Questions and Answers #1 Matrix Transformations

Questions and Answers Sheet 1 Matrix Transformations Question #1 Is division of matrices possible? Is it possible to divide a matrix by another? If yes, What will be the result of 𝐴 𝐵 if A = ( a b w x ), B = ( ) c d y z ? Answer: There is a way to performa sort of division, but I am not sure if it is the way you are looking for. For motivation, consider the ordinary real numbers R. We have that for two real numbers, x/y is really the same as multiplying x and 𝑦 −1 = 1/𝑦. We call 𝑦 −1 the inverse of y, and note that it has the property that yy-1=1. The same goes for different algebraic structures. That is, for two elements x, y in this algebraic structure we define x/y as 𝑥𝑦 −1 (under some operation). Most notably, we have a notion of division in any division ring (hence the name!) . It turns out that if you consider invertible 𝑛 × 𝑛 matrices with addition and ordinary matrix multiplication, there is a sensible way to define division since every invertible matrix has well, an inverse. So just to help you grip what an inverse is, say that you have a 2x2 matrix The inverse of A is then given by 1  d −b   −c a  𝐴−1 = (𝑎𝑑−𝑏𝑐) [  and you should check that 𝐴𝐴−1 = 𝐸, the identity matrix. Now, for two matrices B and A, 𝐵/𝐴 = 𝐵𝐴−1 . Question #2 How to determine which of the following transformations are linear transformations? Determine which of the following transformations are linear transformations A. The transformation 𝑇1 defined by 𝑇1 (𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 , 0, 𝑥3 ) B. The transformation 𝑇2 defined by 𝑇2 (𝑥1 , 𝑥2 ) = (2𝑥1 − 3𝑥2 , 𝑥1 + 4,5𝑥2 ). C. The transformation 𝑇3 defined by 𝑇3 (𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 , 𝑥2 , −𝑥3 ) D. The transformation 𝑇4 defined by 𝑇4 (𝑥1 , 𝑥2 , 𝑥3 ) = (1, 𝑥2 , 𝑥3 ) E. The transformation 𝑇5 defined by 𝑇5 (𝑥1 , 𝑥2 ) = (4𝑥1 − 2𝑥2 , 3|𝑥2 |). I believe that it could be A and E. How can I determine this? If someone could show me one I could figure out the rest. Answer: To test whether T is a linear transformation, you need to check that for some vectors a and b and some constant c T(a+b)=T(a)+T(b) T(ca)=cT(a) T(0)=0 So for example, A. 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 , 0. 𝑥3 ) 𝑇(𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , 𝑥3 + 𝑦3 ) = (𝑥1 + 𝑦1 , 0(𝑥2 + 𝑦2 ), 𝑥3 + 𝑦3 ) = 𝑇(𝑥1 , 0, 𝑥3 ) + 𝑇(𝑦1 , 0, 𝑦2 ) 𝑇(𝑐𝑥1 , 𝑐𝑥2 , 𝑐𝑥3 ) = 𝑇(𝑐𝑥1 , (𝑐)0, 𝑐𝑥3 ) = 𝑐𝑇(𝑥1 , 0, 𝑥3 ) 𝑇(0,0,0) = 0 B. 