What multiplication keeps from ordinary algebra, and the two things it loses

0:00Matrix multiplication keeps three habits of ordinary algebra and loses two. The three it keeps are the reason most of your algebra still works. The two it loses are permanent, and nothing later buys either one back. Start with the losses, and start with the one everybody has heard of: the order of the two factors matters. But before that can even be a question, both products have to exist.

0:31Take every order from one by one up to four by four. Sixteen of them, and two hundred and fifty-six ordered pairs. Sixty-four of those pairs give a product one way round. Only sixteen give one both ways round. And only four of those sixteen land on the same order twice. Those four are exactly the pairs where both factors are square and of the same order. That is the only situation in which the question can be asked position by position at all.

1:08So here is the easy half of the failure. Take a two by three holding one, minus two and three above minus four, two and five. And a three by two holding two and three; four and five; two and one. One way round the inner pair is three and three, so the product exists, and the outer pair says it is two by two. It comes to zero and minus four above ten and three.

1:35The other way round the inner pair is two and two, so that product exists too. But the outer pair now says three by three. It comes to minus ten, two and twenty-one; minus sixteen, two and thirty-seven; minus two, minus two and eleven. Four numbers against nine. There is no position where you could even start comparing them. That is a real failure, but a student can shrug it off. Different sizes, they will say. Of course they differ.

2:10So take the shrug away. Same order, both ways, and still different. One and zero above zero and minus one. Against zero and one above one and zero. Both are two by two, so both products are two by two, and every position of one faces a position of the other. One way round you get zero and one above minus one and zero. The other way you get zero and minus one above one and zero.

2:41They agree at the top left and at the bottom right. Both are zero. They disagree at the top right and at the bottom left, and there the entries are opposite in sign: one against minus one, and minus one against one. Two positions out of four. Not a rounding difference, not a size difference. The same four positions, and two of them come out with the sign turned round.

3:10Hold that picture. It is the whole of the first loss. Now the correction, because students over-learn that failure immediately. Having seen one pair that does not swap, they start saying two products are never equal. That is a different claim and it is false. Take one and zero above zero and two. Against three and zero above zero and four. Both products come to three and zero above zero and eight. The same matrix, either way round.

3:45Nothing has gone wrong. Swapping is not forbidden; it is just not guaranteed. And there is a whole family where it is guaranteed. Matrices with zeros off the diagonal multiply their diagonals position by position, and ordinary numbers do not care about order. Of the eighty-one two by twos built from minus one, zero and one, nine are diagonal. That gives eighty-one ordered pairs of them, and not one fails to swap.

4:20Across all eighty-one matrices there are six thousand five hundred and sixty-one ordered pairs. Eight hundred and seventeen of them swap. Five thousand seven hundred and forty-four do not. So the failure is the common case, and the success is not rare either. And there is a third case that is neither. Take six arrangements of assorted orders and their thirty-six ordered pairs. Two swap, two refuse, and thirty-two are pairs where the question could not be put at all, because one of the two products does not exist.

4:59Those thirty-two are not counterexamples. Counting them as failures would be counting silence as an answer. That was the first loss. Here is the second, and it is the one that actually breaks calculations. In ordinary numbers, if a product is zero then one of the two factors is zero. There is no way out of that and you have relied on it for years. Take zero and minus one above zero and two. Neither row is zero, so the matrix is certainly not the zero matrix.

5:38Take three and five above zero and zero. Not the zero matrix either. Multiply them in that order and every one of the four entries comes out zero. The zero matrix, out of two matrices that are not. It is not symmetric, either. The other way round you get zero and seven above zero and zero, which is not the zero matrix. And this is not one curiosity. Among the eighty-one two by twos over minus one, zero and one, there are two hundred and fifty-six ordered pairs of non-zero matrices whose product vanishes.

