Introduction to the square root of a 2 by 2 matrix

[Pages:6]Introduction to the square root of a 2 by 2 matrix

Yue Kwok Choy

The square root of a 2 by 2 matrix A is another 2 by 2 matrix B such that = , where

stands for the matrix product of B with itself. We write = . In general, there can be no, two, four

or even an infinite number of square root matrices.

For example: 13 24 13 24 = 175 1202

If we take

= 175

1202

, then

=

175

1202 = 13

24

Note that = 13 24 is one of the square roots of = 175 1202 .

There may be another root(s) for .

Questions (See answers at the end of this file)

1. Evaluate : 11 11 11 11. Hence find two square roots of (a) 22 22 (b) 11 11 .

2. Find four values of 40 09 . Hence find four values of 0a 0b , where a, b 0 .

3.

List all possible matrices = ac

db

, where a, b, c and d can be either 0 or 1, such that

does not exist.

(A) Solving equations method

Suppose we like to find the square root of = 12 25. So we write: ac bd ac bd = 12 24 ca(a++bdc) bc(ba++dd) = 12 25 We get four equations:

a + bc = 1 ... (1) b(a + d) = 2 ... (2) c(a + d) = 2 ... (3) cb + d = 5 ...(4) (2) - (3), (b - c)(a + d) = 0 Since a + d 0 (otherwise (2) or (3) give absurdity), b - c = 0, b = c. Hence we get: a + b = 1 ... (5)

b(a + d) = 2 ... (6)

1

b + d = 5 ... (7)

(6),

b(a + d) = 4 ,

b

=

()

... (8)

(8) (5),

a

+

()

= 1,

... (9)

(7) - (5), d - a = 4 ... (10)

d - a = ... (11) (a + d 0)

(11) (9), a + = 1, a + ad + d - a = a + d, a + ad - 2a = 0

a(a + ad - 2) = 0 a = 0 or d = ... (12)

(12) (10), a = 0, d = ?2 or a = ? , d = ?

The rest can be solved easily, we therefore have:

21

25 = ?01

??21 ,

?

?

? ?

3. Prove that 00 10 does not exist. List, without prove, all possible matrices = ac bd , where a, b, c and d can be either 0 or 1,

such that has no real solution(s).

4. Use algebraic method to find 10 21. 5. Check that the square root of the identity matrix is given by:

10

01

=

?d c

?d

,

d

c d

,

where ? ?01 01 , ?c1 01 , ?01 c1 are limiting cases.

6.

(a) Let

= 12

24,

find ||

and

.

Hence find .

(b) = ac bd, where || = 0. Given that tr() = a + d > 0.

Show that = tr() . Hence find .

2

(B) Diagonalization of Matrrix

Solving equation method in finding the square root of a matrix may not be easy. It involves

solving four non-linear equations with four unknowns. You may try this: 4112 1324, and soon may give up.

We note that the square root of a diagonal matrix can be found easily:

0a

0b

=

a 0

0 , -a b 0

0 , a b 0

0 , -a -b 0

0 . -b

If a matrix is NOT a diagonal matrix, we devise a method called diagonalization to help us.

We proceed with the finding of the eigenvalue(s) and eigenvector(s) of A .

A real number is said to be an eigenvalue of a matrix A if there exists a non-zero column vector v such that Av = v or (A - I)v = 0 .

We like to find 4383 5274 (1) Eigenvalues

A = 3438 2547 ,

x v = y,

(A - I)v = 334-8

24 57 -

x y

=

0

Now, (A - I)v = 0 has non-zero solution, |A - I| = 0

334-8

24 57 -

=

0

(33 - )(57 - ) - 48 ? 24 = 0 - 90 + 729 = 0 ( - 81) ( - 9) = 0 = 9 or = 81 , and these are the eigenvalues.

(2) Eigenvectors We usually would like to find the eigenvector corresponding to each eigenvalue. The process is called normalization.

For = 9, 3438 2547 yx = 9 xy

4383xx

+ +

24y 57y

= =

9x 9y

Choose for convenience 24x + 24y = 0 x + y = 0

yx = -11 , which is a eigenvector.

For = 81,

3438

2547 yx = 81 yx

3438xx

+ +

24y 57y

= =

81x 81y

3

Choose -48x + 24y = 0 2x - y = 0 yx = 12 , which is another eigenvector.

