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.(adsbygoogle = window.adsbygoogle || []).push({});

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):

π_{1}(x_{1}, y_{1}) = x_{1}, π_{1}(x_{2}, y_{2}) = x_{2}

π_{1}(x_{1}, y_{1}) + π_{1}(x_{2}, y_{2}) = x_{1}+ x_{2}= π_{1}(x_{1}+ x_{2}, y_{1}+ y_{2}).

Linearity (scalar multiplication):

π_{1}(αx_{1}, y_{1}) + π_{1}(αx_{2}, y_{2}) =

α(x_{1}+ x_{2}) = π_{1}(α(x_{1}+ x_{2}), y_{1}+ y_{2}).

Ker(π_{1}) = {0, y}, Im(π_{1}) = ℝ^{2}; dimension 2.

(b) T: ℝ^{2}→ ℝ

(x, y) [itex] \mapsto[/itex] T(x, y) = [itex]\sqrt x[/itex].

(c) Yes. This condition will be satisfied if each element of ℝ^{2}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 ℝ^{2}.

(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, A^{4}= I, ... Since 2017 is a odd number, A^{2017}= A, and tr(A^{2017}) = (1 x -1 x -1) = 1.

3.

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

(b)

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# I Matrices and linear transformations. Where did I go wrong?

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