6:19Thirty-two different matrices appear as the left factor of one. Now the consequence, which is bigger than the curiosity. You have a habit: if A times B equals A times C, cancel the A and conclude that B equals C. That habit is exactly the vanishing-product fact wearing different clothes. It says A times B minus C is zero, therefore B minus C is zero. The counterexample above kills it. And again, not as a curiosity.

6:56Of the eighty non-zero two by twos in that population, thirty-two have two different partners that give it the same answer. Counted as ordered triples, there are fourteen thousand eight hundred and forty-eight places where cancelling would have been wrong. The worst single left factor is zero and one above zero and one. It cannot tell six hundred and forty-eight ordered pairs of matrices apart. Put that same matrix on the right instead and the number is two hundred and eighty. Even how blind it is depends on which side you put it.

7:36So write the cancellation step down once and then cross it out. It is not a shortcut you are being asked to avoid. It is a step that is false. Two losses. Now the three things that survive, and they survive fully. The first is regrouping. In a triple product you may work out the first two and then multiply by the third, or work out the last two first.

8:04It has to be said plainly that this is stated here and not proved. The argument is a bookkeeping exercise over indices and it is left out. But stated is not the same as unchecked, and it is easy to see it doing real work. Take a three by three, a three by two and a two by four. Bracket the first two together and the middle step is a three by two.

8:31Bracket the last two together and the middle step is a three by four instead. Completely different intermediate matrices, and the same three by four at the end. Here is why that matters more than it looks. What is a matrix to the fourth power? Written out, it is four factors in a row, and there are five different ways to bracket them. Over the sixteen two by twos built from zero and one, all five bracketings give one answer every time. Not one matrix produces two.

9:07Now change the rule for building an entry, so that instead of adding products you add differences. Under that rule fifteen of the sixteen give more than one answer, and the worst gives five different fourth powers. So the fourth power would not be a thing you could write. Regrouping is what makes a power mean anything at all. The second survivor is expanding a bracket. A factor on the left spreads across a sum. A factor on the right spreads across a sum. Both hold.

9:43But those are two laws, not one, and it is worth seeing why. Over the sixteen small two by twos there are four thousand and ninety-six ordered triples. The real multiplication spreads correctly on all four thousand and ninety-six, from either side. Now a rule that simply keeps its left factor and throws the right one away. From the right it spreads perfectly: all four thousand and ninety-six. From the left it spreads on two hundred and fifty-six, and fails on the rest.

10:19One rule, obeying one of the two and breaking the other. That is why they are counted as two. And a rule that takes the largest of the sums instead of adding them spreads on two hundred and two triples from either side, and breaks both. So when the real multiplication comes back four thousand and ninety-six out of four thousand and ninety-six, twice, that is a measurement rather than a formality.

10:50The third survivor is the smallest and the easiest to take for granted. There is a matrix that changes nothing. Ones down the diagonal, zeros everywhere else, of the matching order. Multiply a square matrix by it from either side and you get the square matrix back. All sixteen of the small two by twos come back unchanged, from both sides. Compare that with the rule that multiplies facing entries. Under that rule the identity matrix leaves only four of the sixteen alone.

11:28Under the largest-of-the-sums rule it leaves none. Under the differences rule it leaves none. And the rule that keeps its left factor leaves exactly one, which is not much of a property. So the identity is not a definition tidying itself up. It is a fact about this multiplication that other multiplications do not have. With regrouping in hand, powers are safe, and a polynomial in a matrix becomes something you can write.

12:01Take a three by three: one, two and three; three, minus two and one; four, two and one. Its square is nineteen, four and eight; one, twelve and eight; fourteen, six and fifteen. Its cube is sixty-three, forty-six and sixty-nine; sixty-nine, minus six and twenty-three; ninety-two, forty-six and sixty-three. Now take the cube, subtract twenty-three times the matrix, and subtract forty times the identity. Every one of the nine positions comes out zero.

