Addition and scalar multiplication done entry by entry, and why orders must agree

0:00Two workshops make the same three grades of a product, in two sizes. The first workshop's output, written as three rows and two columns: eighty and sixty on the top row, seventy-five and sixty-five on the middle, ninety and eighty-five on the bottom. The second workshop's, in exactly the same arrangement: ninety and fifty, seventy and fifty-five, seventy-five and seventy-five. How much was made altogether? Before working anything out, write down the shape of the answer. The top left total is eighty plus ninety. Beside it, sixty plus fifty. Then seventy-five plus seventy, sixty-five plus fifty-five, ninety plus seventy-five, and eighty-five plus seventy-five.

0:55Six positions, six additions, and not one of them knows about any of the others. That is the whole rule. Work them out and you get one hundred and seventy, one hundred and ten; one hundred and forty-five, one hundred and twenty; one hundred and sixty-five, one hundred and sixty. Three rows and two columns in, three rows and two columns out. The order of the answer is the order both of them already had, because every position of the answer came from one position of each.

1:32That generalises with no effort whatever. Take two arrays of the same order. The entry of their sum at any position is the sum of the two entries sitting at that position, and the order of the result is the order they shared. Notice how little that rule does. It never multiplies two entries together. It never looks at a row as a whole, or a column as a whole. It does the same one-line thing at every position independently, and it does not care in the slightest what kind of number is sitting there.

2:10Here is an instance that makes that last point sharply. Two arrays of two rows and three columns. The first holds root three, one and minus one, above two, three and zero. The second holds two, root five and one, above minus two, three and one half. Add them position by position. Top row: two plus root three. One plus root five. And minus one plus one, which is nothing at all.

2:41Bottom row: two plus minus two, which is also nothing. Three plus three is six. And zero plus one half is one half. Two of the six positions cancelled to zero. Two hold a root. And here is the thing worth pausing on: root three and root five are still sitting there separately, in different positions, and they will never meet, because addition never brings two entries from different positions into contact. Two distinct roots appear in the answer and neither has been turned into a decimal.

3:18So the arithmetic in this operation is almost free. Which raises the obvious question: if the rule is that cheap, where is the content? It is in the question of when you are allowed to apply it at all. Here is a two by two: one, two, over three, four. And here is a two by three: one, two, three, over four, five, six. Try to add them. The top left of the first faces the top left of the second, fine. The next position, fine. And then the first array runs out. The second has a third column and there is nothing whatever facing it. Two positions, dangling, with no partner.

4:06So what is the sum? There is no sum. Not a smaller one, not a partial one, not a longer one with something guessed. The operation does not apply, and when it does not apply it produces nothing at all. That is worth being very precise about, because 'nothing' is a strange thing to be asked to write down, and there are two extremely natural wrong answers. Wrong answer number one: pad the shorter array out with zeros until they match, then add.

4:40Do that here and you get a perfectly respectable two by three array, whose third column reads three and six - which is to say, the second array's third column copied across unchanged, because it was added to zeros that were never there. Wrong answer number two: add on the positions they do share and quietly drop the rest. That gives a two by two array, and it throws away exactly the same two entries, three and six, that the first wrong answer invented partners for.

5:17Both of those are rules. Both are perfectly well defined. Neither of them is this operation. And that is the point of insisting on it. If refusing were the only thing that could possibly happen, saying so would be a formality. It is not a formality: there are two obvious alternatives, both of which hand you an answer, and the operation we actually have declines to be either of them. The refusal is a decision.

5:51How much of a restriction is that, really? Let us count. Build every array you can make out of the two entries zero and one, at any of these four orders: one by one, one by two, two by one, and two by two. That is twenty-six arrays. Put each of them against each of them, in order, and you have six hundred and seventy-six ordered pairs. Of those six hundred and seventy-six pairs, how many have a sum?

6:21Two hundred and ninety-two. Which leaves three hundred and eighty-four pairs with no sum at all. More than half. Take two arrays at random from that collection and the likelier outcome is that you cannot add them. This is not an edge case hiding at the boundary of the definition; it is the ordinary situation, and the definition's first clause is doing most of the work. An operation like that has a name in plain words: it is partial. It applies to some pairs and not others, and the pairs it applies to are decided before any entry is looked at.

