Matrix transformations #5 - Questions and Answers

Question 1 How can I find the dimension of the eigenspace? The matrix 𝐴 = [ 9 − 1 1 7 ] A=[ 9 1 ​ −1 7 ​ ] has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace. So we found the eigenvalue by doing 𝐴 − 𝜆 𝐼 A−λI to get: 𝜆 = 8. λ=8. But how exactly do we find the dimension of the eigenspace? Answer 1 The dimension of the eigenspace is given by the dimension of the nullspace of 𝐴 − 8 𝐼 = [ 1 − 1 1 − 1 ] , A−8I=[ 1 1 ​ −1 −1 ​ ], which one can row reduce to [ 1 − 1 0 0 ] , [ 1 0 ​ −1 0 ​ ], so the dimension is 1. Note that the number of pivots in this matrix counts the rank of 𝐴 − 8 𝐼 A−8I. Thinking of 𝐴 − 8 𝐼 A−8I as a linear operator from 𝑅 2 → 𝑅 2 R 2 →R 2 , the dimension of the nullspace of 𝐴 A is given by dim ⁡ ( 𝑅 2 ) − rank ( 𝐴 ) = 2 − 1 = 1 dim(R 2 )−rank(A)=2−1=1 by the so-called rank-nullity theorem. Of course, one can be more explicit: it is straightforward to see that the nullspace of 𝐴 − 8 𝐼 A−8I is spanned by the vector ( 1 , 1 ) (1,1), and hence has dimension 1. Answer 2 By definition, an eigenvector 𝑣 v with eigenvalue 𝜆 λ satisfies 𝐴 𝑣 = 𝜆 𝑣 Av=λv, so we have 𝐴 𝑣 − 𝜆 𝑣 = 𝐴 𝑣 − 𝜆 𝐼 𝑣 = 0 , where  𝐼  is the identity matrix. Av−λv=Av−λIv=0,where I is the identity matrix. Thus, ( 𝐴 − 𝜆 𝐼 ) 𝑣 = 0 (A−λI)v=0, and 𝑣 v is in the nullspace of 𝐴 − 𝜆 𝐼 A−λI. Since the eigenvalue in your example is 𝜆 = 8 λ=8, to find the eigenspace related to this eigenvalue we need to find the nullspace of 𝐴 − 8 𝐼 A−8I, which is the matrix [ 1 − 1 1 − 1 ] . [ 1 1 ​ −1 −1 ​ ]. We can row-reduce it to obtain [ 1 − 1 0 0 ] . [ 1 0 ​ −1 0 ​ ]. This corresponds to the equation 𝑥 − 𝑦 = 0. x−y=0. So 𝑥 = 𝑦 x=y for every eigenvector associated to the eigenvalue 𝜆 = 8 λ=8. Therefore, if ( 𝑥 , 𝑦 ) (x,y) is an eigenvector, then ( 𝑥 , 𝑦 ) = 𝑥 ( 1 , 1 ) (x,y)=x(1,1), meaning that the eigenspace is 𝑊 = span { ( 1 , 1 ) } W=span{(1,1)}, and its dimension is 1. Question 2 Discuss the cases of Gaussian elimination method. Answer 1 First write down the given equations in the form of a matrix. Next, by applying the row transformations, let’s reduce it in the form of an upper triangular matrix, i.e., the elements below the diagonal are zero. Next, let’s form an equation from the upper triangular matrix. By back substitution, we will find the values of 𝑥 x. Answer 2 To perform row reduction on a matrix, one uses a sequence of elementary row operations to modify the matrix until the lower left-hand corner of the matrix is filled with zeros as much as possible. There are three types of elementary row operations: Swapping two rows. Multiplying a row by a nonzero number. Adding a multiple of