Symmetric and skew symmetric, and splitting any square matrix into one of each

0:00Two conditions, and they differ by one minus sign. A square matrix is symmetric when it equals its own transpose. It is skew symmetric when it equals minus its own transpose. That is the whole of both definitions. One equation each, between a matrix and its flip. And from those two lines the rest of this video follows, including something the second condition forces that nobody put into it. Take the first condition entry by entry.

0:36The entry at row i, column j has to equal the entry at row j, column i. Which is to say the matrix is unchanged when you reflect it in the diagonal running from the top left corner down to the bottom right. Here is one. The square root of three, two and three; two, minus three halves and minus one; three, minus one and one. Check the three off-diagonal pairs across the diagonal. Two against two. Three against three. Minus one against minus one.

1:12It is symmetric, and notice what the condition did not say. The diagonal itself is unconstrained. Setting the two indices equal gives the entry equal to itself, which asks nothing. The square root of three sits there quite happily, and so does a fraction, and so would anything else. Remember that. The other condition is about to do something very different in the same place. Now the second condition. The entry at row i, column j has to equal MINUS the entry at row j, column i.

1:51Same reflection in the same diagonal, but every pair comes back with its sign turned over. The clearest instance is one built from letters rather than numbers. Zero, x and y in the first row. Minus x, zero and z. Minus y, minus z and zero. Every position above the diagonal has its own negative below it, and the condition holds whatever the three letters happen to be. So this is not one example. It is the shape of every three by three of this kind.

2:28And it already shows you the thing worth deriving. Look at the diagonal. The zeros on that diagonal were not a choice. They are forced, and the argument is three lines. Take a matrix of nine distinct letters and impose the condition, position by position. Each position gives one equation: the entry, plus the entry across the diagonal, is zero. Off the diagonal that pairs two different letters, and there are six of those.

3:02On the diagonal the two positions are the same position. So the equation reads: twice that entry is zero. Halve it, and the entry is zero. Three positions, three zeros, and no example was needed anywhere. Run the same code with the other condition and the diagonal equations come out as zero equals zero. Three constraints against none. That is what it means to say a definition FORCES something. Nobody wrote the zeros down. They fell out.

3:40There is a step in that argument worth stopping on, because it is the step people skip. Going from twice something is zero, to that something is zero, means dividing by two. That is free when your entries are ordinary numbers. It is not free in general, and the way to see it is to take it away. Work with entries that are only zero and one, where adding wraps around, so that one plus one is zero.

4:10There, two is zero. And minus one is one. So the two conditions stop being different conditions. Of the sixteen two by twos, eight equal their transpose and the same eight equal minus their transpose. And the diagonal is not forced at all. Six of those eight have a non-zero entry sitting on it. Go up to three by threes and it is the same story: of five hundred and twelve, sixty-four satisfy the condition and fifty-six of them have a live diagonal.

4:47Over the ordinary numbers, that count is zero. Not small. Zero. The difference between those two answers is exactly the halving step, and that is why it is a step. Both definitions begin with the words square matrix, and it is worth asking why. The reason is the order, and it settles the question before a single entry is looked at. A matrix with two rows and three columns transposes into one with three rows and two columns.

5:22Those two cannot be equal, or negatives of each other, or compared at all. There are positions in one with no counterpart in the other. So neither condition is false for a rectangular matrix. It cannot even be asked. And that is not a general fact about conditions on matrices. Ask instead whether a matrix equals its own left-to-right mirror. That mirror keeps the order, so a rectangular matrix can perfectly well satisfy it, and sixteen of the sixty-four here do.

5:59The same question asked with the transpose is answered by none of them. The transpose swaps the order. That single fact is doing the work again. Now the first of two results that are actually proved rather than checked. Take any square matrix and add its own transpose to it. The answer is always symmetric. Four lines, and each one names the rule it uses. Transpose the sum. By the sum law, that is the transpose of the first piece plus the transpose of the second.

6:37The first piece was the matrix, so its transpose is the transpose. The second piece was the transpose, so its transpose is the matrix back again, by the double-transpose law. That gives the transpose plus the matrix. And addition does not care about order, so that is the matrix plus its transpose. Which is what we started with. So the sum equals its own transpose, and that is the definition of symmetric.

7:09Try it on the smallest interesting case. One and five above six and seven. Add the transpose and you get two and eleven above eleven and fourteen. Symmetric, with the eleven appearing twice. Swept over five hundred and ninety-three square matrices, that construction lands symmetric in all five hundred and ninety-three. It lands SKEW in four of them, and those four are the ones where the sum came out as the zero matrix.

7:43The second half of the same result. Subtract the transpose instead, and the answer is always skew. The proof runs exactly the same way, and it is worth saying which rules answer its two gaps. The first gap is the sum law applied to a difference, which is the same law with a scaling by minus one in front of it. The second is the double-transpose law again. On the same two by two, the difference is zero and minus one above one and zero.

8:17Skew, and there is the forced diagonal, sitting at zero without anyone arranging it. Over the same five hundred and ninety-three, the difference is skew every time, and its diagonal is zero every time. It comes out symmetric in ninety-one of them, and again those are the cases where the difference vanished. One caution about both proofs. They lean entirely on the transpose laws from the previous topic, and those laws are stated without proof.

8:51So these are real derivations standing on stated ground. That is worth knowing about, and it is not a defect. Put those two together and you get the result the whole topic is aimed at. Every square matrix splits into a symmetric part plus a skew part, and there is a recipe. Half of the matrix plus its transpose. Plus half of the matrix minus its transpose. The first bracket is symmetric by what we just proved. The second is skew.

9:26And halving does not disturb either, because scaling passes through a transpose. That law is load-bearing here. Now the halves. They look like a convenience and they are not. Drop them, and the two pieces are still perfectly good. One is still symmetric, the other is still skew, in all five hundred and ninety-three cases. But add them up. The transposes cancel and you are left with twice the matrix you started with. Every single time.

10:00So the unhalved recipe returns the right matrix in exactly two cases out of five hundred and ninety-three, and both of them are the zero matrix, where twice nothing is nothing. That is the whole argument for the halves. They are not a choice, they are the only thing that makes the pieces add back. Run the recipe once, properly, on a three by three, in five stages. The matrix: two, minus two and minus four; minus one, three and four; one, minus two and minus three.

10:38Its transpose: two, minus one and one; minus two, three and minus two; minus four, four and minus three. Add and halve. The symmetric part is two, minus three halves and minus three halves; minus three halves, three and one; minus three halves, one and minus three. Check it. Across the diagonal, minus three halves faces minus three halves, and one faces one. Symmetric. Subtract and halve. The skew part is zero, minus one half and minus five halves; one half, zero and three; five halves, minus three and zero.

11:21Check that too. The diagonal is zero, as it has to be, and every off-diagonal pair is a sign apart. Now the stage everybody skips. Add the two parts back up. Two minus nothing is two. Minus three halves and minus one half make minus two. Minus three halves and minus five halves make minus four. It is the matrix we started with, position by position. That last stage is not a formality. It is the only step that shows the halves were necessary.

12:00Two cases test whether you trust the recipe or your eyes. First, take a matrix that is already skew. The one built from three letters, from earlier. Its symmetric part is half of the matrix plus its transpose. But its transpose is minus the matrix, so that sum is the zero matrix. So the symmetric part is zero and the skew part is the whole thing. The recipe still works, and it hands back one empty half.

12:32Second, the mirror image of that. A matrix that is already symmetric. Its transpose is itself, so the difference vanishes. The symmetric part is the whole matrix and the skew part is the zero matrix. Neither of these is a failure of the recipe. Both are the recipe telling you the truth about a matrix that was already one of the two kinds. If you are expecting two interesting halves and get a zero, nothing has gone wrong.

13:06It is worth asking what the split actually needed, because it is less than you might think and also more. Take any rule that turns a matrix into another matrix of the same order, and run the same recipe with it in place of the transpose. Half of the matrix plus its flip, plus half of the matrix minus its flip. Those two always add back to the matrix. That is just algebra, and it holds for all four rules we tried, on all eighty-one two by twos.

13:41But adding back is not the point. The point is that the two pieces are of the right KINDS. For turning rows into columns, both pieces come out right, eighty-one times out of eighty-one. For mirroring left to right, also eighty-one. For reflecting in the other diagonal, eighty-one again. For a quarter turn, nine. The difference is that a quarter turn does not undo itself. Do it twice and you have a half turn, not the matrix you began with, and only nine of the eighty-one survive that.

14:18So three of the four rules give a proper splitting and one does not, and what separates them is whether the flip is its own undoing. The transpose is. That, and not anything special about symmetry, is what the split is standing on. One more question, and it reaches straight back to the reversal law. Multiply two symmetric matrices together. Is the answer symmetric? Transpose the product and the factors turn around. Both are symmetric, so each transpose is itself, and you are left with the product the other way round.

14:59So the product is symmetric exactly when the two factors commute. Not usually. Exactly then. Count it. Of the seven hundred and twenty-nine ordered pairs of symmetric two by twos here, two hundred and sixty-five commute. And two hundred and sixty-five have a symmetric product. Not roughly the same number. The same pairs, with zero disagreements between the two verdicts. The flat claim, that a product of two symmetric matrices is symmetric, is wrong in four hundred and sixty-four of those pairs.

15:39Two related facts fall out of the same move. The difference of the two products is skew, in all seven hundred and twenty-nine cases, and the sum of the two products is symmetric in all seven hundred and twenty-nine. So the non-commutativity that looked like a nuisance two topics ago is the exact obstruction here, and the two products differ by a skew matrix. Last question. Can a matrix be both kinds at once?

16:11If it is, it equals its transpose and it equals minus its transpose. So it equals its own negative. Add it to itself: twice the matrix is zero. Halve, and the matrix is zero. Over the whole population of nineteen thousand seven hundred and sixty-four square matrices, exactly two satisfy both conditions, and they are the two zero matrices, one of each order. That is the same halving step as the diagonal argument, doing the same job. Over the wrapping carrier, where two is zero, all eight matrices that are one kind are also the other.

16:56And one last item, because it shows the same machinery answering a different question. Take the turn matrix, with the cosine and the sine, and ask when it plus its transpose equals the identity. The sum is twice the cosine on both diagonal positions and zero elsewhere, so the condition is that twice the cosine is one. Searched over twenty-three candidate values, exactly one works: the cosine is one half, which is a third of a half turn.

17:31The same question asked of the DIFFERENCE has no answer at all, because the difference has a zero diagonal whatever the angle. Two conditions, one minus sign apart. One of them forces its diagonal to zero and the other says nothing about it, and every square matrix in the world is one of each, added together, in exactly one way.