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I 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 write my answers.

    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].

    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.

    (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.

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

    Last edited: Feb 2, 2017
  2. jcsd
  3. Feb 3, 2017 #2


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    I'm thinking that breaking this into chunks would be good. For whatever reason I'm thinking it'd make sense to work backward.

    3b) is certainly diagonalizable. The simplest reason is that it is symmetric, and in Reals, ALL symmetric matrices are diagonalizable. (You could also notice the trace is 2, yet it is rank 1, hence you have eigs = 2 and 0...)

    3a) You're right that the only unique eigenvalue = 1. This is a pre-requisite, but by no means is sufficient, for a matrix to be defective. What they'd like to hear is probably the how many linearly independent eigenvectors are associated with this eigenvalue... (you'll hear terms like geometric multiplicity here)
  4. Feb 3, 2017 #3


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    As mentioned above, this is too much for one post. To take question 1:

    b) ##\sqrt{x}## is not linear.

    c) The example you gave is not 1-1, which is what injective means.
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