Questions and Answers #4 Matrix Transformations

Questions and Answers Sheet 4 Matrix Transformations Question #1 Is a matrix multiplied with its transpose something special? In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together. Is 𝐴𝐴𝑇 something special for any matrix A? Answer: The main thing is presumably that 𝐴𝐴𝑇 is symmetric. Indeed (𝐴𝐴𝑇 )𝑇 = (𝐴𝑇 )𝑇 𝐴𝑇 = 𝐴𝐴𝑇 . For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real. Moreover if A is invertible, then 𝐴𝐴𝑇 is also positive definite, since 𝑥 𝑇 𝐴𝐴𝑇 𝑥 = (𝐴𝑇 𝑥)𝑇 (𝐴𝑇 𝑥) > 0 Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors. Last but not least if one is interested in how much the linear map represented by A changes the norm of a vector one can compute √⟨𝐴𝑥, 𝐴𝑥⟩ = √⟨𝐴𝑇 𝐴𝑥, 𝑥⟩ which simplifies for eigenvectors x to the eigenvalue λ to √⟨𝐴𝑥, 𝐴𝑥⟩ = √𝜆√⟨𝑥, 𝑥⟩, The determinant is just the product of these eigenvalues. Question #2 How can I find the dimension of the eigenspace? The matrix 9 −1 A=  has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of 1 7   the eigenspace. So we found the eigenvalue by doing 𝐴 − 𝜆 I to get: 𝜆= 8 But how exactly do we find the dimension of the eigenspace? Answer: The dimension of the eigenspace is given by the dimension of the nullspace of one can row reduce to 1 −1 A − 8I = ( ) , which 1 −1 1 −1 ( )], so the dimension is 1. 0 0 Note that the number of pivots in this matrix counts the rank of A-8I. Thinking of A-8I as a linear operator from 𝑅2 to 𝑅2 , the dimension of the nullspace of A is given by dim (𝑅2 ) -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-8I is spanned by the vector (1,1), and hence has dimension 1. Question #3 Let A be an invertible 𝑛 × 𝑛 matrix, and let B be an 𝑛 × 𝑝 matrix. Show that the equation AX=B has a unique solution 𝐴−1 𝐵. Answer: A is an invertible 𝑛 × 𝑛 matrix and B is an 𝑛 × 𝑝 matrix. A is invertible, therefore, matrix 𝐴−1 𝐵 satisfies AX=B. Observe the following for justification: 𝐴(𝐴−1 𝐵) = 𝐴𝐴−1 𝐵 =IB =B Now, let X be any solution of AX = B. Then by left multiplication of both sides by 𝐴−1 we have: 𝐴−1 (𝐴𝑋) = 𝐴−1 𝐵 𝐼𝑋 = 𝐴−1 𝐵 𝑋 = 𝐴−1 𝐵 Hence AX=B has a unique solution. Result: AX=B has a unique solution. Question #4 If a 4 by 4 matrix has det 𝑎 = 1/2, find det(2𝐴) , det(−𝐴) , det(𝐴2 ), and det(𝐴−1 ). Answer: Evaluate using