7:05But now watch what happens if you stop letting the order vary. Take only the two by two arrays out of that collection. There are sixteen of them, which gives two hundred and fifty-six ordered pairs. How many of those have a sum? All two hundred and fifty-six. And every single one of the answers is itself a two by two array. That is a completely different kind of object. Inside the collection of all arrays of one fixed order, addition is a rule that takes any two of them and hands you back a third one of the same collection, always, with no exceptions and no preconditions left to check.

7:50That is a genuinely useful notion and it is worth having the plain-words version of it: a rule that combines two things from a collection and returns a thing from that same collection. Addition of matrices is one of those - but only once you have fixed the order. Across all orders at once it is not, and the three hundred and eighty-four pairs we just counted are the reason.

8:17Hold on to that, because the second operation of this video behaves in precisely the opposite way. Suppose the first workshop doubles its output. Every grade, every size. Write the new figures before working them out, exactly as we did with the sums. Two times eighty, two times sixty. Two times seventy-five, two times sixty-five. Two times ninety, two times eighty-five. Six positions, six multiplications, each by the same single number, and again no position knows about any of the others. Work them out: one hundred and sixty and one hundred and twenty; one hundred and fifty and one hundred and thirty; one hundred and eighty and one hundred and seventy.

9:12That middle display, with the six products written and not yet resolved, is there for one reason. It is very easy to believe that multiplying an array by two means doubling the first entry, or doubling the top row. It means all six. The single number is called a scalar, and the operation is called scaling. Here is one on a bigger array, to show that nothing about it changes with size. Three rows and three columns: three, one and three halves; root five, seven and minus three; two, zero and five. Multiply the whole thing by three.

9:54Nine, three and nine halves. Three root five, twenty-one and minus nine. Six, zero and fifteen. Notice the root again. Three times root five is three root five - it is not evaluated, it is not approximated, it is simply multiplied like everything else. And the fraction stayed a fraction. Now the question we asked about addition. When is scaling not allowed? Never. There is no condition. Take the same twenty-six arrays, and five numbers to scale by - minus one, zero, one, two and one half. That is one hundred and thirty number-and-array pairs. How many of them produce an array?

10:43One hundred and thirty. All of them. Every array of every order, multiplied by every number, gives an array of exactly the order it started with. Put the two counts side by side. Addition: two hundred and ninety-two out of six hundred and seventy-six. Scaling: one hundred and thirty out of one hundred and thirty. One operation has a precondition that excludes most pairs, and the other has no precondition at all.

11:17That asymmetry is the actual content of this topic. The arithmetic in both rules is trivial. The difference between them is entirely a difference of domain. One caution about the scalar. It is a number sitting to the left of an array, and it is not a one by one array. Those are different kinds of object, and offering an array where a number belongs is not something this operation accepts.

11:49Scaling by minus one deserves its own name, and gets one: the negative of an array. Here is a two by two holding three and one on the top row, and minus five and x underneath - a letter, deliberately. Multiply the whole thing by minus one. Three becomes minus three. One becomes minus one. Minus five becomes five. And x becomes minus x. The letter is not a special case. There is nothing to evaluate and nothing to look up; it is an entry, it gets multiplied by minus one like every other entry, and the answer holds minus x in that position.

12:34And it is worth saying plainly that the negative is not a third operation. Across all twenty-six arrays, the negative and scaling by minus one disagree exactly zero times, because they are the same instruction written two ways. Which sets up the neatest move of the lot. You want to subtract one array from another. You could define a fourth operation: at every position, take the first entry less the second. That works, and then you would owe an account of when it applies, and what order the answer has, and whether regrouping is safe.

13:17Or you could refuse to define anything new, and simply write the difference as the sum of the first array with the negative of the second. A scaling, and then an addition. Both already understood. Do those two things agree? Over all six hundred and seventy-six ordered pairs, the assembled version and the entry-by-entry version disagree zero times. But agreeing on answers is the smaller half of it. The important question is where each one applies, and there the count is this: the number of pairs one of them works on and the other does not is zero. Both are defined on exactly two hundred and ninety-two pairs, which is exactly the set addition was defined on.

