Professor Joana Amorim, jamorim@bu.edu What is on this week Vector spaces (continued). Null space and Column Space of a matrix............................. Null Space...........................................2 Column Space....................................... 3..3 Kernel and Range of Linear Transformations...................... 3.2 Basis, dimension and Rank.................................... 4.2. Definition of basis and dimension............................ 4.2.2 Basis for the spanning set of a set of vectors...................... 6.2.3 Basis for NulA...................................... 7.2.4 Basis for ColA...................................... 7.2.5 The rank Theorem.................................... 8 2 Eigenvalues and Eigenvectors 0 2. Definitions............................................. 2.2 How to find eigenvalues?..................................... 3 3 Complex eigenvalues 6 Vector spaces (continued). Null space and Column Space of a matrix In the book: Section 4.2 pages 98 207 (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.
3 5 3 Example 3 (Ex., section 4.2) Determine if w = 3 is in NulA where A = 6 2 0. 4 8 4 Example 4 (Ex. 3 and 5, section 4.2) Find an explicit description of NulA by listing vectors that span the Null space. [ ] 2 4 0 a. 0 3 2 b. 4 0 2 0 0 0 5 0 0 0 0 0 2 2
..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 = 3 2 9 6. 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
.2 Basis, dimension and Rank In the book: Section 4.3 pages 208 23, Exercises 20. Section 4.5 pages 225 228, Exercises 8. Section 4.6 pages 230 236 (ignore row space), Exercises 6..2. 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
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,, 0 0 3 2 2. 0,, 3 4 3 3 0 0 3. 3, 7, 0, 3 0 0 0 5 2 6 4. 3, 0 5 5
Example 6 (Section 4.3, Ex. ) Find a basis for the plane in R 3 defined by x 3y + 2z = 0..2.2 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, 0 3 5 6
.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. 0 2 2. 0 4 3 7 3 7
2. 2 5 0 0 2 0 0 8 0 6.2.5 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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2 Eigenvalues and Eigenvectors In the book: Section 5. pages 266 270, Exercises 20. Section 5.2 pages 273 276. 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
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? 3 8 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
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 = 5. 3 4 2. B = 2 3 2, λ = 5. 3 3 2 2
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. [ ] [ ] 4 9 2 A =, B = 6 2 5 3
C = [ 4 ] 4 2 Example 30 Find the eigenvalues and a basis to the corresponding eigenspaces of 4 0 D = 0 4 0 2 4
Example 9 (cont.) 5
3 Complex eigenvalues In the book: Section 5.5 pages 295 300, 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
Example 3 ( 5.5, Ex. 2, 4) Find the eigenvalues and a basis for each eigenspace in C 2. [ ] [ ] 3 3 2 A =, B = 3 3 3 7
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
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