𝑇(𝑥1 , 𝑥2 ) = (2𝑥1 − 3𝑥2 , 𝑥1 + 4,5𝑥2 ) 𝑇(𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 ) = (2(𝑥1 + 𝑦1 ) − 3(𝑥2 + 𝑦2 ), (𝑥1 + 𝑦1 ) + 4,5(𝑥2 + 𝑦2 )) = (2𝑥1 + 2𝑦1 − 3𝑥2 − 3𝑦2 , 𝑥1 + 𝑦1 + 4,5𝑥2 + 5𝑦2 ) 𝑇(𝑥1 , 𝑥2 ) + 𝑇(𝑦1 , 𝑦2 ) = (2𝑥1 − 3𝑥2 , 𝑥1 + 4,5𝑥2 ) + (2𝑦1 − 3𝑦2 , 𝑦1 + 4,5𝑦2 ) = (2𝑥1 − 3𝑥2 + 2𝑦1 − 3𝑦2 , 𝑥1 + 𝑦1 + 8,5𝑥2 + 5𝑦2 ) ≠ 𝑇(𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 ) So B is not a linear transformation. Question #3 Find k such that the following matrix M is singular. [[−3, −3,3], [6,7, −6], [11 + 𝑘, 16, −15]] Answer: [[−3, −3,3], [6,7, −6], [11 + 𝑘, 16, −15]] Matrix M is Singular So determinant of M=0 [[−3, −3,3], [6,7, −6], [11 + 𝑘, 16, −15]] = 0 −3[−15 × 7 − (−6 × 16)] + 3[6 × (−15) − (−6𝑥(11 + 𝑘))] + 3[16 × 6 − 7(11 +)] = 0 ⇒ −3[−105 + 96] + 3[−90 + 66 + 6𝑘] + 3[96 − 77 − 7𝑘] = 0 = −3[−11] + 3[−24 + 6𝑘] + 3[19 − 7𝑘] = 0 = 33 − 72 + 18𝑘 + 57 − 21𝑘 = 0 ⇒ 90 − 72 − 3𝑘 = 0 3𝑘 = 18 18 𝑘= =6 3 𝑘=6 Question #4 The reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding to the given matrix. Use x, y;x,y; or x, y, z;x,y,z; or 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution. 1 0 0 0 | 1  0 1 0 0 | 2    0 0 1 2 | 3  Answer: The goal of the is to write a system of equations that corresponds to a given matrix. And then determine whether the system is consistent or not and if so give a solution. The system that corresponds to a given matrix is: 1? x1 + 0? x2 + 0? x3 + 0? x4 = 1  0? x1 + 1? x2 + 0? x3 + 0? x4 = 2 0? x + 0? x + 1? x + 2? x = 3 1 2 3 4   x1 = 1   x2 = 2 x + 2x = 3 4  3 The system is consistent. 𝑥1 = 1, 𝑥2 = 2, 𝑥3 + 2𝑥4 = 3, 𝑥3 = 3 − 2𝑥4 𝑥4 is real number Question #5 Assume that T is a linear transformation. Find the standard matrix of T. 𝑇: 𝑅𝟚 ⇒ 𝑅𝟜 , 𝑇(𝑒1 ) = (3,1,3,1) and 𝑇(𝑒2 ) = (−5,2,0,0), where 𝑒1 = (1,0) and 𝑒2 = (0,1). Answer: We know that 1  e1 =   0  and 0  e2 =   1  are the columns of the 2 × 2 identity matrix. The is given  3  −5 1 2   T (e1 ) = and T (e2 ) =    3 0     1 0 Theorem says that there exists a unique matrix A for the linear transformation T for which it holds T(u)= Au for all u and a is a form 𝐴 = [𝑇(𝑒1 ), … , 𝑇(𝑒𝑛 )], where 𝑒𝑖 , 𝑖 = 1,2, ... are vectors from the identity matrix, respectively to columns. Matrix A is the standard matrix for the linear transformation T. So, the standard matrix A for this linear transformation T is 3 −5 1 2  . A = [T (e1 )T (e2 )] =  3 0    1 0  Result: 3 −5 1 2   A= 3 0    1 0  Question #6 Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.  1 3 −2   −4 h 8    Answer: The given matrix is  1 3 −2   −4 h 8    Reduce matrix to row echelon form 3 −2  1 3 −2 1 −4 h 8  R2 + 4R1  R2 0 h + 12 0  .     If h=-12 the equations are row equivalent and will have infinitely many solutions. If h is any other number, the lines are not parallel and will meet at one point so there will be one solution. Therefore all h is possible. Result: All h is consistent. Question #7 Assume that T is a linear transformation. Find the standard matrix of T. 