14:09So subtraction has no domain of its own. It inherited addition's, because it is made of addition. Nothing further has to be proved about it, ever - not its precondition, not the order of its result, not anything. That is what building an operation out of the ones you have buys you. Let us run a combined one slowly, because there is a trap in the notation. Twice one array, less another. The first is one, two, three above two, three, one. The second is three, minus one, three above minus one, zero, two.

14:50Three stages, kept apart. Stage one: twice the first array. Two, four, six above four, six, two. Stage two: the negative of the second. Minus three, one, minus three above one, zero, minus two. Stage three: add those two. Minus one, five, three above five, six, zero. Now the trap. Twice the first array less the second is not the same as twice their difference. Take their difference first and double it and you get minus four, six, zero above six, six, minus two - which is a different array in five of its six positions.

15:33And it is not a near miss that happens to bite here. Over the pairs of our collection that have a sum at all, the two readings differ on two hundred and sixty-six of the two hundred and ninety-two. They agree only on the twenty-six pairs where the second array holds nothing but zeros, because those are the only ones where doubling it changes nothing. The scaling binds to the array it is written against, and to that array only.

16:03Three small ones to finish the arithmetic. Two, four over three, two, plus one, three over minus two, five, gives three, seven over one, seven. The first less the second gives one, one over five, minus three. And three times the first, less minus two, five over three, four, gives eight, seven over six, two. One last thing these rules do that is easy to miss: the entries do not have to be numbers, and sometimes what comes out is smaller than what went in.

16:41Take a two by two whose entries are sums of squares - a squared plus b squared, b squared plus c squared, a squared plus c squared, a squared plus b squared. Add to it an array holding twice a b, twice b c, minus twice a c, and minus twice a b. Add position by position, exactly as before. Nothing clever happens during the addition. But look at what each position now holds: a squared plus two a b plus b squared. That is the square of a plus b. Three terms collapsing into one.

17:22All four positions do it. The answer is the square of a plus b, the square of b plus c, the square of a minus c, and the square of a minus b. The collapse is not part of the addition - it happened afterwards, to the entries, once they had been assembled. Now the same shape with trigonometry. A two by two holding cosine squared on both diagonal positions and, off the diagonal, minus sine times cosine above sine times cosine. Add an array holding sine squared on the diagonal and the same two off-diagonal terms with their signs reversed.

18:07The two off-diagonal positions cancel outright - no identity needed, just a number and its negative. Both diagonal positions become cosine squared plus sine squared. And cosine squared plus sine squared is one. Check it over seven hundred and twenty-one angles, a whole turn either side of zero: all seven hundred and twenty-one satisfy it and none fail. Run the same sweep on the two squares SUBTRACTED, set equal to one, and only five of the seven hundred and twenty-one hold - the half turns - while seven hundred and sixteen fail. So the sweep can come out either way, and the first result means something.

18:53Apply it, and both diagonal positions become one. The answer is one, zero, over zero, one - the identity array, arriving here as the answer to a question rather than as a definition handed down. Two operations, and one rule between them: do the same small thing at every position, independently. Everything interesting is in what they ask before they will do it. Addition demands that the two orders be identical - both counts, in order - and if they are not, it produces nothing whatever. Not a padded answer, not the overlap. Over twenty-six small arrays, only two hundred and ninety-two of six hundred and seventy-six pairs have a sum at all. Fix one order and that becomes all of them, and only then is it a rule that takes two things from a collection and returns a third from the same collection.

19:55Scaling demands nothing. One hundred and thirty of one hundred and thirty. Any number, any array, every entry multiplied, and the order comes out exactly as it went in. The negative is scaling by minus one, disagreeing with it zero times. The difference is a sum with a negative wrapped round the second array, disagreeing with the entry-by-entry version zero times and defined on exactly the same pairs. Neither of those two needs an account of its own, and that is the whole reason for building them out of the other two rather than defining them fresh.

20:38And because addition is a single independent operation at each position, regrouping three arrays cannot possibly change anything. Over five hundred and twelve triples, the two ways of bracketing disagree zero times. The arithmetic in all of this is nothing. Check the orders first, then do the obvious thing at every position. What you have to hold on to is which of the two operations is allowed to refuse - and that everything built out of them inherits the answer.