one row to another row. Using these operations, a matrix can always be transformed into an upper triangular form, and in fact, one that is in row echelon form. Once all of the leading coefficients (the leftmost nonzero entry in each row) are 1, and every column containing a leading coefficient has zeros elsewhere, the matrix is said to be in reduced row echelon form. This final form is unique; in other words, it is independent of the sequence of row operations used. For example, in the following sequence of row operations (where two elementary operations on different rows are done at the first and third steps), the second and fourth matrices are the ones in row echelon form, and the final matrix is the unique reduced row echelon form: [ 1 3 1 − 1 3 1 5 0 − 2 − 8 8 − 3 ] → [ 1 3 1 − 1 0 − 8 2 3 0 2 3 − 1 ] → ⋯ → [ 1 0 0 − 2 0 1 0 4 0 0 1 − 3 ] . ​ 1 3 −2 ​ 3 1 −8 ​ 1 5 8 ​ −1 0 −3 ​ ​ → ​ 1 0 0 ​ 3 −8 2 ​ 1 2 3 ​ −1 3 −1 ​ ​ →⋯→ ​ 1 0 0 ​ 0 1 0 ​ 0 0 1 ​ −2 4 −3 ​ ​ . Using row operations to convert a matrix into reduced row echelon form is sometimes called Gauss-Jordan elimination. Question 3 Find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations. The line of 𝑦 = − 2 𝑥 3 y=− 3 2x ​ all in 𝑅 2 R 2 . Answer 1 So you have a vector ( 𝑥 1 , 𝑦 1 ) 𝑇 (x 1 ​ ,y 1 ​ ) T , and you need to locate a matrix 𝑀 M such that ( 𝑥 2 , 𝑦 2 ) 𝑇 = 𝑀 ( 𝑥 1 , 𝑦 1 ) 𝑇 (x 2 ​ ,y 2 ​ ) T =M(x 1 ​ ,y 1 ​ ) T is the reflection across 𝑦 = − 2 𝑥 3 y=− 3 2x ​ . What does the reflection mean? It means that the “middle” of the two points is on the line of reflection, and the line between those points is perpendicular to the reflection line. The first condition can be written as 𝑦 1 + 𝑦 2 2 = − 2 3 𝑥 1 + 𝑥 2 2 . 2 y 1 ​ +y 2 ​ ​ =− 3 2 ​ 2 x 1 ​ +x 2 ​ ​ . The second condition means that the slope of the line between the two points is − 1 𝑚 − m 1 ​ , where 𝑚 m is the slope of the reflection line: 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = − 1 2 3 = − 3 2 . x 2 ​ −x 1 ​ y 2 ​ −y 1 ​ ​ =− 3 2 ​ 1 ​ =− 2 3 ​ . Now type 𝑥 2 x 2 ​ and 𝑦 2 y 2 ​ in terms of 𝑥 1 x 1 ​ and 𝑦 1 y 1 ​ : 𝑥 2 = 𝑚 11 𝑥 1 + 𝑚 12 𝑦 1 , 𝑦 2 = 𝑚 21 𝑥 1 + 𝑚 22 𝑦 1 . x 2 ​ =m 11 ​ x 1 ​ +m 12 ​ y 1 ​ ,y 2 ​ =m 21 ​ x 1 ​ +m 22 ​ y 1 ​ . Next, the matrix you are looking for has these coefficients: 𝑀 = [ 𝑚 11 𝑚 12 𝑚 21 𝑚 22 ] . M=[ m 11 ​ m 21 ​ ​ m 12 ​ m 22 ​ ​ ]. Answer 2 Rotating the plane to make the line appear straighter is one tactic. If the line is 𝑦 = 0 y=0, apply the reflection in the 𝑥 x-axis, and then rotate back. First, tan ⁡ 𝜃 = 3 2 tanθ= 2 3 ​ . Then the required matrix is [ cos ⁡ 𝜃 sin ⁡ 𝜃 sin ⁡ 𝜃 − cos ⁡ 𝜃 ] . [ cosθ sinθ ​ sinθ −cosθ ​ ].