Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone Text: Linear Algebra With Applications, Second Edition Otto Bretscher

Monday, Mar 3 Chapter 4.2 Page 164 Problems 6,14,16 Main Idea: Dejavu! We are doing everything all over again. Key Words: Kernel, Image, Linear Transformation, rank, nullity Goal: We want to expand the ideas of Matrices into a broader context.

Previous Assignment The Leontief Problem The production of the plants R, S, and T for some period of time is given below. R S T Consumer Total R S T The Leontief input-output model for the open model is: X = AX + D.

(1) What is the matrix A? (2) Solve the equation X = AX+D for X | 145 | when D = | 290 | | 145 |

A | 10/80 10/60 30/100 | | 10/80 20/60 10/100 | | 60/80 20/60 10/100 |

Check with original system: A X + D = X |10/80 10/60 30/100|| 80| |30| | 80| |10/80 20/60 10/100|| 60| + |20| = | 60| |60/80 20/60 10/100||100| |10| |100| It checks.

Calculating X the new system where consumer | 145 | demands is | 290 | | 145 | AX+D=X D = (I-A)X (I-A) -1 D = X

|| | |10/80 10/60 30/100| | |145| |616| || | - |10/80 20/60 10/100| | |290| =|690| || | |60/80 20/60 10/100| | |145| |930|

| | | - -(--) -(---) | | | | | | | | 145 | X = | -(-) -- -(---) | | 290 | | | | 145 | | | | | | -(-) -(--) --- | | |

| | | | | | | | | | | 145 | X = | | | 290 | | | | 145 | | | | | | | | |

| 616 | X = | 690 | | 930 |

Previous Assignment Page 157 Problem 4 Which of the subsets of P 2 given in Exercises 1 through 5 are subspaces of P 2 ? Find a basis for those that are subspaces. t=1 { p(t) | INT p(t) dt = 0} t=0

t=1 It is closed under addition since if INT p(t) dt = 0 t=1 t=0 and INT q(t) dt = 0, t=0 t=1 t=1 t=1 then INT(p(t) + q(t)) dt = INT p(t) dt + INT q(t) dt t=0 t=0 t=0 = = 0 So p(t) + q(t) is also in the subspace.

It is closed under scalar multiplication since if c is a number and t=1 t=1 t=1 INT p(t) dt = 0, then INT c p(t) dt = c INT p(t) dt = t=0 t=0 t=0 c 0 = 0 so c p(t) is also in the subspace.

Now we want to find a basis of the subspace. {1,x,x 2 } of P 2. If p(x) = a + bx + cx 2 is any element of p 2, | | x=1 then p(x) is in the space if |ax+bx 2 /2 +c x 3 /3|= 0 | | x=0 That is, if and only if a + b/2 + c/3 = 0

This is a linear system [ 1 1/2 1/3 | 0 ] in Row Canonical Form. |a| |-1/2| |-1/3| |b| = u | 1 | + v | 0 | |c| | 0 | | 1 | So the basis is -1/2 + x, and -1/3 + x 2.

Page 157 Problem 10 Which of the subsets of R 3x3 given in Exercises 6 through 11 are subspaces of R 3x3. | 1 | The 3x3 matrices A such that vector | 2 | is in the kernel of A. | 3 |

The set is closed under addition since if AV = 0, and BV = 0, then (A+B)V = AV + BV = 0 so A+B is also in the set. The set is closed under scalar multiplication since if c is a number and AV = 0, then (cA)V = c(AV) = c0 = 0 so cA is also in the subset.

A basis of the space is | a b c | | 1 | | 0 | | d e f | | 2 | = | 0 | | g h i | | 3 | | 0 | a + 2 b + 3 c = 0 d + 2 e + 3 f = 0 g + 2 h + 3 i = 0

x 1 x 2 x 3 x 4 x 5 x 6 a b c d e f g h i RHS | | | | | | |a| |-2| |-3| | 0| | 0| | 0| | 0| |b| | 1| | 0| | 0| | 0| | 0| | 0| |c| | 0| | 1| | 0| | 0| | 0| | 0| |d| | 0| | 0| |-2| |-3| | 0| | 0| |e| = x 1 | 0| + x 2 | 0| + x 3 | 1| +x 4 | 0| +x 5 | 0|+x 6 | 0| |f| | 0| | 0| | 0| | 1| | 0| | 0| |g| | 0| | 0| | 0| | 0| |-2| |-3| |h| | 0| | 0| | 0| | 0| | 1| | 0| | i| | 0| | 0| | 0| | 0| | 0| | 0|

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Page 157 Problem 20 Find a basis for each of the spaces in Exercises 16 through 31 and determine its dimension. The space of all matrices A = | a b | in R 2x2 such that a+d = 0. | c d |

The basis written as 2x2 matrices is: | 0 1 | | 0 0 | |-1 0 | | 0 0 | | 1 0 | | 0 1 | The dimension is 3.