properties of determinants: 1 det(2𝐴) = 24 ∗ det 𝐴 = 16 ∗ = 8 2 1 det(−𝐴) = (−1)4 ∗ det 𝐴 = 1 ∗ = 1 1 2 2 2 det(𝐴2 ) = det 𝐴 ∗ det 𝐴 = ∗ = det(𝐴−1 ) = 1 det 𝐴 = 1 1/2 1 1 2 4 =2 Result: 1 1 8, , , 2 2 4 Question #5 Detemine if b is a linear combination of 𝑎1 , 𝑎2 , 𝑎3  2  −4  −5 13        a1 = 0  , a2 =  3  , a3 =  8  , b =  −4  2  −4  4   9  Choose the correct answer below A. Vector b is a linear combination of 𝑎1 , 𝑎2 , 𝑎3 . The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the third column. B. Vector b is not a lincar combination of 𝑎1 , 𝑎2 , 𝑎3 C. Vector b is a linear combination of 𝑎1 , 𝑎2 , 𝑎3 . The pivots in the corresponding echelon matrix are in the first entry in the first column and the third entry in the second column, and the third entry in the third column. D. Vector b is a linear combination of 𝑎1 , 𝑎2 , 𝑎3 . The pivots in the corresponding echelon matrix are in the first entry in the first column, the second entry in the second column, and the third entry in the fourth column. Answer: Consider the following vectors: 𝑎1 = [[2], [0], [2]], 𝑎2 = [[−4], [3], [−4]], 𝑎3 = [[−5], [8], [4]], 𝑏 = [[13], [−4], [9]] First, check whether the vector b is in linear combination of the other three vectors or not. [[13], [−4], [9]] = 𝑐1 [[2], [0], [2]] + 𝑐2 = [[−4], [3], [−4]] + 𝑐3 [[−5], [8], [4]] = [[2𝑐1 − 4𝑐2 − 5𝑐3 ], [3𝑐2 + 8𝑐3 ], [2𝑐1 − 4𝑐2 + 4𝑐3 ]] The followng equations can be expressed in matrix form as follows: = [[2, −4, −5, |, 13], [0,3,8, |, −4], [2, −4,4, |, 9]] Apply 𝑅3 ⇒ 𝑅3 − 𝑅1 = [[2, −4, −5, |, 13], [0,3,8, |, −4], [0,0,9, |, 9]] Apply 𝑅2 ⇒ 1/3𝑅2 = [[2, −4, −5, |, 13], [0,1,8/3, |, −4/3], [0,0,9, |, 9]] Apply 𝑅1 ⇒ 1/2𝑅1 = [[1, −2, −5/2, |, 13/2], [0,1,8/3, |, −4/3], [0,0,9, |, 9]] Since, every row has a pivot position there extra a unique solution to the system. Hence there is a solution to the system [[13], [−4], [9]] = [[2𝑐1 − 4𝑐2 − 5𝑐3 ], [3𝑐2 + 8𝑐3 ], [2𝑐1 − 4𝑐2 + 4𝑐3 ]]. Clearly, the pivot positions are in the first entry in the first column, second entry n the second column, and the third entry in the third column as shown below: 5 13   1 −2 − 2 ∣ 2    8 4  = 0 1 ∣ −  3 3   9 ∣ 9  0 0   Hence, option A is correct. Question #6 Find the characteristic equation and the eigenvalues (and corresponding eigenvectors) of the matrix.  