Eigenvectors and Eigenvalues Tutorial Sheet, Sheet #4B
Learning targets
- Understand and produce visual representations of eigenvectors and eigenvalues
- Find the eigenvectors and eigenvalues for 2 x 2 and 3 x 3 matrices
Recap definitions
- Eigenvector: a vector which when operated on by a given operator gives a scalar multiple of itself.
- Eigenvalue: any number such that a given matrix minus that number times the identity matrix has zero determinant.
Additional Resources
Tutorials
- Linear Algebra Playlist : The same playlist as mentioned last week.
Software
- Eigen-calculator and Visualisation : Shows the eigs on a graph and calculates them for you - get an idea of where things move!
- Matlab Documentation : Information on eigenvectors and values on Matlab. Getting familiar with matlab will be a great advantage.
Problem sheet
Skill Building Questions
Problem 1.
Find the eigenvalues and the associated eigenvectors for the transformations represented in the following figures. (Note: The red vector shown is the result of transforming the blue vector. Be careful with the difference between the two arrowheads )
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Problem 2.
Find the eigenvalues and the associated eigenvectors of the matrices:
A quick trick for finding the eigenvalues of a 2x2 matrix just by looking at it. This is a good way for double checking your answer. Remember what you learned last week with Matlab - it’ll make your life a lot easier…
(a) $\begin{pmatrix}1&3\\ 2&2\end{pmatrix}$
For $\boxed{\lambda_1 = 4, \begin{pmatrix}1,1\end{pmatrix}},$
For $\boxed{\lambda_2 = -1, \begin{pmatrix}-3,2\end{pmatrix}}.$
(b) $\begin{pmatrix}1&3&1\\ 0&5&0\\ 1&3&0\end{pmatrix}$
For $\lambda_1 = 5$ $\Rightarrow{}\quad \begin{pmatrix}-4&3&1\\\ 0&0&0\\\ 1&3&-5\end{pmatrix} \begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \Rightarrow{}\quad \begin{cases} \begin{aligned} -4x_1 +3x_2 +x_3 =& 0\\\ 0x_1 +0x_2 +0x_3 =& 0\\\ x_1 +3x_2 -5x_3 =& 0 \end{aligned} \end{cases}\\\ \Rightarrow{}\quad \begin{cases} x_1 = \frac{6x_2}{5}\\\ x_2 = \frac{19x_3}{15} \end{cases} \quad\Rightarrow{}\quad \text{ in parametric form } x_1=\frac{6t}{5},\quad x_2=\frac{19t}{15},\quad x_3=t \text{ for any } t \in R\\\ \Rightarrow{}\quad \text{ if } t=15, \text{ eigenvector of } \lambda_1:\quad (18,19,15)$
For $\lambda_2 = \frac{1+\sqrt{5}}{2}$ $\Rightarrow{}\quad \begin{pmatrix} 1 - \frac{1+\sqrt{5}}{2} & 3 & 1\\\ 0 & 5 - \frac{1+\sqrt{5}}{2} & 0 \\\ 1 & 3 & - \frac{1+\sqrt{5}}{2}\end{pmatrix} \begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \Rightarrow{}\quad \begin{cases} \begin{aligned} \bigg(1 - \frac{1+\sqrt{5}}{2}\bigg)x_1 + 3x_2 + x_3 =& 0\\\ \bigg(5 - \frac{1+\sqrt{5}}{2}\bigg)x_2 =& 0\\\ x_1 + 3x_2 -\bigg(\frac{1+\sqrt{5}}{2}\bigg)x_3 =& 0 \end{aligned} \end{cases}\\\ \Rightarrow{}\quad \begin{cases} x_2 = 0\\\ x_1 = \bigg(\frac{1+\sqrt{5}}{2}\bigg)x_3 \end{cases} \quad\Rightarrow{}\quad \text{ in parametric form } x_1=\bigg(\frac{1+\sqrt{5}}{2}\bigg)t,\quad x_2=0,\quad x_3=t \text{ for any } t \in R \\\ \Rightarrow{}\quad \text{ if } t=1, \text{ eigenvector of }\lambda_2:\quad \big(\frac{1+\sqrt{5}}{2}, 0, 1\big)$
