Trace, Commuting and Orthogonal Matrices

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge before we start. Find two square matrices, A and B, such that A B minus B A is the identity matrix. Any size you like. Looks innocent, right? But today we will prove, in two short lines, that no such pair exists. The tool is one tiny number: the trace. We also meet commuting and orthogonal matrices, which come up often in JEE questions. So, let's begin.

0:31Trace: add the main diagonal

So, what is the trace? Take a square matrix A, of order n by n; the cross is read as by. The trace is the sum of the entries on its main diagonal. The small letters t r stand for trace. Here a one one, a two two, up to a n n are those diagonal entries: the small numbers below are subscripts, row and column, and the three dots mean and so on. Right? Now tell me. Take a two by two matrix, with first row a, b and second row c, d. What is its trace? Yes, exactly! It is a plus d. The square brackets hold the matrix, and the entries b and c play no part at all. And remember: only square matrices have a trace. Clear?

1:18Rules of trace

Now the rules, for square matrices A and B of the same order, and a scalar k. The trace of A plus B is the trace of A plus the trace of B: the diagonals simply add. And the trace of k A is k times the trace of A, okay? Next, the trace of A transpose equals the trace of A itself. The small capital T up top is the superscript for transpose. Why? Transpose turns rows into columns, but the main diagonal stays exactly where it is. Same diagonal, same trace. With me? Now the big one. The equals sign with a slash through it means not equal: usually, A B and B A are different matrices. So their traces must be different too, right? No, no, careful! The trace of A B always equals the trace of B A, even when the matrices differ. The entries can be completely different, but the diagonal sums match. Keep this line in mind; it solves our challenge. Got it? One more fact. The trace of A transpose A, and of A A transpose, is the sum of the squares of all the entries, okay? The big Greek sigma means sum, over every row i and every column j, and the small two up top means squared. So every entry counts, not just the diagonal. Clear?

2:41Challenge solved: AB-BA=I is impossible

Now, back to our challenge. Suppose A B minus B A equals the identity matrix. The trace of the left side is the trace of A B, minus the trace of B A, okay? Those are equal, so the left side has trace zero. Now the right side. I sub n is the identity matrix of order n: ones on the diagonal, zeros everywhere else. What is its trace? Yes, exactly! There are n ones on the diagonal, so the trace is n, and n is at least one, so it is never zero. Right? So the left side has trace zero, and the right side does not. Contradiction! No square matrices A and B can give A B minus B A equal to the identity. Two short lines, as promised!

3:30Commuting matrices: when AB=BA

Now, commuting matrices. In general, A B is not equal to B A: the order of multiplication matters. But when A B does equal B A, we say A and B commute. Okay? Some pairs always commute. Any square matrix commutes with the identity matrix of its order: A I and I A are both just A itself. And powers of one matrix commute: A to the m, times A to the n, is A to the m plus n, either way. With me? Now a trap. What is A plus B, the whole squared? It is A squared, plus two A B, plus B squared, just like in algebra, right? Oh no, careful! Multiply it out: A squared, plus A B, plus B A, plus B squared. Matrices keep their order, so A B and B A are separate terms. Clear? They combine into two A B only when A B equals B A, okay? So the school formula works only for commuting matrices. This is a very common mistake. Don't make it!

4:35Orthogonal matrices

Now our last idea: orthogonal matrices. We call a square matrix A orthogonal if A times A transpose, and A transpose times A, both equal the identity matrix, okay? Square, and both products: remember both conditions. How do we check this quickly? Look at the entry in row i, column j of A times A transpose: that is what the small i j below the bracket means. The columns of A transpose are the rows of A, so this entry is row i of A, dotted with row j, right? The dot means the dot product: multiply matching entries, and add. Now, A times A transpose must be the identity: ones on the diagonal, zeros everywhere else. So what does that tell you about the rows of A? Yes, exactly! Each row dotted with itself gives one, so every row has length one. And two different rows dotted give zero, so they are perpendicular. The same holds for the columns. This row test is the fastest check. Got it? And one fact to remember: rotation matrices are orthogonal. You will check one yourself in the practice problems.

5:46Example 1: trace of 2A - 3B transpose

Let's start with an easy one. The trace of A is three, and the trace of B is minus two. Find the trace of two A minus three B transpose, okay? Pause the video, and try it first. Done? Okay, let's check. Using the sum rule and the scalar rule, split it: two times the trace of A, minus three times the trace of B transpose. Right? And the trace of B transpose is just the trace of B: minus two. Same diagonal, remember? So, two times three is six, and then minus six, so the answer is zero, right? No, no, wait! The cross means times, and minus three times minus two is plus six. So it is six plus six: twelve. Watch the signs; that is where marks are lost!

6:35Example 2: count 2x2 orthogonal matrices with -1, 0, 1

Now a harder one, JEE pattern. Each entry of a two by two matrix is minus one, zero or one; the curly brackets hold the allowed values. How many such matrices are orthogonal? Pause the video and try it. Hint: use the row test. Done? Okay, let's check. Length one means the squares of the two entries add up to one. So one entry is plus or minus one, and the other is zero; the plus-minus sign means either sign. In round brackets: plus or minus one, zero, or zero, plus or minus one. Four choices for row one. Okay? Row two also has length one, so again four choices, and four times four is sixteen, right? Oh no, careful! Row two must also be perpendicular to row one. If row one is one, zero, then row two needs zero in the first place: zero, plus or minus one. Only two choices. The same happens for every choice of row one. Clear? So the total is four times two: eight. Only eight orthogonal matrices. See how fast the row test makes it?

7:50Quick revision: trace rules, commuting, orthogonal matrices

So, quick revision. Trace: the sum of the diagonal entries, and the traces of A B and B A are always equal. Commuting: A B equals B A, and only then does A plus B, the whole squared, expand the school way. Orthogonal: A A transpose equals A transpose A equals the identity, so the rows are perpendicular, of length one. You did really well today. Now go and try the practice problems on this page. All the best!