Matrices and linear transformations. Where did I go wrong?

[davidge]

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

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

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

2. Consider the matrix

A = \begin{pmatrix}1&-2&8\\0&-1&0\\0&0&-1\end{pmatrix}

(a) Calculate the eigenvalues and eigenspaces of A.

(b) Is A a diagonalizable matrix? Explain.

(c) Calculate tr(A^{2017}).

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

(a) \begin{pmatrix}1&1\\0&1\end{pmatrix}.

(b) \begin{pmatrix}1&1\\1&1\end{pmatrix}.

Homework Equations

The Attempt at a Solution

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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) \mapsto T(x, y) = \sqrt x.

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

 

[Ray Vickson]

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

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

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

2. Consider the matrix

A = \begin{pmatrix}1&-2&8\\0&-1&0\\0&0&-1\end{pmatrix}

(a) Calculate the eigenvalues and eigenspaces of A.

(b) Is A a diagonalizable matrix? Explain.

(c) Calculate tr(A^{2017}).

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

(a) \begin{pmatrix}1&1\\0&1\end{pmatrix}.

(b) \begin{pmatrix}1&1\\1&1\end{pmatrix}.

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!

 

[davidge]

Thanks Ray Vickson and StoneTemplePython. What about my answers on (b) and (c) on question 1.?