Questions and Answers #7 Matrix Transformations

Questions and Answers Sheet 7 Matrix Transformations Question #1 How can you find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations? The line of: 𝑦 = − 2𝑥 3 , all in 𝑅𝟚 Answer: Step 1 So you have a vector (𝑥1 , 𝑦1 )𝑇 , and you want to find a matrix M such that (𝑥2 , 𝑦2 )𝑇 = 𝑀(𝑥1 , 𝑦1 )𝑇 is the 2𝑥 reflection across 𝑦 = − . 3 So what does the reflection means? 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 𝑦1 + 𝑦2 2 𝑥1 + 𝑥2 =− 2 3 2 The second condition means that the slope of the line between the two points is − 1 𝑚 where m is the slope of the reflection line: 𝑦2 −𝑦1 𝑥2 −𝑥1 =− 1 −2/3 = 3 2 You now write 𝑥2 and 𝑦2 in terms of 𝑥1 and 𝑦1 : 𝑥2 = 𝑚11 𝑥1 + 𝑚12 𝑦1 𝑦2 = 𝑚21 𝑥1 + 𝑚22 𝑦1 Then the matrix you are looking for has these coefficients. m M =  11  m21 m12   m22  Question #2 Given basis B: 𝑏1 = (1 2)𝑇 and 𝑏2 = (2 1)𝑇 and A: 𝑎1 = (1 2)𝑇 , 𝑎2 = (2 7)𝑇 . Find the transformation matrix T from the basis B into basis A? Answer: Step 1 The change of basis matrix from basis 2 (R , ) to basis 2 is the matrix of the identity map from (R , ) to Let 𝑇𝐵 be the matrix with column vectors 𝑏1 and 𝑏2 : it's the change of basis matrix from the canonical basis to basis Similarly the matrix 𝑇𝐴 with colum,n vectors 𝑎1 and 𝑎2 is the change of basis matrix from the canonical basis to basis . The commutative diagram: = (R 2 , Can) (R 2 , ) = = 2 (R , ) shows at once that 𝑇 = 𝑇𝐵−1 𝑇𝐴 . Question #3 Why is the Jacobian matrix equal to the matrix associated to a linear transformation? Given the linear transformation f, we can construct the matrix A as follows: on the i-th column we put the vector 𝑓(𝑒𝑖 ) where 𝐸 = (𝑒1 , ⋯ , 𝑒𝑛 ) is a basis of n . Answer: If 𝑓: 𝑅𝑛 → 𝑅𝑝 is linear, then you know that for all 𝑎 ∈ 𝑅𝑛 , you have that Df(a)=f. As Jf(a) is the matrix which represents f in the canonical bases 𝐶𝑛 and 𝐶𝑝 of 𝑅𝑛 and 𝑅𝑝 , you will have that 𝐽𝑓(𝑎) = [𝑓]𝒞𝓅𝒞𝓃 = 𝐴, with your definition of A. Question #4 a) Find the standard matrix of the linear operator 𝑇: 𝑅2 → 𝑅2 given by the orthogonal projection onto the vector (1,-2). b) Given the linear transformation 𝑇: 𝑅2 → 𝑅3 such that Find T(2,3). Answer: a) The projection of v on the vector w is 1 5 2 5 v.w w ‖ w ‖2 So, the projection of (1,0) is ( , − ) and the projection of (0,1) is ( − 2 4 , ) . So, the matrix is 5 5 2  1  5 − 5  . − 2 4   5 5  b) Note that (2,3)=3(1,1)-(1,0) . Therefore, T(2,3)=3T(1,1)T(1,0) . Question #5 1) Calculate the transformation matrix 2) Calculate the dimension of the kernel of the transformation, justify. 𝑇: 𝑅𝟛 → 𝑅𝟛 is a linear transformation such that: 𝑇(𝑒⃗⃗⃗1 ) = ⃗⃗⃗ 𝑒1 − ⃗⃗⃗ 𝑒2 + 2𝑒⃗⃗⃗3 𝑇(𝑒⃗⃗⃗1 + ⃗⃗⃗ 𝑒2 ) = 2𝑒⃗⃗⃗3 𝑇(𝑒⃗⃗⃗1 + ⃗⃗⃗ 𝑒2 + ⃗⃗⃗ 𝑒3 ) = −𝑒⃗⃗⃗2 + ⃗⃗⃗ 𝑒3 Answer: Step 1 For the first question, you almost solve it the only thing remaining it's to find the matrix. So we have 𝑇(𝑒⃗⃗⃗1 ) = ⃗⃗⃗ 𝑒1 − ⃗⃗⃗ 𝑒2 + 2𝑒⃗⃗⃗3 𝑇(𝑒⃗⃗⃗2 ) = −𝑒⃗⃗⃗1 + ⃗⃗⃗ 𝑒2 + 0𝑒⃗⃗⃗3 𝑇(𝑒⃗⃗⃗3 ) = 0𝑒⃗⃗⃗1 − ⃗⃗⃗ 𝑒2 − ⃗⃗⃗ 𝑒3 Then using the coefficients of the first equation for the numbers in the first column in the matrix associated with T. We have  1 −1 0  AT =  −1 1 −1  2 0 −1 Then making Gauss elimination you need to find the number of free variables and that number is the dimension of the kernel; in this case dim(𝐾𝑒𝑟𝐴 𝑇 ) = 0 , in other words 𝑅𝑎𝑛𝑘(𝐴 𝑇 ) = 3 . And remember that 𝐾𝑒𝑟(𝑇) = 𝐾𝑒𝑟(𝐴 𝑇 ), dim(𝐾𝑒𝑟𝑇 ) = dim(𝐾𝑒𝑟𝐴 𝑇 ) , and the same with the rank of T. And for the last part we have to choose two basis for 𝑅3 one for the