What is on this week. 1 Vector spaces (continued) 1.1 Null space and Column Space of a matrix

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1 Professor Joana Amorim, What is on this week Vector spaces (continued). Null space and Column Space of a matrix Null Space Column Space Kernel and Range of Linear Transformations Basis, dimension and Rank Definition of basis and dimension Basis for the spanning set of a set of vectors Basis for NulA Basis for ColA The rank Theorem Eigenvalues and Eigenvectors 0 2. Definitions How to find eigenvalues? Complex eigenvalues 6 Vector spaces (continued). Null space and Column Space of a matrix In the book: Section 4.2 pages (only R n examples). Exercises section 4.2: all with R n examples but specially the ones similar to the homework... Null Space Definition Let A be a m n matrix. NulA = {x in R n such that Ax = 0} = set of all solutions of the homogeneous equation Ax = 0. Theorem 2 Let A be a m n matrix. Then NulA is a subspace of R n. Proof : Page 200.

2 3 5 3 Example 3 (Ex., section 4.2) Determine if w = 3 is in NulA where A = Example 4 (Ex. 3 and 5, section 4.2) Find an explicit description of NulA by listing vectors that span the Null space. [ ] a b

3 ..2 Column Space Definition 5 Let A be a m n matrix. ColA = {b such that b = Ax for some x in R n } = set of all linear combinations of columns of A. This is a subset of R m Theorem 6 Let A be a m n matrix. Then ColA is a subspace of R m. 6 4 Example 7 (Ex. 7, section 4.2) Let A = Find k such that NulA is a subspace of Rk. 9 6 Find s such that ColA is a subspace of R s. Example 8 (Ex. 23, section 4.2) Let A = NulA. [ 2 ] 4 2 and w = [ ] 2. Determine if w is in ColA or in..3 Kernel and Range of Linear Transformations Definition 9 Let T : R n R m be a linear transformation. The kernel T (ker(t )) are the vectors x in R n such that T (x) = 0. The range of T (range(t )) are the vectors b in R m for which there is a x in R n such that T (x) = b. Proposition 0 Let T : R R m be a linear transformation and A its standard matrix. Then Remark Hence if T (x) = Ax T is one-to-one iff ker(t ) = NulA = {0} T is onto iff range(t ) = ColA = R m ker(t ) = NulA and range(t ) = ColA. Example 2 Let T : R 2 R 2 be the linear map that does a projection onto the x-axis followed by an expansion by a factor of 3. Find ker(t ) and range(t ). 3

4 .2 Basis, dimension and Rank In the book: Section 4.3 pages , Exercises 20. Section 4.5 pages , Exercises 8. Section 4.6 pages (ignore row space), Exercises Definition of basis and dimension The definition we saw for Linearly Independent set also applies in general: Let v,..., v p be a set of vectors in a vector space V. They are linearly independent iff the only solution of a v a p v p = 0 is the trivial solution. Equivalently, the vectors are linearly dependent iff at least one of them can be written as a linear combination of the rest. Definition 3 Let V be a vector space. A set of vectors B = {v,..., v n } is called a basis for V if B is a linearly independent set; B spans V, ie, V = span B. Important remarks: All basis of a vector space have the same number of elements. The dimension of a vector space is the number of elements in a basis. If a vector space has dimension n there are at most n linear independent vectors in that space. If you have n LI vectors in a space of dimension n then they automatically span the set. Example 4 R n is a vector space of dimension n, dim(r n ) = n. The standard basis of R n is the set of coordinate vectors {e,..., e n }. 4

5 Example 5 (Section 4.3, Exercises, 3, 5 and 7) Determine whether the following sets are bases for R 3. Of the sets that are not basis determine which ones are LI and which ones span R 3.. 0,, ,, , 7, 0, , 0 5 5

6 Example 6 (Section 4.3, Ex. ) Find a basis for the plane in R 3 defined by x 3y + 2z = Basis for the spanning set of a set of vectors Let s now see how do we find the basis for the spanning set of a set of vectors. Let V = span{v,..., v p }. If the vectors are LI then they are a basis. Otherwise we need to remove the vectors that are linear combination of others. How? Consider the matrix that has the vectors as columns [v,..., v p ] and do row operation to get it to echelon form. Look for the pivots. The original columns corresponding to pivot columns are LI (and hence a basis). Example 7 Find a basis for span 2, 4,