2 −2 3  0 3 −2    0 −1 2  a) The characteristic equation 𝜆3 − 7𝜆2 + 14𝜆 − 8 = 0 b) The eigenvalues (𝜆1 , 𝜆2 , 𝜆3 ) = (1,2,4) The corresponding eigenvectors 𝑥1 = −7,1,1? 𝑥2 = 1,0,0 𝑥3 = 13, −4,2? Answer: Step 1 Explanation: Given that,  2 −2 3  0 3 −2    0 −1 2  a. Characteristic equation for given matrix 𝜆3 − 7𝜆2 + 14𝜆 − 8 = 0 b. Eigenvalues are: 𝜆3 − 7𝜆2 + 14𝜆 − 8 = 0 (𝜆 − 1)(𝜆 − 2)(𝜆 − 4) = 0 𝜆−1=0 𝜆=1 𝜆−2=0 𝜆=2 𝜆−4=0 𝜆=4 𝜆1 = 1 𝜆2 = 2 𝜆3 = 4 Step 2 c) Find the eigenvectors To find the eigenvectors, we will set up the equation: Solve (𝐴 − 𝜆𝐼)  2 −2 3  1 A − 1 I =  0 3 −2  − 1 0  0 −1 2  0  2 −2 3  1 [ A − 1I =  0 3 −2  − 0  0 −1 2  0 0 0 1 0  ] 0 1  0 0  1 −2 3  1 0  = 0 2 −2  0 1  0 −1 1  Now reduce this matrix using row echelon form. Reduce 1 −2 3  1 −2 3  0 2 −2  ]R  R + R2 [= 0 2 −2  ] 3   3   2 0 −1 1  0 0 0  1 −2 3  R2  R2  [= 0 1 −1 ] 2 0 0 0  1 R1  R1 + 2 R2 [= 0 0  x  1 0 [( A − ! I )  y  = 0 1  z   0 0 0 1 1 −1 ] 0 0  1   x −1  y  = 0 0   z  𝑥 + 𝑧 = 0 ⇒ 𝑥 = −𝑧 𝑦−𝑧 =0⇒𝑦=𝑧  x  − z   y =  z       z   z  Suppose z=1  −1 1    1  First eigenvector is 𝑥1 = −1,1,1 Similarly we will find the second and third eigenvector. Eigenvector for 𝜆2 = 2 −2 3  1 0 0   3 −2  − 2 0 1 0  ]  0 0 1  −1 2  −2 3   2 0 0  0 −2 3  3 −2  −  0 2 0  = 0 1 −2  ] −1 2   0 0 2  0 −1 0  0 −2 3  0 1 0  Reduce[ 0 1 −2  = 0 0 0  ] 0 −1 0  0 0 1   x  0 1 0   x  [( A − 2 I )  y  = 0 0 0   y  = 0  z  0 0 1   z  2 A − 2 I =  0  0 2 [ A − 2 I =  0  0 y=0 z=0 Here x=1 1  0    0  Similarly we will find the second and third eigenvector. 𝑥2 = 1,0,0 For third eigenvector : Eigenvector for 𝜆3 = 4 2 A − 3 I = 0 0 2 A − 4 I = 0 0 −2 3  1  3 −2 − 4 0 0 −1 2  −2 3   4 0 3 −2 − 0 4 −1 2  0 0 0 0 1 0 0 1  0  −2 −2 3 0  = 0 −1 −2 4 0 −1 −2 Reduce −7   1 0  −2 −2 3   2  0 −1 −2  = 0 1 2  ]      0 −1 −2  0 0 0      −7   1 0  x   2   x     [( A − 4 I )  y  = 0 1 2   y  = 0  z  0 0 0   z      7 7 2 2 𝑥− 𝑧=0⇒𝑥= 𝑧 𝑦 + 2𝑧 = 0 ⇒ 𝑦 = −2𝑧 let z=2 x=7 y=-4 7 −4 2 Third eigenvector : 𝑥3 = 7, −4,2 Question #7 Let 𝑎1 = [[1], [3], [−1]], 𝑎2 = [[−6], [−14], [3]], 𝑏 = [[5], [3], [ℎ]] . for what value(s) of h is b in the plane spanned by 𝑎1 and 𝑎2 Answer: b in the plane spanned by 𝑎1 and 𝑎2 if b is the linear combination of 𝑎1 and 𝑎2 . 