For $\lambda_3 = \frac{1-\sqrt{5}}{2}$ $\Rightarrow{}\quad \begin{pmatrix}1-\frac{1-\sqrt{5}}{2} & 3 & 1 \\\ 0 & 5-\frac{1-\sqrt{5}}{2} & 0 \\\ 1 & 3 & -\frac{1-\sqrt{5}}{2}\end{pmatrix} \begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix}\\\ \Rightarrow{}\quad \begin{cases} x_2 = 0\\\ x_1 = \bigg(\frac{1-\sqrt{5}}{2}\bigg)x_3 \end{cases} \quad\Rightarrow{}\quad \text{ in parametric form } x_1=\bigg(\frac{1-\sqrt{5}}{2}\bigg)t,\quad x_2=0,\quad x_3=t \text{ for any } t \in R$ $\Rightarrow{}\quad \text{ if } t=1, \text{ eigenvector of } \lambda_3:\quad \big(\frac{1-\sqrt{5}}{2}, 0, 1\big)$
To summarise:
$\boxed{\text{For } \lambda_1 = 5, \begin{pmatrix}18\\\ 19\\\ 15\end{pmatrix} \\\ \text{For } \lambda_2 = \frac{1+\sqrt{5}}{2}, \begin{pmatrix}\frac{1+\sqrt{5}}{2}\\\ 0\\\ 1\end{pmatrix} \\\ \text{For } \lambda_3 = \frac{1-\sqrt{5}}{2}, \begin{pmatrix}\frac{1-\sqrt{5}}{2}\\\ 0\\1\end{pmatrix}}$
(c) $\begin{pmatrix}3&1&1\\ 2&4&2\\ 1&1&3\end{pmatrix}$
$\Rightarrow{}\quad$ eigenvalues: $\lambda_1=2,\quad \lambda_2=2,\quad \lambda_2=6$
For $\lambda_1 = 2$ and $\lambda_2 = 2$ $\Rightarrow{}\quad \begin{pmatrix}1&1&1\\\ 2&2&2\\\ 1&1&1\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad x_1 + x_2 + x_3 = 0 \quad\Rightarrow{}\quad x_1 = -x_2 -x_3$ $\Rightarrow{}\quad$ in parametric form $x_1=-t-u,\quad x_2=t,\quad x_3=u$ for any $t,u \in R$ $\Rightarrow{}\quad$ if $t=0$ and $u=1$, eigenvector of $\lambda_1: (-1,0,1)$ $\Rightarrow{}\quad$ if $t=1$ and $u=0$, eigenvector of $\lambda_2: (-1,1,0)$
For $\lambda_3 = 6$ $\Rightarrow{}\quad \begin{pmatrix}-3&1&1\\\ 2&-2&2\\\ 1&1&3\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad \begin{cases} -3x_1 + x_2 + x_3 = 0\\\ 2(x_1 - x_2 + x_3) =0\\\ x_1 + x_2 - 3x_3 =0 \end{cases}\quad\Rightarrow{}\quad \begin{cases} x_1 = x_3\\ x_2 = 2x_3 \end{cases}$ $\Rightarrow{}\quad$ in parametric form $x_1=t,\quad x_2=2t,\quad x_3=t$ for any $t \in R$ $\Rightarrow{}\quad$ if $t=1$, eigenvector of $\lambda_3: (1,2,1)$
To summarise:
$\boxed{\text{For } \lambda_1 = 2 \text{ and } \lambda_2 = 2, \begin{pmatrix}-1\\\ 0\\\ 1\end{pmatrix}, \begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix}. \\\ \text{For } \lambda_3 = 6, \begin{pmatrix}1\\\ 2\\\ 1\end{pmatrix}}$
(d) $\begin{pmatrix}1&-1&-1\\ 1&-1&0\\ 1&0&-1\end{pmatrix}$
$\Rightarrow{}\quad$ eigenvalues: $\lambda_1 = -1, \quad \lambda_2 = i, \quad \lambda_3 =-i$
For $\lambda_1 = -1$ $\Rightarrow{}\quad \begin{pmatrix}2&-1&-1\\\ 1&0&0\\\ 1&0&0\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad \begin{cases} 2x_1 -x_2 -x_3 = 0\\\ x_1 = 0 \end{cases}$ $\Rightarrow{}\quad \begin{cases} x_2= -x_3\\\ x_1 = 0 \end{cases}$
$\Rightarrow{}\quad$ in parametric form $x_1=0,\quad x_2=t, \quad x_3=-t$ for any $t \in R$ $\Rightarrow{}\quad$ if $t=1$, eigenvector $(0,1,-1)$
For $\lambda_2 = i$ $\Rightarrow{}\quad \begin{pmatrix}1-i&-1&-1\\\ 1&-1-i&0\\\ 1&0&-1-i\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad \begin{cases} x_1(1-i) -x_2 -x_3 =0\\\ x_1 -(1+i)x_2 =0\\\ x_1 -(1+i)x_3 =0 \end{cases}$ $\Rightarrow{}\quad \begin{cases} x_1 =(1+i)x_2\\\ x_2 = x_3 \end{cases}$
$\Rightarrow{}\quad$ in parametric form $x_1=(1+i)t,\quad x_2=t,\quad x_3=t$ for any $t \in R$ $\Rightarrow{}\quad$ if $t=1$, eigenvector $(1+i,1,1)$
For $\lambda_3 = -i$ $\Rightarrow{}\quad \begin{pmatrix}1+i&-1&-1\\\ 1&-1+i&0\\\ 1&0&-1+i\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad \begin{cases} x_1(1+i) -x_2 -x_3 =0\\\ x_1 -(1-i)x_2 =0\\\ x_1 -(1-i)x_3 =0 \end{cases}$ $\Rightarrow{}\quad \begin{cases} x_1 =(1-i)x_2\\ x_2 = x_3 \end{cases}$
$\Rightarrow{}\quad$ in parametric form $x_1=(1-i)t,\quad x_2=t,\quad x_3=t$ for any $t \in R$ $\Rightarrow{}\quad$ if $t=1$, eigenvector $(1-i,1,1)$
To summarise: $\boxed{\text{For } \lambda_1 = -1, \begin{pmatrix}0\\\ 1\\\ -1\end{pmatrix}, \\\ \text{For } \lambda_2 = i. \begin{pmatrix}1 + i \\\ 1\\\ 1\end{pmatrix}, \\\ \text{For } \lambda_3 = -i,\begin{pmatrix}1 - i \\\ 1\\\ 1\end{pmatrix}}$
Problem 3.
Let $A$ be a $2\times2$ matrix whose trace equals 2 and determinant equals -2. Find the eigenvalues of $A$. (The trace of a square matrix is the sum of the terms along it’s leading diagonal.)
$\Rightarrow{}\quad$ eigenvalues: $\boxed{\lambda_1 = 1 + \sqrt(3), \quad \lambda_2 = 1 - \sqrt(3)}$
Problem 4.
$A=\begin{pmatrix}4&-6&2\\ -1&9&-2\\ -4&12&-2\end{pmatrix}$
Given that 6 is one of the eigenvalues and its determinant is 36. Find the other eigenvalues of $A$.
$\Rightarrow{}\quad p(\lambda): \lambda^3 -11\lambda^2 +36\lambda -36 = 0$ $\Rightarrow{}\quad$ if $\lambda_1 = 6$, then $(\lambda - 6)$ is a factor of the polynomial $p(\lambda)$ then dividing the polynomial $p(\lambda)$ by $(\lambda -6)$
$\Rightarrow{}\quad \lambda^3 -11\lambda^2 +36\lambda -36 = (\lambda -6)(\lambda^2 -5\lambda + 6) = 0$
$\Rightarrow{}\quad$ eigenvalues $\boxed{\lambda_2 = 2, \quad \lambda_3 = 3}$
Problem 5.
Given that $\begin{pmatrix}-2\\ 1\\ k\end{pmatrix}$ is an eigenvector of the matrix $\begin{pmatrix}1&2&-1\\ 1&0&1\\ 4&-4&5\end{pmatrix}$ find $k$ and the corresponding eigenvalue.
$\Rightarrow{}\quad \begin{pmatrix}1-\lambda&2&-1\\\ 1&-\lambda&1\\\ 4&-4&5-\lambda\end{pmatrix}\begin{pmatrix}-2\\\ 1\\\ k\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix}$
$\Rightarrow{}\quad \begin{pmatrix} -2(1-\lambda) + 2 - k = 0\\\ -2 -\lambda + k = 0\\\ -8 -4 +k(5 -\lambda)=0 \end{pmatrix}$ $\Rightarrow{}\quad$ evaluating these three equations with $\boxed{\lambda_2 = 2}$
$\Rightarrow{}\quad \boxed{k = 4}$
Problem 6.