𝑇: 𝑅𝟚 ⇒ 𝑅𝟚 is a vertical shear transformation that maps 𝑒1 into 𝑒1 − 2𝑒2 but leaves the vector 𝑒2 unchanged. Answer: 𝑇(𝑒1 ) = 𝑒1 − 2𝑒2 and 𝑇(𝑒2 ) = 𝑒2 Here, 𝑒1 and 𝑒2 are the unit vectors, as you can see in the next step 1  0  1  0 T (e1 ) =   − 2   =   and T (e2 ) =   . Writing the above as a matrix yields these equation 0 1  −2 1   1 0 T =  . According to theorem 10, the transformation matrix equals the transformation of the unit  −2 1  vectors, each occupying one column in this matrix. Result:  1 0 T =   −2 1  Question #8 Show that if 𝐴2 is the zero matrix, then the only eigenvalue of A is 0. Answer: 𝐴2 𝑣 = 𝐴(𝐴𝑣) = 𝐴(𝜆𝑣) = 𝜆𝐴𝑣 = 𝜆2 𝑣 Let 𝜆 be any eigenvalue of A. Then there exists a nonzero vector v such that 𝐴𝑣 = 𝜆𝑣 . Using this the equation on the left implies that 𝐴2 𝑣 = 𝜆2 𝑣. 0 = 𝜆2 𝑣 ⇒ 𝜆 = 0 Since 𝐴2 is the zero matrix, 𝐴2 𝑣 = 0 which implies that 𝜆2 𝑣 = 0 . Since v is nonzero 𝜆2 = 0which implies 𝜆 = 0 . Therefore, 𝜆 = 0 is the only eigenvalue of A. Question #9 Find k such that the following matrix M is singular.  −1 M =  0  −12 + k −1 −2  −1 −4  −2 −2 k=? Answer:  −1  0 Consider the matrix: M =   −12 + k −1 −2  −1 −4  −2 −2 A square matrix is singulare if and only if it's determinant is 0 Now, the determinant is  −1  0   −12 + k −1 −2  −1 −4  0 −1 −4 = 0(−1)  − (−1)    −2 −2  −12 + k −2 −2 −4  0 + (−2)   −2  −12 + k −1 =0 −2 (-1)(2-8)+1(0+4(-12+k))-2(0+1(-12+k))=0 6+4(-12+k)-2(-12+k)=0 6-48+4k+24-2k=0 -18+2k=0 2k=18 Hence, the required value of k for the singular matrix M is k=9 Question #10 Consider the 3 × 3 matrices with real entrices. Show that the matrix forms a vector space over R with respect to matrix addition and matrix multiplication by scalars? Answer: Let 𝑀3 (𝑅) be the set of all 3 × 3 matrices of real entries. Then we have to show that the set 𝑀3 (𝑅) of all 3 × 3 matrices form a vector space with respect to the usual matrix addition and usual scalar multiplication. We will show that all 10 properties are satisfied for 𝑀3 (𝑅) to be a vector space as follows; Suppose A, B, 𝐶 ∈ 𝑀3 (𝑅), where  a11 a12 a13  A =  a21 a22 a23   a31 a32 a33   b11 b12 b13  B = b21 b22 b23  b31 b32 b33   c11 c12 c13  C = c21 c22 c23   c31 c32 c33  and 𝑎𝑖𝑗 , 𝑏𝑖𝑗 , 𝑐𝑖𝑗 ∈ 𝑅 for all 1 ≤ 𝑖, 𝑗 ≤ 3 (i) Closure:  a11 a12 a13   b11 b12 A + B =  a21 a22 a23  + b21 b22  a31 a32 a33  b31 b32  a11 + b11 a12 + b12 a13 + b13  =  a21 + b21 a22 + b22 a23 + b23   a31 + b31 a32 + b32 a33 + b33  b13  b23  b33  Now we know that 𝑎𝑖𝑗 + 𝑏𝑖𝑗 ∈ 𝑅 for all 1 ≤ 𝑖 < 𝑗 ≤ 3 So (a) A+B is also ∈ 𝑀3 (𝑅) (ii) Associativity:  a11 ( A + B) + C = (  a21  a31  (a11 + b11 ) + c11 = (a21 + b21 ) + c21  (a31 + b31 ) + c31 a12 a22 a32 a13   b11 b12 a23  + b21 b22 a33  b31 b32 (a12 + b12 ) + c12 (a22 + b22 ) + c22 (a32 + b32 ) + c32 b13   c11 c12 b23  ) + c21 c22 b33   c31 c32 c13  c23  c33  (a13 + b13 ) + c13  (a23 + b23 ) + c23  (a33 + b33 ) + c33   a11 + (b11 + c11 ) a12 + (b12 + c12 ) a13 + (b13 + c13 )  =  a21 + (b21 + c21 ) a22 + (b22 + c22 ) a23 + (b23 + c23 )   a31 + (b31 + c31 ) a32 + (b32 + c32 ) a33 + (b33 + c33 )  A+(B+C) Therefore, (A+B)+C=A+(B+C). (iii) Commutativity:  a11 A + B =  a21  a31  a11 + b11 =  a21 + b21  a31 + b31 a12 a22 a32 a13   b11 b12 a23  + b21 b22 a33  b31 b32 a12 + b12 a22 + b22 a32 + b32 a13 + b13  a23 + b23  a33 + b33   b11 + a11 b12 + a12 = b21 + a21 b22 + a22 b31 + a31 b32 + a32 b13 + a13  b23 + a23  b33 + a33  b13  b23  b33  =B+A Therefore, A+B=B+A (iv) Identity: 0 0 0    Consider the zero matrix 0, where 0 = 0 0 0 . Now since 𝑎𝑖𝑗 + 0 = 0 + 𝑎𝑖𝑗 = 𝑎𝑖𝑗 for all 1 ≤ 𝑖, 𝑗 ≤ 3,   0 0 0 So we can write A+0=0+A=A (v) Inverse: For each 𝐴 ∈ 𝑀3 (𝑅) , define  −a11 −a12 − A =  −a21 −a22  −a31 −a32 −a13  −a23  −a33  Now since 𝑎𝑖𝑗 + (−𝑎𝑖𝑗 ) = (−𝑎𝑖𝑗 ) + 𝑎𝑖𝑗 = 0 for all 1 ≤ 𝑖, 𝑗 ≤ 3 So we can write A+(-A)=(-A)+A=0 (vi) Closure (Scalar Multiplication): k is scalar and A is a 3 × 3 matrix then by part (a) k 𝐴 ∈ 𝑀3 (𝑅) since 𝑘𝑎𝑖𝑗 is also real number for all 1 ≤ 𝑖, 𝑗 ≤ 3 (vii). 0A=0 , since 0. 𝑎𝑖𝑗 = 0 for all 1 ≤ 𝑖, 𝑗 ≤ 3 and 1A=A since 1. 𝑎𝑖𝑗 = 𝑎𝑖𝑗 for all 1 ≤ 𝑖, 𝑗 ≤ 3 (viii) c(dA)=cdA are follows from the definition of scalar multiplication, here c and d are scalars. (ix) Distributivity 1:  a11 a12 a13   b11 b12 k ( A + B ) = k (  a21 a22 a23  + b21 b22  a31 a32 a33  b31 b32  a11 + b11 a12 + b12 a13 + b13  = k (  a21 + b21 a22 + b22 a23 + b23  )  a31 + b31 a32 + b32 a33 + b33   ka11 + kb11 ka12 + kb12 ka13 + kb13  =  ka21 + kb21 ka22 + kb22 ka23 + kb23   ka31 + kb31 ka32 + kb32 ka33 + kb33   ka11 ka12 ka13   kb11 =  ka21 ka22 ka23  +  kb21  ka31 ka32 ka33   kb31  a11 a12 a13   b11   = k  a21 a22 a23  + k b21  a31 a32 a33  b31 kb12 kb22 kb32 b12 b22 b32 b13  b23  ) b33  kb13  kb23  kb33  b13  b23  b33  =kA+kB So, k(A+B)=kA+kB (x) Distribitivity 2: Suppose k and 1 are scalars.  a11 a12 ((k + 1) A) = (k + 1)  a21 a22  a31 a32  (k + 1)a11 (k + 1)a12 = (k + 1)a21 (k + 1)a22  (k + 1)a31 (k + 1)a32  a11 a12 = k  a21 a22  a31 a32 a13  a23  a33  (k + 1)a13   ka11 + 1a11 ka12 + 1a12 (k + 1)a23   ka21 + 1a21 ka22 + 1a22 (k + 1)a33   ka31 + 1a31 ka32 + 1a32 a13   a11 a12  a23  + 1k  a21 a22 a33   a31 a32 =(kA+1A) So, (k+1)A=kA+1A Hence 𝑀3 (𝑅) is a vector space. a13  a23  a33  ka13 + 1a13  ka23 + 1a23  ka33 + 1a33 