LECTURE 1 I. Inverse matrices We return now to the problem of solving linear equations. Recall that we are trying to find x such that IA = A


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1 LECTURE I. Inverse matrices We return now to the problem of solving linear equations. Recall that we are trying to find such that A = y. Recall: there is a matri I such that for all R n. It follows that I = IA = A for all n n matrices A. For the rest of the day, suppose A is an n n matri. Definition: Let A be an n n matri. We say a matri B is an inverse for A if AB = BA = I. Notation: If B is an inverse for A, we write B = A, and we say A is invertible. Note that it is also true that B = A (check!) Theorem (Invertibility Theorem I): Suppose A has an inverse A. Then the equation has a unique solution, namely A = y = A y. Remark: The converse of this theorem is also true, but we can t prove it yet. We ll do it later in the week when we discuss linear transformations. Proof. We have from which it follows that and by definition of inverse, A = y A A = A y, I = A y,
2 and by definition of the identity, = A y. So if there s a solution, this is it. To verify that this is a solution, check which was the desired result. A(A y) = (AA ) y = I y = y Theorem: Suppose B is inverse to A. Then B is the only inverse to A. Proof. Suppose B and B were two different inverses to A. Then we would have but on the other hand so B = B. BAB = (BA)B = IB = B BAB = B(AB ) = BI = B Theorem: Suppose A and B are both invertible. Then AB is invertible, and its inverse is B A. Proof. We observe that (AB)(B A ) = A(BB )A = AIA = AA = I. Left as eercise to check the other direction. [ ] Not all matrices have inverses. For instance Eample: A = does not have an inverse. Because the equation A = [ ] has solutions for any y; in particular, the solution is not unique, so if y A were invertible it would violate the theorem above. Some large classes of matrices have inverses. Eample: Let D be a diagonal matri, i. e. one which has all its nonzero entries on the diagonal. Suppose that D has no zeroes on the diagonal. Then D is invertible. 2
3 In fact, its inverse can be written in a very simple form. For eample: = /2 /5 /4 Eercise: check this fact. Eample: Let A = 3. I claim that A has an inverse: 2 8 namely, /7 5/4 /4 A = 3/4 9/28 /28 3/4 /28 3/28 To verify this, one needs only check that AA = A A = I. But how did I find the inverse? II. Gaussian elimination and the inverse In order to get this down, we begin with a definition. Definition: An elementary matri is an n n matri with s on the diagonal and eactly one nonzero entry off the diagonal. Eample: E = 3 Observation: E is the vector obtained by adding 3 the first coordinate of to the second coordinate. Thus, if A is a 3 p matri, EA is the matri obtained from A by adding 3 the first row to the second row. Many of you observed this in the groupwork on Friday. Note that E is invertible; its inverse is E = 3 Eercise: check this. Theorem Every elementary matri is invertible. Proof left as an eercise. Use the form of E above as a hint.. 3
4 Write So we have E = 3 3, E 2 = = 3 2 8, E 3 = /3 Then E 3 E 2 E A is the matri obtained from A by subtracting 3 first row from second row, first row from third row, and adding /3 second row to third row. But that s eactly what we did when we did Gaussian elimination. In other words, we have E 3 E 2 E A = 3 28/3 Write U for that latter, upper triangular matri. Now we can write E 4 = 3/28 so that E 4 U = 3 28/3 And likewise, E 5 and E 6 will be elementary matrices chosen to eliminate the other columns. Finally, one gets E 6 E 5 E 4 E 3 E 2 E A = D = 3 28/3 But we know that D is invertible, by the theorems. So can write D E 6 E 5 E 4 E 3 E 2 E A = I.. 4
5 Warning: We haven t really shown that D E 6 E 5 E 4 E 3 E 2 E = A, because we haven t shown that you get I when you multiply in the other order! See Strang for a way around this problem, or (better) look forward to our proof of the following theorem: Theorem: Suppose A, B are square matrices with AB = I. Then BA = I. LECTURE 2 I. Subspaces of R n. This week we re really going to get into the basic theoretical notions of linear algebra vector spaces and linear transformations. Let s jump right in. Definition: A subspace of R n is a subset V R n such that: If v and w are in V, then so is v + w If v is in V, then so is a v for any a R. Eample: Let V R 3 be the set of vectors of the form. y Then v = and w = are elements of V, we have v + w = + y V and a v = a V. Eample: Let V R 3 be the set of vectors of the form. 5
6 This set is not a subspace. Because take, for eample, v =, w =. Then v and w are in V, but their sum v + w is not. So, by definition, V is not a subspace. 2 minute contemplation: With a neighbor, think of all the subspaces of R 3 you can. Regroup, discuss. Remark: a subspace is an eample of a more general notion of a vector space, which section 2. of Strang discusses but which we, for the moment, shall not. Definition: Let v,..., v n be a set of vectors in R m. Then the span of v,..., v n is the set of vectors v + 2 v n v n. Theorem: The span of v,..., v n is a subspace of R m. Proof. Let V be the span of v,..., v n. Let v, w V. Then and for some choices of the i and y i. So v = v + 2 v n v n w = y v + y 2 v y n v n v + w = ( + y ) v ( n + y n ) v n which is also in V. The check of the other condition is left as an eercise. Eample: The span of and is the set of all vectors of the 2 form ; 6