domain and the other for the codomain. The easiest one is the canonical basis {𝑒⃗⃗⃗1 , ⃗⃗⃗ 𝑒2 , ⃗⃗⃗ 𝑒3 } but note that you can choose a different basis for both. Question #6 Let 𝑇: 𝑅3 → 𝑅3 be the linear transformation such that 1  −2 1  4 1  1  T 1 =  5  , T 1  = 1  , T 0 =  −1 1  −2 0  4 0  1  a) Find a matrix A such that T(x)=Ax for every 𝑥 ∈ 𝑅3 b) Find a linearly independent set of vectors in 𝑅3 that spans the range of T Answer: Step 1 Presumably your book wants the answer in terms of the standard basis. For ease of typing, I will use row vectors instead of column vectors. Due to linearity, we can see: 𝑇(0,0,1) = 𝑇(1,1,1) − 𝑇(1,1,0) = (−2,5, −2) − (4,1,4) = (−6,4, −6) Similarly, we can see 𝑇(0,1,0) = 𝑇(1,1,0) − 𝑇(1,0,0) = (4,1,4) So then −(1, −1,1) = (3,2,3) matrix for T with respect to the standard basis is:  1 3 −6  −1 2 4  .    1 3 −6 The range of this transformation will be spanned by the column vectors of this matrix. We must find a linearly independent subset of them in order to answer your question. Question #7 Different way to determine rank of 3 × 3 -matrix with parameter by using row transformations a   2 − a 1− a   A =  1− a 1− a 0 .  2 − 2a 1 − a −(1 − a)    Answer: You can simplify algorithm, borrowing techniques used for the computation of determinants: a   2 − a 1− a  1− a 1− a 0    2 − 2a 1 − a −(1 − a)  0   1− a 1− a  2 − 2a 1 − a −(1 − a)     2 − a 1 − a a  Now, if 𝑎 ≠ 1 this matrix has the same rank as 1 0  1  0 −1 −1   2 − a 1 − a a  1 1 0  0 1 1    0 −1 a  0  1 1 0 1 1   0 0 a + 1 You can now conclude (not forgetting to examine the case a=1 ). Question #8 Let (x, y) be the co-ordinates of a point P referred to a set of rectangular axes OX, OY. Then its coordinates (x', y') referred to OX', OY', obtained by rotating the former axes through an angle 𝜃 are given by 𝑥 ′ = 𝑥 cos 𝜃 + 𝑦 sin 𝜃 𝑦 ′ = −𝑥 sin 𝜃 + 𝑦 cos 𝜃 How do we derive the result for (x', y') from (x, y)? Answer: Let 𝐴𝜃 the matrix of the rotation transformation: A = ( cos  sin  − sin  ) cos  so if 𝑢 = (𝑥, 𝑦)𝑇 and 𝑢 ′ = (𝑥 ′ , 𝑦 ′ )𝑇 is the vector obtained by rotating the vector u by an angle 𝜃 then 𝑢 ′ = 𝐴𝜃 × 𝑢 Question #9 The matrix in question is:  −1 1  p=   −1 0  acting on vectors  x    y in the unit square. Is there an intuitive way to interpret this transformation geometrically? Or summarize it with a nice pictorial mapping from [0,1]2 → 𝑅𝟚 Answer: 𝑖̂ = (1,0) → (−1, −1) 𝑗̂ = (0,1) → (1,0) So, 𝑖̂ is rotating 135 degrees clockwise, and stretching by a factor of √2 𝑗̂ is rotating 90 degrees clockwise, with no stretch. This may be sufficient for you, but you could then try to break this into rotations and shears - A 90 degree clockwise rotation followed by a horizontal shear. Areas are preserved: | det (𝐴)| = 1 Question #10 Find the basis for kernel of a matrix transformation Let 𝜓 : 𝑀𝑎𝑡2×2 (𝑅) → 𝑀𝑎𝑡2×2 (𝑅) (𝜓: ( 𝑎 𝑐 𝑎+𝑏 𝑏 )→( 𝑎+𝑐 𝑑 𝑎−𝑐 )) 𝑏−𝑐 Find basis for ker 𝜓 Answer: Assuming that 𝑀𝑎𝑡2 × 2(𝑅) are being thought of as additive groups, the kernel is made up of all matrices that are send to the zero matrix. So you need a+b=0, a-c=0, a+c=0 and b-c=0 . Well, a+b=0 if and only if a=-b . Let's substitute that into the remaining three equations: -b-c=0, -b+c=0 and b-c=0 . Well, -b-c=0 if and only if b=-c . Let's put that into the two remaining equations: -(-c)+c=0 and -(-c)-c=0. Well, 2c=0 if and only if c=0 . Hence a=-b, b=-c and c=0 meaning that a=b=c=0 . You're free to choose d. So the matrix 0 0 0 1  spans the kernel.  