12:38And notice what that last term had to be. Forty times the identity — not the number forty. This is where a polynomial in a matrix stops behaving like a polynomial in a number. The constant term has to be a matrix, and the only sensible matrix is a multiple of the identity. Read it the other way, adding minus forty to every entry, and the diagonal still comes out zero. It looks like it worked.

13:09But the six off-diagonal positions come out at minus forty each. Six of the nine positions are wrong, and the three that look right are a coincidence of the diagonal. There is a second version of that trap, and it is worth a minute. Take three and minus two above four and minus two. Its square is one and minus two above four and minus four. You are asked for the number k with the square equal to k times the matrix, less twice the identity.

13:44Position by position: the top left says k is one. The top right says one. The bottom left says one. The bottom right says one. Four positions, one answer, and the identity holds. But suppose you had checked the top left and stopped. Over the six hundred and twenty-five two by twos with entries from minus two to two, six hundred and twenty-four have at least one position that names a value of k.

14:15Only ninety-six have all four positions agreeing on it. Five hundred and twenty-eight would have misled you. Checking one position is not a light version of the check. It is not the check. One more thing regrouping buys, and it is the clearest evidence that it is doing real work. Take the two by two that turns the plane: cosine and sine on the top row, minus sine and cosine below.

14:47The claim is that its nth power turns the plane n times: cosine of n theta and sine of n theta, minus sine of n theta and cosine of n theta. You prove that by induction, and the induction is a chain of products regrouped. The base case is the matrix itself. The step assumes the formula at k and multiplies by one more copy. And the step is not really a matrix fact at all. It is the compound-angle formulas, collapsing four entries into four.

15:25So checking the step means checking that two turns compose into one. Twenty-one different turns were tried, giving four hundred and forty-one ordered pairs. Four hundred and twenty-seven of those pairs have a sum that can still be written down, and in every one of the four hundred and twenty-seven the two turns compose into the sum. The three by three version, with a fixed third direction, behaves identically on all four hundred and twenty-seven.

15:58Then the powers themselves: one hundred and fifty-four inductive steps were taken, from the first power up to the eighth, and not one failed. There is a companion identity in the same family, about half a turn, and it holds at every one of the twenty-one angles. Turn the sines round the other way and it holds at exactly one angle, which is the one where the turn is nothing at all.

16:27That is what it looks like when a check could have failed and did not. Last piece. An unknown matrix in a product equation. The habit from ordinary algebra is to move things to whichever side is convenient. You cannot, because the two sides are different operations. Here is the move to make first, before writing down a single entry: work out the order of the unknown. The known factor is two by three. The right-hand side is three by three.

17:02If the unknown sits on the left, its columns must match the known factor's rows, so it has two columns. And its rows must give the answer's three, so it has three rows. Three by two, on the left. That is forced. Put it on the right instead and the shapes do not even meet. There is nothing to solve, because there is nothing to write. With the order settled, the nine entries of the right-hand side become nine equations in six unknowns, and they have exactly one answer.

17:38One and minus two; two and zero; zero and one. The same discipline settles an applied question. Suppose contacts are counted for two cities across three methods, and each method has a cost per contact. Telephone forty, house call one hundred, letter fifty, in the smaller unit. The counts are a two by three and the costs are a column of three, so the counts have to be on the left. The other order does not exist.

18:11That gives one total per city: three hundred and forty thousand, and seven hundred and twenty thousand. Three thousand four hundred and seven thousand two hundred, once you divide by a hundred. The side was never a choice. It was decided by the shapes before any arithmetic happened. So: three kept, two lost. Kept: you may regroup a triple product, so powers mean something. Kept: you may expand a bracket, from either side, and those are two separate permissions.

18:47Kept: there is a matrix that changes nothing. Lost: you may not swap two factors. Some pairs do swap and you can check whether yours is one of them, but you may never assume it. Lost: you may not conclude anything from a product being zero, and therefore you may not cancel a common factor. Every solved matrix equation you will meet is solved by rearranging with the three that survive. Not one of them cancels.

19:18If you remember only one thing, remember which list cancellation is on.