(3) Diagonalization of matrix We place two eigenvectors together. Let = yx xy = -11 12 Consider another matrix: = = -11 12 3438 2547 -11 12

=

21

-11 3438

2547 -11

12

= 90 801

At last we get a matrix B which is diagonal with eigenvalues as entries in the main

diagonal.

=

90

801 = 30

09 , -03

-09 , 30

-09 , -03

09

Note:

The diagonalization of a matrix may not be a simple subject since |A - I| = 0

may have equal roots or even complex roots. Although most matrices are not

diagonal, they can be diagonalized. Not all square matrices can be diagonalised.A

thorough study of diagonalization of a matrix is not discussed here.

(4) Finding the square root of the original matrix A

Since = , we have =

= () = = =

So =

(a)

=

=

-11

12 30

09 -11

12 = -11

12 30

09

21

-11

= -33

198

21

-11 = -11

36 21

-11 = 54

27

(b)

A

=

PBP

=

-11

12 -03

09 -11

12 = -11

12 -03

09

21

-11

= -33

198

21

-11 = -11

36 21

-11 = 18

45

4383 5274 = , -- -- , , -- --

4

Questions

7. Real matrix may have irrational square roots. Check by multipication:

10

23

=

1 0

3 - 1 , 1 3 0

-3 - 1 , -1

-3

0

3 + 1 , -1

3

0

-3 + 1 -3

8. A simple real matrix may have complex square roots. Check by multipication:

01

10

=

11

+ -

i i

1 1

- +

ii

,

11

- +

i i

1 1

+ -

ii

,

--11

- +

i i

-1 -1

+ -

ii

,

--11

+ -

i i

-1 -1

- +

ii

9. A matrix can have both integral and fractional square roots. Use Diagonalization Method to show:

-01 54 = ? 21 --43 , ? -

Answers

1. 11 11 11 11 = 22 22 , Hence:

(a) 22 22 = 11 11 , --11 --11

(b) 11

11 11

11 = 22

22 = 2 11

11 11

11

=

?

11

11

?

11

11

11

11

=

?

11

11 =

-

,

-

- -

2. 04 90 = 20 03 , -02 03 , 20 -03 , -02 -03

0a

0b

=

a 0

0 , -a b 0

0 , a b 0

0 , -a -b 0

0 -b

3. ac bd ac bd = 00 10 ca(a++bdc) bc(ba++dd) = 00 10

a + bc = 0 = cb + d a = ?d

(a) Since b(a + d) = 1, a + d 0, and so a = d 0.

(b) Lastly, since c(a + d) = 0 c = 0 and so a + bc = a = 0, contradicts with (a).

00 10 , 01 00 does not exist

01 11 11 10 has no real solution(s).

4. ac bd ac bd = 10 21 ca(a++bdc) bc(ba++dd) = 10 21

5

We get four equations: a + bc = 1 ... (1) b(a + d) = 2 ... (2) c(a + d) = 0 ... (3) cb + d = 1 ...(4)

From (3), c = 0 ... (5) (a + d = 0 gives contradiction in (2)) From (1), a = 1, a = ?1. From (3), d = 1, d = ?1

Hence, 10 21 = 10 11 , -01 --11

6. (a) = 12 24, || = 0 , = 12 24 12 24 = 150 1200 = 5 12 24 = 5.

Hence

= =

.

=

=

12

24 =

.

(b) = ac bd, || = ad - bc = 0 ad = bc ... (1)

= ac bd ac bd = ca(a++bdc) bc(ba++dd) = ca(a++add) ba(da++dd) , by (1).

=

ac((aa

+ +

d) d)

b(a d(a

+ +

dd))

=

(a

+

d)

ac

bd = tr()

=

()

=

()

,

where

tr() > 0.

Hence,

=

.

()

9. Eigenvalues : 4, 1

Eigenvectors: 11 , 41

= yx

xy = 11

41

,

=

-11

-41

, = 40 01

=

40

01 = 20

01 , -02

-01 , 20

-01 , -02

01

(a)

=

=

11

41 20

01 11

41 = 11

41 20

01

-11

-41 = -

(b)

=

=

11

41 -02

01 11

41 = 11

41 -02

01

-11

-41 = 21

--43

-01 54 = ? 21 --43 , ? -

6

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