7 as we discussed, this is a plane. Key eample of span: Once again, consider the equation A = y. Can we solve this equation? Well, observe that the set of all is just the set of all A. n v + 2 v n v n as the i range over R; that is, A = y has a solution if and only if y is in the span of the columns of A. (see 2A in Strang). Definition: The column space of A is the span of the columns of A. It is a subspace of R m. Definition: The nullspace of A is the set of vectors R n such that A =. (Eercise: check that the nullspace is a subspace of R n. see Strang, p. 68) II. Linear transformations Let f : R n R m be a function. For instance, a function from R 3 to R 3 might be f y z = 2 y 2 z 2 or a function from R to R 2 might be [ ] f() = sqrt + 2 In multivariable calculus, we studied the general properties of functions of this type or at least those which were nice enough to have derivatives. Now we are going to restrict ourselves to a very special class of function, which turns out to be quite important. Definition: A function f : R n R m is called a linear transformation if (and only if) f( + y) = f( ) + f( y) f(a ) = af( ) for all a f(). 7
8 Eamples:. The functions above are not linear transformations. For instance, note that in the first case f y z + 2 y 2 z 2 = y 2 + y2 2 z 2 + z2 2 2 y 2 z 2. The identity transformation i : R n R n given by f( ) = 2. The zero transformation z : R n R m given by z( ) = + (Possibly skip eample 3) 3. The averaging transformation a : R 3 R 3 given by ( + y + z)/3 a( y ) = ( + y + z)/3 z ( + y + z)/3 Observe: the set of vectors such that a = is vec : + y + z =. What is the set of vectors y such that a( ) = y has a solution? Maybe take a oneminute contemplation here? Geometrically speaking: a projects y z onto the line t 4. Given any m n matri A, the function T A : R n R m defined by is linear. Proof: and T A ( ) = A T A ( + y) = A( + y) = A + A y = T A ( ) + T A ( y) T A (a ) = Aa = aa = at A ( ) Note that 3 is an eample of 4. We can write ( + y + z)/3 /3 /3 ( + y + z)/3 = /3 + y /3 + z ( + y + z)/3 /3 /3 8 /3 /3 /3 2 2 y 2 2 z 2 2 =.. /3 /3 /3 /3 /3 /3 /3 /3 /3 y z.
9 So a = T A where A = /3 /3 /3 /3 /3 /3 /3 /3 /3 ( Feb 2: I got to this point with only minutes left, so briefly talked about the electrical network eample but didn t break into groups.) Now we re going to do a groupwork. We draw a picture of a 4node 6 edge network. (Triangular is best). Point out that we can describe the state of such a network (that is, the charge on each verte) by a vector in R 4. The potential differences (that is, the potential difference on each edge) can be described by a vector in R 6. For eample, if the potentials on the nodes are 3, 4, 5, 6, the potential differences on the edges are 2,,, 3, 2,.. 9
10 GROUPWORK: Linear transformations in an electrical network Suppose is the vector of potentials. Define a function C : R 4 R 6 by the rule that C( ) is the vector of potential differences corresponding to the vector of potentials. Eample: C = What is C( ) for an arbitrary vector of potentials = Show that C is a linear transformation. Is there a matri A such that C = T A? In other words, is there a matri A such that for all R 4? A = C( ) For which R 4 do we have C( ) =? For which y R 6 does C( ) = y have a solution? In other words, which y are possible vectors of potential differences? y z w?
11 That should take 225 minutes. Regroup, have students present solutions. Let them know that there s a dicussion of this kind of system in Strang, 2.5 (optional reading). Observe that we have not discussed the transpose this week leave it to Strang. If there s time, I want to prove the following theorem, which may not be surprising by now. Theorem. Suppose T : R n R m is a linear transformation. Then there eists an m n matri A such that T ( ) = A for all in R m. In other words, all linear transformations come from matrices. And the study of matrices is one and the same as the study of linear transformations. Proof. For each i, let e i be the matri with in the i th place and elsewhere; i. e. e =, e 2 =,.... vdots Now given a linear transformation T, define A = T ( e )... T ( e n ). We claim this is the desired matri. Check: We did it! A. n = T ( e ) + 2 T ( e 2 ) n T ( e n ) = T ( e ) + T ( 2 e 2 ) T ( n e n ) = T T. n = T. n.
12 Using this fact, we can prove a better version of our invertibility theorem from Monday. Theorem (Invertibility Theorem II): Suppose A is an n n matri such that A = y has a unique solution for all vecy R n. Then A is invertible. Proof. We need to prove that there is a matri B such that AB = BA = I. Define a function U : R n R n by the rule: U( y) is the unique such that A = y. Claim: U is a linear transformation. To check: U( y + z) = U( y) + U( z). But what is U( y + z)? It is the unique vector w such that A w = y + z. But note that A[U( y) + U( z)] = AU( y) + AU( z) = y + z. So U( y + z) can be nothing other than U( y) + U( z). Proof of good behavior under scalar multiplication left to reader. Now, by the Theorem above, there eists an n n matri B such that U( y) = B y for all y R n. Claim: BA = I. Proof. Let be some vector in R n. (BA) = B(A ) = B(T ( )) = U(T ( )) which is the unique vector w such that T ( w) = T ( ). That is, it is! So we ve shown (BA) = for all R n ; in other words, BA = I. The proof that AB = I is very similar and is left as an eercise. 2
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