𝑏 = 𝜃1 𝑎1 + 𝜃2 𝑎2 ⇒ [[5], [3], [ℎ]] = 𝛼1 [[1], [3], [−1]], 𝛼2 = [[−6], [−14], [3]] ⇒ 5 = 𝛼1 − 6𝛼2 − − − 1 3 = 3𝛼1 − 14𝛼2 − − − 2 ℎ = −𝛼1 − 3𝛼2 − − − 3 1 × 3 ⇒ 15 = 3𝛼1 − 18𝛼2 2 ⇒ 3 = 3𝛼1 − 14𝛼2 Subtract, 12 = −14𝛼2 ⇒ 𝛼2 = −3 Sub 𝛼2 = −3 𝑖𝑛 1,5 = 𝛼1 + 18 ⇒ 𝛼1 = −13 Sub 𝛼1 = −13 and 𝛼2 = −3 in 3, ℎ = 13 + 3(−3) = 13 − 9 = 4 i.e ℎ = 4 Question #8 Assume that T is a linear transformation. Find the standard matrix of T. T: 𝑅𝑅2 → 𝑅𝑅4 , 𝑇(𝑒1 ) = (3,1,3,1), and 𝑇(𝑒2 ) = (−5,6,0,0), where 𝑒1 = (1,0) and 𝑒2 = (0,1) A= Answer: Let 𝑥 ∈ 𝐹 𝑛 . Here, [[𝑥1 ], [⋮], [𝑥𝑛 ]] 𝑇(𝑥) = 𝑇(∑𝑗 𝑒𝑗 𝑥𝑗 ) = ∑𝑗 𝑇(𝑒𝑗 )𝑥𝑗 = ∑𝑗 𝐴𝑗 𝑥𝑗 = 𝐴𝑥 So, here 𝐴 = [𝑇(𝑒1 ) … 𝑇(𝑒𝑛 )] According to question 𝑇: 𝑅𝑅2 → 𝑅𝑅4 , 𝑇(𝑒1 ) = (3,1,3,1), and 𝑇(𝑒2 ) = (−5,6,0,0) So, the standart matrix of T is [𝑇(𝑒1 )𝑇(𝑒2 )] = [[3, −5], [1,6], [3,0], [1,0]] - Answer Question #9 Find matrix of linear transformation A linear transformation 𝑇: 𝑅𝟚 → 𝑅𝟚 is given by 𝑇(𝑖) = 𝑖 + 𝑗 𝑇(𝑗) = 2𝑖 − 𝑗 Answer: If 𝑇(𝑖) = (1,1) and 𝑇(𝑗) = (2, −1) and 𝑒1 = 𝑖 − 𝑗 = (1, −1) and 𝑒2 = 3𝑖 + 𝑗 = (3,1) , then 𝑇(𝑒1 ) = 𝑇(𝑖 − 𝑗) = 𝑇(𝑖) − 𝑇(𝑗) = (1,2) = 𝑎1 𝑒1 + 𝑏1 𝑒2 (say) and 𝑇(𝑒2 ) = 𝑇(3𝑖 + 𝑗) = 3(1, 1) + (2, − 1) =(5,\ 2) = 𝑎2 𝑒1 + 𝑏2 𝑒2 (say) Thus the matrix of T w.r.t. the new basis {𝑒1 , 𝑒2 } is ((𝑎1 , 𝑎2 ), (𝑏1 , 𝑏2 )) and you need to find the values of 𝑎1 , 𝑎2 , 𝑏1 and 𝑏2 . The above systems of equations reduces to 𝑎1 + 2𝑏1 = −1 −𝑎1 + 𝑏1 = 2 and 𝑎2 + 3𝑏2 = 5 −𝑎2 + 𝑏2 = 2 Solve these equations to obtain 7 𝑎1 = − 𝑏1 = 1 4 𝑎2 = − 4 1 4 and 7 𝑏2 = 4 Question #10 Matrix of singular transformation A=(0,1) to A'=(0,0), B=(2,0) to B'=(0,1) and C=(2,1) to C'=(0,2) find the matrix representation r? Answer: The map can be expressed as [[𝑥 ′ ], [𝑦 ′ ]] = [[𝑎11 , 𝑎12 ], [𝑎21 , 𝑎22 ]][[𝑥], [𝑦]] + [[𝑏1 ], [𝑏2 ]] For the non-singular case, you should be able to solve for the 6 unknowns (𝑎𝑖𝑗 , 𝑏𝑖 ) using the 6 equations that you have: 0 = 𝑎11 (0) + 𝑎12 (1) + 𝑏1 0 = 𝑎21 (0) + 𝑎22 (1) + 𝑏2 0 = 𝑎11 (2) + 𝑎12 (0) + 𝑏1 1 = 𝑎21 (2) + 𝑎22 (0) + 𝑏2 0 = 𝑎11 (2) + 𝑎12 (1) + 𝑏1 2 = 𝑎21 (2) + 𝑎22 (1) + 𝑏2 In matrix form: [[0], [0], [0], [1], [0], [2]] = [[0,1,0,0,1,0], [0,0,0,1,0,1], [2,0,0,0,1,0], [0,0,2,0,0,1], [0,0,2,1,0,1]][[𝑎11 ], [𝑎12 ], [𝑎21 ], [𝑎22 ], [𝑏1 ], [𝑏2 ]] For the singular case you will need to compute a pseudoinverse to find the unknowns. For the data you have provided an inverse appears to exist and you don't need a pseudoinverse.