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Matrices and linear transformations. Where did I go wrong?

  1. Feb 2, 2017 #1
    Hi everyone. Excuse me for my poor English skills. I did an exam today and my exam result was 13 of 40. I dont understand why it was my result, because while doing the exam I though I was doing it well, then the result was a surprise for me. I will write down the questions and after show my answers.

    1. The problem statement, all variables and given/known data

    1. (a) Let [itex] \pi_1: \mathbb{R} ^{2} \longrightarrow \mathbb{R} [/itex] such that [itex]\pi_1 (x,y) = x[/itex]. Show that [itex]\pi_1[/itex] is a linear transformation. Calculate the kernel of [itex]\pi_1[/itex]. What is the dimension of its image? Explain your reason.

    (b) Give an example of a linear transformation [itex] T: \mathbb{R} ^{2} \longrightarrow \mathbb{R} [/itex] which is not surjective.

    (c) There can be a injective linear transformation [itex] T: \mathbb{R} ^{2} \longrightarrow \mathbb{R} [/itex]? Explain your reason.

    2. Consider the matrix

    [itex] A = \begin{pmatrix}1&-2&8\\0&-1&0\\0&0&-1\end{pmatrix} [/itex]

    (a) Calculate the eigenvalues and eigenspaces of A.

    (b) Is A a diagonalizable matrix? Explain.

    (c) Calculate [itex]tr(A^{2017})[/itex].

    3. Are the matrices below diagonalizable? If not, explain your reason, if yes, diagonalize it.

    (a) [itex] \begin{pmatrix}1&1\\0&1\end{pmatrix}[/itex].

    (b) [itex] \begin{pmatrix}1&1\\1&1\end{pmatrix}[/itex].

    2. Relevant equations


    3. The attempt at a solution

    My answers:

    1. (a) Linearity (addition):
    π1(x1, y1) = x1, π1(x2, y2) = x2
    π1(x1, y1) + π1(x2, y2) = x1 + x2 = π1(x1 + x2, y1 + y2).

    Linearity (scalar multiplication):
    π1(αx1, y1) + π1(αx2, y2) =
    α(x1 + x2) = π1(α(x1 + x2), y1 + y2).

    Ker(π1) = {0, y}, Im(π1) = ℝ2; dimension 2.

    (b) T: ℝ2 → ℝ
    (x, y) [itex] \mapsto[/itex] T(x, y) = [itex]\sqrt x[/itex].

    (c) Yes. This condition will be satisfied if each element of ℝ2 is mapped into each element of ℝ, e.g. (x, y) [itex] \mapsto[/itex] x.

    2.
    (a) 1; -1. I found these values by setting the determinant of the matrix equal to zero.
    Eigenvectors of A are for λ= 1: t(1,0,0), for λ = -1: (-4α + β, β, α), with α, β, t ∈ ℝ.
    So A has two independent eigenvectors and the eigenspace is ℝ2.

    (b) No. We need three independent eigenvectors to form the square matrix S in SAS-1 = D, and A has only two independent eigenvectors.

    (c) A² = I, A³ = A, A4 = I, ... Since 2017 is a odd number, A2017 = A, and tr(A2017) = (1 x -1 x -1) = 1.

    3.
    (a) The matrix has only one eigenvalue and is not diagonalizable.

    (b)
     
    Last edited: Feb 2, 2017
  2. jcsd
  3. Feb 2, 2017 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    1 (a) ##\text{Im}(\pi_1) = \{ x \in \mathbb{R}\}##, so is 1-dimensional.
    2. (a)##A## has three (linearly independent) eigenvectors.
    (b) So, yes: ##A## is diagonalizable.
    (c) The trace = sum of diagonal elements, not the product!
     
    Last edited: Feb 2, 2017
  4. Feb 3, 2017 #3
    Thanks Ray Vickson and StoneTemplePython. What about my answers on (b) and (c) on question 1.?
     
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