Math Linear algebra, Spring Semester Dan Abramovich

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1 Math 52 - Linear algebra, Spring Semester Dan Abramovich Review We saw: an algorithm - row reduction, bringing to reduced echelon form - answers the questions: - consistency (no pivot on right) - dimension (count free variables) - write all solutions: free variables arbitrary basic variable x i appearing as pivot in row k: the equation says reduced echelon... x i + a k i+ x i a kn x n = b k x i = b k (a k i+ x i a kn x n ).

2 people asked if we get the same thing!! we may! and which basic 2 Theorem said: reduced echelon form is unique. Say x j is a free variable. Set x j = and all other free variables to be. This determines all the basic variables. We get the equation x i a kj ++ = b k, x i + a kj = b k, which means a kj = b k x i. This determines U, given which variables are free! To determine that it is best to think in terms of vectors.

3 Vectors A column vector is a matrix with one column. Here is a 3-vector: a b c The order of the entries does matter. 3

4 do example do example 4 Basic operations:. addition - you add two vectors of the same size and get again a vector of the same size: a b + c e f = g a + e b + f c + g 2. rescaling, or multiplication by a scalar: a da d b = db c dc

5 5 Notation issues: different texts and different people use different notations. I grew up with b. a c Vectors are also denoted in different ways: x, x, x or x. Get used to it!

6 examples! but we will imagine 6 Given a coordinate system, 2-vectors correspond to points in the plane. 3- vecotrs to points in space. We can interpret the basic operations of addition and rescaling - you may have seen this in calculus. The vector of all zeros is the zero vector, denoted n-vectors? we give up imaging them. Physicists know that they do appear in nature. So do computer scientists.

7 7 The operations have the usual properties (p27): u + v = v + u (u + v) + w = u + (v + w) u + = u = + u c(u + v) = cu + cv (c + d)u = cu + du c(du) = (cd)u u = u. we ll see these as axioms for a vector space

8 a most important notion 8 linear combination. we say that the vector example! c v + + c k v k is a linear combination of the vectors v,..., v k. The set of all linear combinations of v,..., v k is their span. Example: All plane vectors [ ] are linear combinations of e = and e [ ] 2 =

9 [ ] 2 Question: is the vector b = 3 a linear [ ] combination[ of] 3 a = and a =? Namely are there x, x 2 so that we have the vector equation x a + x 2 a 2 = b? 9 do this

10 A vector equations x a + x n a n = b is equivalent to the system of linear equations with augmeted matrix.... a a 2... a n b.... It answer the question: In what ways is b a linear combination of a, a 2,..., a n?

11 Uniqueness of reduced echelon form, one more step: We need to determine basic variables, so pivot columns. In an echelon form U, the k-th column is a pivot column exactly if it is not a linear combination of the previous columns. still need to show: the same holds on A!!

12 justify notation 2 The equation Ax = b Definition Say m n matrix A has columns x a, a 2,..., a n, and x =.. x n Then Ax = x a + + x n a n Do this: 2 [ ] Note: matching n, result is an m vector.

13 3 The system of linear equations with augmeted matrix.... a a 2... a n b.... Is equivalent to the vector equation Do this Ax = b x =

14 columns span R m 4 Theorem. A an m n matrix. The following are equivalent: () Ax = b has a solution for every b R m (2) every b is a linear combination of columns of A (3) The echelon form of A has a pivot in every row. Proof.

15 alternative rules r row rule: A =. then r m r x Ax =. r m x Completely explicitly aj x j Ax = a2j x j. anj x j 5

16 6 linearity of operation Theorem: A(u + v) = Au + Av A(cu) = cau Proof. () [ ] u a a 2... a n. + u n = [ ] a a 2... a n v. v n u + v. u n + v n = (u + v )a + + (u n + v n )a n = (u a + + u n a n ) +(v a + + v n a n ) = Au + Av

17 Vector form of solutions We looked at the system given by augmented matrix which has reduced echelon form The original system is Ax = b where A = ; b =

18 8 Start with the associated homogeneous equation Ax h =. which has reduced echelon form 3 2 and solutions x = 3x 3, x 2 = 2x 3, x 4 =. In vector form: 3x 3 3 x h = 2x 3 x 3 = x 2 3 = x 3 v at least there is always the trivial solution!

19 9 The original system has solutions x = 5+3x 3, x 2 = 3 2x 3, x 4 =. = x = x 3 3 2x 3 x 3 + x 3 =: p + x 3 v 3 2 This is called a paprametric vector solution. Do this: Ax = b with A = [ ], b = [ ].

20 2 x = + x 2 + x 4 Always: particular + general homogeneous. Theorem. If p is one solution of Ax = b, then x is a solution if and only if it is of the form x = p + x h, where x h satisfies Ax h =. Proof.

21 2 Example: a chemical reaction combines a number of Al 2 O 3 molecules and a number of C atoms resulting in an amount of Al and an amount of CO 2. What can we say about these amounts? x x 2 = x 3 Ax =, where A = (Do this) + x

22 22 Reduced echelon: so x = x 4 2/3 4/3 2/3 4/3 To get integers take x 4 = 3 and get 2Al 2 O 3 + 3C 4Al + 3CO 2.

23 netweork: 23 x x x 3

24 24 x + x 2 = x x 3 = 2 x 2 + x 3 = 3 The bottom two are the reduced echelon x x 3 = 2 x 2 + x 3 = 3 2 x = 3 + x 3

25 INDEPENDENCE Consider the vector equation 25 x v + + x n v n =. Definition. The vectors v,..., v n are linearly independent if the only solution is x =, and linearly dependent if there is a nonzero solution. discuss explicitly Examples: dependent or independent? 2 2, 4,? 2 2 2,,? 2

26 26,,? When is v linearly dependent? What does it mean for v, v 2 to be linearly dependent? Similar for v,..., v n to be linearly dependent?

27 Theorem v,..., v n are linearly dependent if and only if at least one of them is in the span of the others. Proof. 27 do it now

28 28 Columns of [ ] 2? 2 do it now Proposition. The columns of A are linearly independent if and only if Ax = b has only the trivial solution. Proof.

29 29 Theorem If v i are m-vectors, and n > m, then v,..., v n are linearly dependent. Proof. do it now