New Material A linear transformation T:V  W requires (1) V and W to be vector spaces. (2) T(V 1 +V 2 ) = T(V 1 )+T(V 2 ). (3) T(c V) = c T(V).

_________ ____________ | V | | W | | | | iiiii | | kk | T |iiiiiiii | | kkkk | >|0iiiiii | | kkk | |iiii | | | | ii | |_________| |____________|

Points of interest. (4) Ker(T) = { V | T(V) = 0}. (5) Im(T) = {T(V) | V is in V} (6) T is invertible is equivalent to (a) T(V) = 0 ==> V = 0. Ker(T) = 0 (b) T(X) = W is solvable for every W. Im(T)=W

Main Theorem: Dim(V) = Dim( IMAGE ) + Dim(KERNEL). Pretty much that is what you would expect. The quantity you started with is how much you have left plus how much was lost.

We now discuss the kernel and the range for various functions.

Example 1. For differentiation, what is the kernel?

The kernel is the set of all functions which map to zero. That is, those functions whose derivative is 0. Those are the constant functions like: y = constant.

The range is the set of all functions which are derivatives of something. This includes all continuous functions.

Is differentiation invertible?

No because you cannot recover the constant. Both y = x 2 and y = x 2 +1 have the same derivative. Thus having been given only y = 2x, you cannot determine the original function. Thus differentiation does not have an inverse.

Example 2. Is integration invertible?

Let INT f be the indefinite integral of f. INT f+g = INT f + INT g INT cf = c INT f So integration satisfies the requirements to be a linear transformation.

There is one small problem. INT is not a function since as it stands, there is not one, but many possible integrals. We can patch up this problem by saying INT f is the function F such that F' = f and F[0] = 0.

Notice that d/dx INT f = f so that INT has a left inverse. However INT does not have a right inverse since INT d/dx f will only equal f(x) - f(0). Since you do not get f(x) back exactly, integration does not have an inverse.

We once remarked that if a matrix has a left inverse, it also has a right inverse and the two inverses are equal.

This theorem requires that the matrix be square and the space be finite dimensional. It is not true for non square matrices nor for infinite dimensional spaces.

Example 3. Consider the function R 2x2 ----> R 2x2 given by |a b|  | 1 2 | | a b | |c d| | 2 4 | | c d | Is this function invertible?

We see if anything is mapped to zero. That is, is there any matrix | a b | | c d | such that | 1 2 | | a b | = 0 ? | 2 4 | | c d |

Thus a+2c = 0 b+2d = 0 2a+4c = 0 2b+4d = 0 a b c d RHS a b c=u d=v RHS | | | | | | | | | | | | | | | | | a | |-2 | | 0 | | b | = u | 0 | + v |-2 | | c | | 1 | | 0 | | d | | 0 | | 1 |

The kernel contains |-2u -2v | | u v | So since some information is lost, the function is not invertible.

Define T[X] = AX-XA for a fixed matrix A. 1. Show that T is a linear transformation. 2. Show that T is never invertible.

We have to first show that T[X+Y] = T[X]+T[Y].

T[X+Y] = A(X+Y) – (X+ Y)A = AX+AY – XA – YA = AX – XA + AY – YA = T[X] + T[Y].

T[cX] = A(cX) + (cX)A = c(AX+XA) = cT[X].

The kernel of T contains the identity matrix I so T is not invertible.

What is the matrix for T if A = | 1 2 | | 3 4 |

|1 0| |0 1| |0 0| |0 0| |0 0| |0 0| |1 0| |0 1| |1 0|  |1 2| - |1 0| = | 0 2| | | | 0 0| |0 0| |3 0| |-3 0| | | | | |0 1|  |3 4| -| 0 1| = | 3 3| | | |0 0| |0 0| | 0 3| | 0-3| | | | | |0 0 |  |0 0| - | 2 0| =|-2 0| | | |1 0 | |1 2| | 4 0| |-3 2| | | | | |0 0|  |0 0| - | 0 2| =| 0 -2| | | |0 1| |3 4| | 0 4| | 3 0| | |

The matrix for T is | | | | | | | |

What is the kernel of T?

| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | /3 0 | | | | | | | | | | | | |

x y z=a w=b | | | /3 0 | | | The kernel is | x | | -1 | | 1 | | y | = a |2/3| + b | 0 | | z | | 1 | | 0 | | w | | 0 | | 1 |

So the elements in the kernel are linear combinations of | -1 2/3 | and | 1 0 | | 1 0 | | 0 1 | I.E. linear combinations of A and I.

Example 4. Suppose we have a linear transformation T which reflects across a line. What is the matrix of T with respect to the standard basis? | 1 | | 0 | | 0 | | 1 |

Case 1. /|\ / | / y = x | /  /|

Case 2. \ /|\ \ | y = -x \ |  | \