The eigenvalues of the matrix $A=\begin{pmatrix}1&1&1\\ -1&3&1\\ -1&1&3\end{pmatrix}$ are 2 and 3. Find the corresponding eigenvectors and write down a matrix which diagonalizes $A$.
$\Rightarrow{}\quad \begin{pmatrix}-1&1&1\\-1&1&1\\-1&1&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$ $\Rightarrow{}\quad -x_1 + x_2 + x_3 = 0$ $\Rightarrow{}\quad x_1 = x_2 + x_3$
$\Rightarrow{}\quad$ in parametric form $x_1=t+u, \quad x_2=t,\quad x_3=u$ for any $t,u \in R$
$\Rightarrow{}\quad$ if $t=0,\quad u=1$, eigenvector of $\lambda_1:\quad (1,0,1)^T$
$\Rightarrow{}\quad$ if $t=1,\quad u=0$, eigenvector of $\lambda_2:\quad (1,1,0)^T$
For $\lambda_3 = 3$ $\Rightarrow{}\quad \begin{pmatrix}-2&1&1\\\ -1&0&1\\\ -1&1&0\end{pmatrix}\begin{pmatrix}x_1\\\ x_2\\\ x_3\end{pmatrix} = \begin{pmatrix}0\\\ 0\\\ 0\end{pmatrix} \quad\Rightarrow{}\quad \begin{cases} -2x_1 + x_2 + x_3 = 0\\\ -x_1 + x_3 = 0\\\ -x_1 + x_2 = 0 \end{cases}$
$\Rightarrow{}\quad \begin{cases} x_1 = x_3\\\ x_1 = x_2 \end{cases}$ $\Rightarrow{}\quad$ in parametric form $x_1=t,\quad x_2=t,\quad x_3=t$ for any $t \in R$ $\Rightarrow{}\quad$ if $t=1$, eigenvector of $\lambda_3:\quad (1,1,1)^T$
$\Rightarrow{}\quad P = \begin{pmatrix}1&1&1\\\ 0&1&1\\\ 1&0&1\end{pmatrix}$ is an invertible matrix that diagonalizes $A$,
so that $P^{-1}AP = \begin{pmatrix}2&0&0\\\ 0&2&0\\\ 0&0&3\end{pmatrix}$
To summarise: For $\lambda_1 = 2$ and $\lambda_2 = 2$, eigenvectors of $\boxed{\lambda_1: (1,0,1)^T}, \boxed{\lambda_2: (1,1,0)^T}$
For $\boxed{\lambda_3 = 3, (1,1,1)^T.}$
Exam Style Questions
Problem 7.
This is the exam question from the Linear Transforms tutorial sheet with an extra part on eigenvectors and eigenvalues. The following figure shows a square in $\mathbb{R}^2$, marked with a circle and cross on its perimeter.
(a) On a single plot, sketch the result of applying the following transformation, A, to the square (including the new locations of the circle and cross).
\(A = \begin{bmatrix} 2 \ 0 \\\ 1 \ 1.5\end{bmatrix}\)
(b) On the same axes draw the spans of any eigenvectors of A and label these spans with their corresponding eigenvalues.
(c) Assuming the area of the initial square is 4, what is the area of this region after the transformation?
Extension Questions
Problem 8.
Let $A$ be a $3\times3$ matrix whose trace equals 5 and determinant equals -12. Given that 3 is an eigenvalue of $A$, find the other eigenvalues.
$\Rightarrow{}\quad \begin{cases} a + e +j = 5\\ a(ej - fh) - b(dj - fg) +c(dh - eg) = -12 \end{cases}\\ \Rightarrow{}\quad \begin{pmatrix}a&b&c\\d&e&f\\g&h&j\end{pmatrix} -\lambda\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}a-\lambda&b&c\\d&e-\lambda&f\\g&h&j-\lambda\end{pmatrix} = B\\ \Rightarrow{}\quad \text{det}(A) = a(ej - fh) +a(\lambda^2 -e\lambda -j\lambda) - \lambda(\lambda^2 - (e+j)\lambda +ej -fh) - \\ b(dj - fg) + bd\lambda +c(dh - eg) + cg\lambda = 0\\ \Rightarrow{}\quad -\lambda^3 + \lambda^2(a + e + j) +\lambda(bd -ae -aj -ej +fh +cg) -12 = 0\\ \Rightarrow{}\quad -\lambda^3 + 5\lambda^2 +\lambda(bd -ae -aj -ej +fh +cg) -12 = 0$
$\Rightarrow{}\quad$ for $\lambda_1 = 3,\quad bd -ae -aj -ej +fh +cg = -2$ $\Rightarrow{}\quad -\lambda^3 +5\lambda^2 -2\lambda -12 = 0 \quad\Rightarrow{}\quad (\lambda -3)(-\lambda^2 + 2\lambda +4) = 0$
$\Rightarrow{}\quad$ Eigenvalues: $\boxed{\lambda_2 = 1 + \sqrt{5},\quad \lambda_3 = 1 - \sqrt{5}}$
Problem 9.
