Fix me or crush me

6575 /www/nrich/html/content/id/6575/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Imagine that you have a pair of vectors ${\bf F}$ and ${\bf Z}$ $$ {\bf F}=\pmatrix{1\cr 1 \cr 0}\quad {\bf Z}=\pmatrix{0\cr 1 \cr 1} $$ Can you construct an example of a matrix $M$, other than the identity, which leaves ${\bf F}$ fixed, in that $M{\bf F}={\bf F}$? How many such matrices can you find? Which is the simplest? Which is the most complicated? Can you construct an example of a matrix $N$, other than the zero matrix, which crushes ${\bf Z}$ to the zero vector ${\bf 0}$, in that $N{\bf Z}={\bf 0}$? How many such matrices can you find? Which is the simplest? Which is the most complicated? Can you find a matrix which leaves ${\bf F}$ fixed and also crushes ${\bf Z}$? Can you find any (many?) vectors fixed or crushed by the following matrices? Give examples or convincing arguments if no such vectors exist. $$ M = \begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}, \begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5\\ \end{pmatrix}, \begin{pmatrix} 1&-2&1\\ 1&1&0\\ -2&1&-2\\ \end{pmatrix} $$ Very hard extension: Imagine that you are given a vector ${\bf F}$ and a vector ${\bf Z}$. Investigate whether you will be able to make a matrix $M$ which both fixes ${\bf F}$ and crushes ${\bf Z}$. <div class="framework"> NOTES AND BACKGROUND Matrices are used to represent transformations of vectors; vectors and matrices are usually studied together as an inseparable pair. Although matrices and the rules of matrix multiplication might seem abstract upon first encounter, they are actually very natural and encode in an entirely meaningful way notions of symmetry and transformation. This problem allows you to explore the effects matrix multiplication has on various vectors. The eigenvectors of a matrix are those vectors whose direction is unchanged by the action of the matrix. The kernel of a matrix is the set of vectors which are squashed to zero. Both concepts are of fundamental importance in higher-level algebra and its applications to science.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Doug writes: For the first part, I took a general matrix $\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{bmatrix}$, and multiplied this with $\mathbf{F}$ to get $\begin{pmatrix} a+b\\ d+e\\ g+h\\ \end{pmatrix} \therefore$ a+b=1, d+e=1, g+h=0, and c,f,i are irrelevant. There are an infinite number of solutions to this. Check the identity matrix is one, yes it is. The simplest non-identity is probably $\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&0\\ \end{bmatrix}$ And a complicated one is $\begin{bmatrix} m&1-m&j\\ n&1-n&k\\ p&-p&l\\ \end{bmatrix}$. To crush $\mathbf{Z}$, I followed the same process. The simplest non-zero matrix is probably $\begin{bmatrix} 1&0&0\\ 0&0&0\\ 0&0&0\\ \end{bmatrix}$ And a complicated one is $\begin{bmatrix} j&m&-m\\ k&n&-n\\ l&p&-p\\ \end{bmatrix}$. To find a matrix that achieves both, I just combined the equations found previously to the simultaneous set: a+b=1,d+e=1,g+h=0,b+c=0,e+f=0,h+i=0 $\therefore$ b=-c,e=-f,h=-i $\therefore$ a-c=1,d-f=1,g-i=0 9 variables, 6 equations, but there is probably not just one unique solution. In fact looking at these, we can quickly jump to $\begin{bmatrix} 1-j&j&-j\\ 1-k&k&-k\\ -l&l&-l\\ \end{bmatrix}$ which satisfies both sets, thus any matrix of this form will fix $\mathbf{F}$ and crush $\mathbf{Z}$. The above matrix must be, and is, singular, i.e. the eigenvectors do not form an independent set. If the eigenvectors formed an independent set then there would be no vector that would be crushed by being transformed by the matrix. For the next part, it seems that $\mathbf{F}$ and $\mathbf{Z}$ have been carefully chosen to have 1s and 0s to make simultaneous fixing & crushing possible. I tried to prove this with the general matrix $\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{bmatrix}$ multiplied with a fixable general vector $\begin{pmatrix} j\\ k\\ l\\ \end{pmatrix}$ and a crushable general vector $\begin{pmatrix} m\\ n\\ p\\ \end{pmatrix}$ so we have 6 simultaneous equations: aj+bk+cl=j dj+ek+fl=k gj+hk+il=l am+bn+cp=0 dm+en+fp=0 gm+hn+ip=0 I proceeded to attempt to solve the simultaneous equations, but did not find a general argument to prove it either way. Steve notes: The first matrix fixes all vectors and crushes no vectors. So, clearly not all matrices crush vectors. I wonder if others fail to crush vectors? Anyway, to determine more generally if a matrix $M$ crushes a vector $\mathbf{Z}$ we need to solve $M\mathbf{Z}= \mathbf{0}$. By putting this into components we get three simultaneous equations in $x, y, z$. For the second matrix, the equations I get are $$x+2y+3z=0, 2x+3y+4z = 0, 3x+4y+5z=0$$ I can solve these by elimination to get the solution as $$ 2x=-y=2z $$ This is a line of solutions - each vector can be written as $(a, -2a, a)$ for any number $a$. To check, we can see that $$ \begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5\\ \end{pmatrix} \begin{pmatrix} a\\ -2a\\ a\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix} $$ To determine if the same matrix fixes a vector, I need to solve $$ \begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5\\ \end{pmatrix} \begin{pmatrix} x\\ y\\ z\\ \end{pmatrix} = \begin{pmatrix} x\\ y\\ z\\ \end{pmatrix} $$ This gives me equations $$ 2y+3z=0, 2x+2y+4z = 0, 3x+4y+4z=0 $$ I tried to solve these, but they proved to be inconsistent with no solution. So the second matrix does not fix any vector. So, there are clearly matrices which don't leave any vector fixed. For the third matrix I saved some time by noticing that the second row is $(1, 1, 0)$. So, if this matrix is to crush any vector then that vector must be of the form $(a, -a, z)$. Multiplying this vector by the matrix gives $(3a+z, 0, -3a-2z)$. This can't be set equal to zero for any choice of $a\neq 0$ and $z$, so the last matrix does not crush any vector. What does it fix? The second row is again key. It implies that $x+y=y$ which implies that $x=0$. The other rows give equations $-2y+z=0$ and $y-2z=z$, which are inconsistent unless $y=z=0$. So, this matrix doesn't fix a vector either! I wasn't expecting that. For the last part, note that if the vector $\mathbf{F}$ has all non-zero components $x, y$ and $z$ then it is always possible to construct a matrix such that $\mathbf{F}$ is fixed $\begin{bmatrix} 1&a& -(x+ay)/z+x/z\\ 1&b& -(x+by)/z+y/z\\ 1&c&-(x+cy)/z+1\\ \end{bmatrix}$ It looks like variable choices $a, b, c$ will be sufficient to make it possible to crush any choice of vector $\mathbf{Z}$ which is not parallel to $\mathbf{F}$, but I'd better make sure. Acting with this matrix on any other vector $(X, Y, Z)$ gives me $$ X+aY+\left(\frac{x-(x+ay)}{z}\right)Z = X+bY+\left(\frac{y-(x+by)}{z}\right)Z =X+cY+\left(\frac{z-( x+cy)}{z}\right)Z=0 $$ I can rewrite this to make the factors of $a, b, c$ more clear as $$ X+\left(\frac{x-x}{z}\right) Z+a\left(Y-\frac{y}{z}Z\right)=X+\left(\frac{y-x}{z}\right) Z+b\left(Y-\frac{y}{z}Z\right)=X+\left(\frac{z-x}{z}\right) Z+c\left(Y-\frac{y}{z}Z\right) $$ Each choice $a, b, c$ premultiplies the same factor. So, if this factor is not zero I can choose $a, b, c$ so that $(X, Y, Z)$ is annihilated. This leaves the case when the factor is zero, which occurs if and only if $$ \frac{Y}{Z} = \frac{y}{z} $$ Geometrically, this occurs when and only when the projections of the two vectors onto the $x=0$ plane are parallel. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> This problem gives students the opportunity to explore the effect of matrix multiplication on vectors, and lays the foundations for studying the eigenvectors and kernel of a matrix, ideas which are very important in higher level algebra with applications in science. <h3>Possible approach</h3> <div>Start by asking students to work with the vector ${\bf F}$ to find a matrix which fixes it. Initially, let students find their own methods of working - some may choose to try to fit numbers in the matrix, some may straight away work with algebra. Once students have had a chance to try the task, allow some time to discuss methods, as well as the simplest and most complicated examples of matrices they have managed to find.</div> <div>Repeat the same process to find a matrix which crushes the vector $\bf Z$.</div> <div>The last part of the problem asks students to seek vectors which are fixed or crushed by each of the three matrices given. This works well if students are first given time to explore the properties of the matrices and to construct the conditions needed for a vector to be fixed or crushed by them. Then encourage discussion of their findings, particularly focussing on justification for matrices where appropriate vectors can't be found.</div> <h3>Key questions</h3> <div>What properties must a matrix have if it fixes $\bf F$? Or if it crushes $\bf Z$?</div> <div>What is the simplest matrix with these properties?</div> <div>What is the most general matrix you can write down?</div> <div>What properties must a vector have to be fixed or crushed by the three matrices given?</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This question becomes much more meaningful and straightforward if you consider the geometrical interpretation of the action of a matrix on a vector. For example, a rotation matrix will leave a line of points invariant. A matrix which crushes a vector is a type of 'projection'. Clearly a projection must send a non-zero vector to the origin. This problem uses mathematical concepts found in the later Further Pure Mathematics A Level modules. </mdoxml> 2 4 0 0 0 0 1 Fix me or crush me Can you make matrices which will fix one lucky vector and crush another to zero? 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