CHAPTER 5 REVIEW Throughout this note, we assume that V and W are two vector spaces with dimv = n and dimw = m. T : V W is a linear transformation.. A map T : V W is a linear transformation if and only if T (c v + c v ) = c T (v ) + c T (v ), for all v, v V and all scalars c, c. Every linear transform T : R n R m can be expressed as the matrix product with an m n matrix: T (v) = [T ] m n v = [ T (e ) T (e ) T (e n ) ] v, for all n-column vector v in R n. Then matrix [T ] m n is called the matrix of transformation T, or the matrix representation for T with respect to the standard basis. Remark.. More generally, given arbitrary basis B = {v,, v n } of R n, T (v) = [T B ] m n v = [ T (v ) T (v ) T (v n ) ] c., c where v = c v + + c n v n. c n c c can be considered as the coordinates of v. c n with respect to the basis {v,, v n }. Example.. Let T : P 3 P be a linear transformation defined by T (a + a t + a t + a 3 t 3 ) = (a + a 3 ) + (a + a )t + (a + a + a + a 3 )t. Then the matrix of T relative to the bases. If we choose the standard bases {t 3, t, t, }, {t, t, } for P 3 and P, respectively. Then [T ] =.
CHAPTER 5 REVIEW. How to find the image of a vector under a linear transformation. Example.3. Let T : R R be a linear transformation given by [ [ ] [ [ 3 4 T ( ) =, T ( ) =. ] 3 ] ] [ 4 Find T ( ). 3] Solution. We first try to find constants c, c such that [ [ ] [ ] 4 = c 3] + c. It is not a hard job to find out that Therefore, c =, c =. [ [ [ ] 4 3 4 T ( ) = = 3] 3 ] [ ]. 4 Example.4. T is a linear transformation from P to P, and T (x ) = x + x 3, T (x) = 4x, T (3x + ) = x + 6. Find T (), T (x), and T (x ). Solution. We identify T as a linear transformation from R 3 to R 3 by the map a ax + bx + c b. c By the given conditions, we have T (,, ) = (,, 3), T (,, ) = (, 4, ), T (, 3, ) = (,, 6). We immediately have T (,, ) = T (,, ) = (,, ). Let Solving A = 3. A x =,
CHAPTER 5 REVIEW 3 we get x = 3 4. Therefore, T (,, ) = 4 3 4 =. 3 6 3 Finally, T (,, ) = T (,, ) + T (,, ) =. Now we restore this result back to the space P and obtain T () = x + 3, T (x) = x, T (x ) = x x. Summary: Suppose the images of a basis B = {v,, v n } are given, i.e., we known T (v ),, T (v n ). For a given vector v,. we identify T as a linear transformation from R n to R m ;. write down the representation matrix [T B ]; 3. find the coordinates of v, i.e., v = c v + + c n v n ; 4. T (v) = [T B ] c. c n ; 5. restore the result in R n and R m to the original vector spaces V and W. If you are asked to find the images for all vectors in V,. we identify T as a linear transformation from R n to R m ;. follow the steps given above to find T (e ), T (e n ); 3. write down the representation matrix [T ]; 4. T (v) = [T ]v; 5. restore the result in R n and R m to the original vector spaces V and W. 3. Kernel and Range
4 CHAPTER 5 REVIEW Example.5. T is a linear transformation from P to P. Moreover, T (a + bx) = (a 3b) + (b 5a)x + (a + b)x. Find Ker(T ) and Rng(T ). Solution. We identify T as a linear transformation from R to R 3. By the given conditions, we have [ [ T ( =, T ( ) = 5. ] ] 3 So the representation matrix [T ] of T is 5. 3 Ker(T ) = Null space of [T ]. For any m n matrix A, So Ker(T ) = {}. On the other hand, rank(a) + dim(null space of A) = n. Rng(T ) = {T (ax + b) : a, b R} = {(a 3b) + (b 5a)x + (a + b)x : a, b R} = {a( 5x + x ) + b( 3 + x + x ) : a, b R} = span{ 5x + x, 3 + x + x }. dim(rng(t )) =. Summary: Kernel. we identify T as a linear transformation from R n to R m ;. find the representation matrix [T ] = [ T (e ) T (e n ) ] ; 4. Ker(T ) is the solution space to [T ]x =. 5. restore the result in R n to the original vector space V. Example.6. Find the range of the linear transformation T : R 4 R 3 whose standard representation matrix is given by 3 A = [T ] = 3 4. 5 4
CHAPTER 5 REVIEW 5 Solution. Rng(T ) = colspace([t ]). Note that the colspace([t ]) consists of x all the vectors v = y such that the homogeneous linear system [T ]x = v z is consistent. 3. x 9. 4x y 3 4. y 7 7. y 3x. 5 4. z. y + z x The above system is consistent if and only if y + z x =, that is, x colspace([t ]) consists of all vectors v = y satisfying y + z x =. z In other words, colspace([t ]) is the plain y + z x =. Remark.7. This example actually gives an algorithm to find colspace(a) and rowspace(a) of a matrix A. Summary: Range. we identify T as a linear transformation from R n to R m ;. find the representation matrix [T ] = [ T (e ) T (e n ) ] ; 3. Rng(T ) = colspace([t ]), which is a subspace of R m ; 4. restore the result in R m to the original vector space W. 4. How to find eigenvalues and eigenvectors/eigenspaces? Example.8. Find the eigenvalues and eigenspaces of the matrix 5 6 A = 3 6. 3 8 Determine A is defective or not. Solution. The characteristic polynomial is given by p(λ) = det(a λi 3 ) = (λ ) (λ + ). So the eigenvalues are λ =, λ =. Their multiplicities are m =, m =. Since 4 A I 3.
6 CHAPTER 5 REVIEW The eigenspace with respect to λ = is 4 E = span{, }. Similarly, the eigenspace with respect to λ = is E = span{ }. We have dime i = m i for i =,. So A is non-defective. Example.9. Find the eigenvalues and eigenspaces of the matrix [ ] 6 5 A =. 5 4 Determine A is defective or not. Solution. The characteristic polynomial is given by p(λ) = det(a λi ) = (λ ). So the eigenvalues are λ = with multiplicity m =. Since [ ] A I. Thus, the eigenspace with respect to λ = is [ ] E = span{ }. We have dime = < m =. So A is defective. Summary:. Factorize the characteristic polynomial p(λ) = det(a λi) = ±(λ λ ) m (λ λ k ) m k. Then λ,, λ are all the eigenvalues of A.. The eigenspace E i with respect to the eigenvalue λ i is the solution space to the homogeneous linear system (A λ i I)x =. All the nonzero vectors in E i are the eigenvectors with respect to the eigenvalue λ i. 3. A is non-defective if and only if dime i = m i for k =,, k. Remark.. An n n matrix A is non-defective if A has n distinct roots. If A has some repeated root(s), then while check defectiveness A, one only need to check the dimension(s) of the eigenspace(s) E i with respect to the eigenvalue(s) λ i with multiplicity m i.