For each of the following matrices, find an invertible matrix which diagonalizes it. (click ‘toggle answer’ to see the definition for a diagonalisable matrix)
(a) $A = \begin{pmatrix}1&3\\ 2&2\end{pmatrix}$
(b) $A = \begin{pmatrix}1&-1\\ 1&3\end{pmatrix}$
(c) $A=\begin{pmatrix}1&3&1\\ 0&5&0\\ 1&3&0\end{pmatrix}$
(d) $A = \begin{pmatrix}1&-1&-1\\ 1&-1&0\\ 1&0&-1\end{pmatrix}$
Problem 10.
Given that for a $(2\times2)$ matrix $M$, one of the eigenvalue $\lambda_1=8$ and its corresponding eigenvector $v_1=(1,1)$, it is also known that matrix $M$ changes point $(-1,2)$ to $(-2,4)$.
(a) Find matrix $M$.
$\Rightarrow{}\quad \begin{pmatrix}a&b\\\ c&d\end{pmatrix}\begin{pmatrix}1\\\ 1\end{pmatrix} = 8\begin{pmatrix}1\\\ 1\end{pmatrix}
\quad\Rightarrow{}\quad \begin{cases} a+b = 8\\\ c+d = 8 \end{cases}$
Also, from the question is is known that:
$\quad \begin{pmatrix}a&b\\\ c&d\end{pmatrix}\begin{pmatrix}-1\\\ 2\end{pmatrix} = \begin{pmatrix}-2\\\ 4\end{pmatrix}$
$\quad\Rightarrow{}\quad \begin{cases} -a+2b = -2\\\ -c+2d = 4 \end{cases}$
Solving simultaneously, we get: $\quad\Rightarrow{}\quad \begin{cases} a=6\\\ b=2\\\ c=4\\\ d=4 \end{cases}$
$\quad\Rightarrow{}\quad \boxed{M=\begin{pmatrix}6&2\\\ 4&4\end{pmatrix}}$
(b) Find $\lambda_2$ and $v_2$.
${det(M)=(\lambda-6)(\lambda-4)-8=\lambda^2-10\lambda+16=(\lambda-2)(\lambda-8)=0}$
Therefore the other eigenvalue for matrix $M$ is $\lambda=2$
For $\lambda_2 = 2, $ $\Rightarrow{}$ let $v_2= \begin{pmatrix} x \\ y \end{pmatrix} \Rightarrow{} Mv_2 =\begin{pmatrix} 6x+2y \\4x+4y\end{pmatrix}=2 \begin{pmatrix}x\\y \end{pmatrix}$ $\Rightarrow{}\quad \boxed{v_2 = \quad \begin{pmatrix} -1 \\ 2 \end{pmatrix}}$
(c) For line $l: x-y+1=0$, find $l’$ after transformed by matrix $M$.
$\Rightarrow{}\quad \begin{pmatrix}6&2\\\ 4&4\end{pmatrix}\begin{pmatrix}x\\\ y\end{pmatrix} = \begin{pmatrix}x'\\\ y'\end{pmatrix}$
$\Rightarrow{}\quad \begin{cases} x = \frac{1}{4}x'-\frac{1}{8}y'\\\ y = -\frac{1}{4}x'+\frac{3}{8}y' \end{cases}$
$\Rightarrow{}\quad x'-y'+2=0$
$\Rightarrow{}\quad l': \boxed{y=x+2}$
Answers
For Printing
Revision Questions
The questions included are optional, but here if you want some extra practice.
- Engineering Mathematics 7th edition, Stroud and Dexter : Pages 489-508, 509-518
- A-Level Exam Questions : Matricies and other integrated vector/transform questions - need to look to the past markschemes for answers
- Matrices Practice : Many practice questions, starting off with eigs but goes on to involve all previous topics.