BA, and then this is just equivalent to CA.
- So now we've seen that the distributive property works
- both ways with matrix-vector products.
- That B plus C times A is equal to BA+CA, and that A times B
- plus C is equal to AB plus AC.
- Now the one thing that you have to be careful of is that
- these two things are not equivalent [UNINTELLIGIBLE].
- We just figured out that this guy is equal to BA plus CA.
- So the distribution works in both ways.
- But when you're dealing with matrices, it's very important
- to keep your order.
- So this is going to be you have the A second here.
- So it is BA plus CA.
- You can't say that this is equal to AB plus AC.
- You can't just switch these up.
- Because we've shown multiple times, or we've talked about
- it multiple times, that matrix products are not commutative.
- You can't just switch the order of the products.
- But we've at least shown in this video that the
- distributive property works both ways.

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