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

1. Feb 2, 2017

### davidge

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 $\pi_1: \mathbb{R} ^{2} \longrightarrow \mathbb{R}$ 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) $\begin{pmatrix}1&1\\1&1\end{pmatrix}$.

2. Relevant equations

3. The attempt at a solution

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)

Last edited: Feb 2, 2017
2. Feb 2, 2017

### Ray Vickson

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
3. Feb 3, 2017

### davidge

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