- #1
davidge
- 554
- 21
Hi everyone. Excuse me for my poor English skills. I did an exam today and my exam result was 13 of 40. I don't 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. Homework Statement
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].
[/B]
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)
1. Homework Statement
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].
Homework Equations
The Attempt at a Solution
[/B]
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: