Powers of a Matrix and Special Matrices

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge first. Take a two by two matrix, and find its tenth power. What would you do? Multiply it by itself, nine times? Slow, and one small slip ruins everything, right? Today, for a special kind of matrix, you will do it in just two lines. Powers of matrices come up often in JEE, so let's begin.

0:26Powers of a square matrix

So, powers. To square a matrix, we multiply it by itself. Now think. Take a two by three matrix; the cross on screen is read as by. Can we square it? No, we cannot! Columns of the first must match rows of the second, so powers need a square matrix, okay? The small number up top is a superscript, the power. So, A squared is A times A, A cubed is A squared times A, and A to the power zero is I, the identity matrix. Clear? Now, A B and B A are usually different; the equals sign with a slash means not equal. But two powers of the same matrix, A to the power m and A to the power n: do they commute? Yes, always! Both are A multiplied by itself m plus n times, so both equal A to the power m plus n. And A to the power m, in round brackets, to the power n, is A to the power m times n. With me?

1:27One identity for every 2×2 matrix

Now the star of this video. Take a two by two matrix with entries a, b, c, d in square brackets. The letters t r mean trace, the sum of the diagonal entries, here a plus d. The vertical bars around A mean the determinant of A, a d minus b c; more on it in Chapter two. Okay? Here is the identity: A squared, minus trace of A times A, plus determinant of A times I, equals O, the zero matrix. Every two by two matrix satisfies this. Got it? Move two terms across, and A squared equals trace of A times A, minus determinant of A times I, okay? So the square of A is built from just A and the identity. Now think. Can A cubed be built the same way? Yes, it can! Multiply both sides by A: now A cubed equals trace of A times A squared, minus determinant of A times A, right? Replace that A squared once more, and only A and the identity remain. Clear? And this never stops. So every power of a two by two matrix can be written as p A plus q I, where p and q are just numbers. Even the fiftieth power! With me?

2:41Example 1: show A satisfies A^2 - 2A + 3I = O

Let's try an easy one. Matrix A has rows two, minus one, and three, zero. Show that A squared, minus two A, plus three I, equals O, okay? Pause the video and try it. Hint: you may not need to multiply. Done? Okay, let's check. Trace: two plus zero, two. Determinant: a d minus b c. Here a d is zero, and b c is one times three, three. So zero minus three, minus three, right? No, no, careful! Look at b again: it is minus one, not one. So b c is minus three, and zero minus minus three is plus three. The determinant is three. This sign slip is a very common mistake. Got it? Now put trace two and determinant three into the identity. We get A squared, minus two A, plus three I, equals O, as required. Not a single matrix multiplied! Perfect! Still, let's check directly; it is a good habit. Row by column, A squared comes out as one, minus two, six, minus three. With me? Now subtract two A and add three I, entry by entry. One minus four plus three, zero. Minus two plus two, zero. Six minus six, zero. Minus three plus three, zero. The zero matrix. Correct!

4:11Idempotent, involutory, nilpotent

Alright, now three special names. First, idempotent: A squared equals A, okay? Squaring such a matrix changes nothing. Second, involutory: A squared equals I, the identity matrix. Multiply it by itself, and you get the identity. Right? Third, nilpotent: some power becomes the zero matrix. If A to the power k is O, but A to the power k minus one is not, it is nilpotent of index k: the first power that gives zero. Clear? Now a question. With numbers, if x squared is zero, x must be zero. For matrices, if N squared is the zero matrix, must N itself be zero? No! Look at this N: zero, three, zero, zero. Not the zero matrix, but multiply it by itself, and every entry comes out zero. So N is nilpotent of index two. Keep it in mind. Got it?

5:15Topper's trick: split off aI

Now the toppers' trick for big powers. Write A as a I plus N: a number a times the identity, plus a matrix N, okay? Can we expand a I plus N, to the power n, by the binomial theorem? Yes, here we can! The binomial theorem needs the two parts to commute, and matrices usually do not. But a I times N and N times a I both equal a N, so they commute. Got it? Now suppose N squared is the zero matrix. Every term with N squared or higher vanishes. So, A to the power n is a to the power n times I, plus n times a to the power n minus one, times N, nothing more. With me? If N cubed is the zero matrix, three terms survive. The third has n C two, the binomial coefficient, written as n above two in round brackets, then a to the power n minus two, times N squared. Clear? This form appears in upper triangular matrices with equal diagonal entries. The diagonal number is a, and the rest of the matrix is N, okay? Spot such a matrix, and think of this trick.

6:29Example 2: A^10 by writing A = 2I + N

Now, remember our challenge, the tenth power? Here it is. Matrix A has first row two, three, and second row zero, two. Find A to the power ten. Pause the video and try it, with the trick. Done? Okay, let's check. Many students simply raise every entry to the power ten: two to the power ten, three to the power ten. Right? Oh no, careful! That is the trap: matrix powers are not taken entry by entry. Look: a zero below the diagonal, and both diagonal entries are two. So A is two I plus N, with N equal to zero, three, zero, zero: our nilpotent friend, whose square is zero. Okay? Now use the trick with a equal to two and n equal to ten. Two to the power ten times I, plus ten times two to the power nine, times N, clear? The small dot just means times. Now the numbers. Two to the power ten is one thousand and twenty four. The corner of N is three, so our corner is ten times two to the power nine times three: ten times five hundred and twelve times three. Fifteen thousand three hundred and sixty. With me? So A to the power ten has one thousand and twenty four on the diagonal, fifteen thousand three hundred and sixty in the corner, and zero below. Two lines, challenge solved! Three to the power ten there would be completely wrong.

8:01Quick revision: A^n = pA + qI, special matrices, aI + N

So, quick revision. For a two by two matrix, A squared equals trace times A, minus determinant times I, so every power is p A plus q I, okay? Idempotent: A squared equals A; involutory: A squared equals I; nilpotent: some power is the zero matrix. And for a I plus N with N squared zero, keep two binomial terms. Now try the practice problems on this page. All the best!