7 .2.3 Basis for N ula Let A be a m n matrix. Remember that To find a basis for NulA one NulA = {x in R n such that Ax = 0}. Solves the system Ax = 0 and gives the solution in vector parametric form (using reduced echelon form). The vectors in the solution are a basis. If Ax = 0 only has the zero solution, there is no basis, since the space is just {0}. Hence dim(nula) = number of free variables in the system Ax = 0. If there are none it means there is only one solution and the dimension is zero in that case..2.4 Basis for ColA Let A be a m n matrix. Remember that ColA = span{columns of A}. So to find a basis for ColA we use the method described above in section.2: see which columns have pivots in echelon form; the basis are the correspondind original columns. Definition 8 The rank of a matrix is the dimension of its column space (which is the number of pivots of the matrix in echelon form). Example 9 (Section 4.3, Ex. 9 and 0) Find bases for the Null and the column spaces of the following matrices. Give their dimension

8 The rank Theorem Theorem 20 (The rank Theorem) Let A be a m n matrix. Then ranka + dim NulA = n. Example 2 (Section 4.6, Ex. and 3) Assume that the matrix A is row equivalent to B. Without calculations list ranka and dim NulA. Then find bases for NulA and ColA. 8

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10 2 Eigenvalues and Eigenvectors In the book: Section 5. pages , Exercises 20. Section 5.2 pages Exercises 8 Goal: dissect the action of a linear map T (x) = Ax into elements that are easily visualized. Eigenvectors and eigenvalues are usefull in discrete dynamical systems, differential equations, etc... [ ] 3 2 Example 22 Let A =. Consider the transformation T (x) = Ax. Let s see how this matrix acts [ ] 0 [ ] 2 on two vectors: v =, v = 0

11 2. Definitions Definition 23 Let A be a n n matrix. An eigenvector is a non-zero vector v such that for some scalar λ. Av = λv The scalar λ is called an eigenvalue. So an eigenvalue is a scalar such that Av = λv has a solution v 0 Av λv = 0 has a solution v 0 (A λi)v = 0 has a non-trivial solution. The subspace Nul(A λi) is called the eigenspace corresponding to the eigenvalue λ. Example 24 ( 5., Ex., 2). Is λ = 2 and eigenvalue of A = [ ] 3 2? Is λ = 3 an eigenvalue of A = [ ] 4? 6 9 Example 25 ( 5., Ex. 4) Is v = [ ] an eigenvector of A = [ ] 5 2? If so find eigenvalue. 3 6

12 Example 26 ( 5., Ex. 0, 5) Find a basis for the eigenspace (and state its dimension) corresponding to each given eigenvalue. [ ] 4 2. A =, λ = B = 2 3 2, λ =

13 2.2 How to find eigenvalues? Proposition 27 The eigenvalues of a triangular matrix are the entries on its main diagonal. In general: Proposition 28 Let A be a n n matrix. A scalar λ is an eigenvalue iff det(a λi) = 0. This equation is the characteristic equation of the matrix A. The multiplicity of an eigenvalue is the number of times that eigenvalue is a zero of the characteristic equation. Example 29 ( 5.2, Ex. 2, 6, 2) Find the characteristic equation and the real eigenvalues of the matrices. [ ] [ ] A =, B =

14 C = [ 4 ] 4 2 Example 30 Find the eigenvalues and a basis to the corresponding eigenspaces of 4 0 D =

15 Example 9 (cont.) 5

16 3 Complex eigenvalues In the book: Section 5.5 pages , Exercises 2. Important application on vibrations, periodic motions,... If a real matrix has a complex eigenvalue λ them its complex conjugate λ its also an eigenvalue for the matrix. If v is an eigenvector corresponding to λ then the complex conjugate of v is an eigenvector corresponding to λ. 6

17 Example 3 ( 5.5, Ex. 2, 4) Find the eigenvalues and a basis for each eigenspace in C 2. [ ] [ ] A =, B =

18 When a real matrix A has complex eigenvalues it means the map T (x) = Ax contains a rotation. Let s look at a special case of this where it is easy to identify such a rotation. [ ] a b Example 32 Let C =, where a, b are non-zero real numbers. Then the eigenvalues of C are b a λ,2 = a ± bi and, if we call r = λ = a a + b 2 and ϕ = arg(λ ), we can write C as [ ] a/r b/r C = r = b/r a/r [ r 0 0 r ] [ cos ϕ sin ϕ sin ϕ cos ϕ ]. 8

19 Example 33 ( 5.5, Ex. 2) Use the example above to list the eigenvalues of the matrix write the map T (x) = Ax as a composition of a rotation and a scaling identify the angle ϕ of rotation (ϕ in ( π, π)) and the scaling factor r [ ] 3 3 A = 3 3 9