# Fix me or crush me

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 ${\bf M}$, other than the identity matrix, which leaves ${\bf F}$ fixed, in that ${\bf M}{\bf F}={\bf F}$? How many such matrices can you find? Which is the simplest? Which is the most complicated?

Can you construct an example of a matrix ${\bf N}$, other than the zero matrix, which crushes ${\bf Z}$ to the zero vector ${\bf 0}$, in that ${\bf N}{\bf Z}={\bf 0}$? 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.

$$

{\bf 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} \phantom{-}1&-2&\phantom{-}1\\ \phantom{-}1&\phantom{-}1&\phantom{-}0\\ -2&\phantom{-}1&-2\\ \end{pmatrix}

$$

*You might find this Matrix Multiplication calculator helpful for testing out your ideas.*

*There are more matrix problems in this feature.*

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 "Fixed" vectors here are eigenvectors with an

*eigenvalue*of $1$. More generally the eigenvectors of ${\bf M}$ satisfy ${\bf M}{\bf F} = \lambda {\bf F}$, where $\lambda$ is the eigenvalue associated with the eigenvector.

The

*kernel*of a matrix is the set of vectors which are squashed to zero, or you can think of them as being the set of eigenvectors with eigenvalue $0$.

Both concepts are of fundamental importance in higher-level algebra and its applications to science.

Nishad from Rugby High School in the UK found the most general forms of matrices to fix and crush the given vectors:

Can you combine these ideas to get a matrix that both fixes $\bf{F}$ and crushes $\bf{Z}$?

Nishad then moved on to the matrices below and found vectors that they fix and crush.

Nishad used the ideas of eigenvectors and eigenvalues, where $\lambda$ is used to represent an eigenvalue of a matrix. Nishad also used determinants, written $\text{det}(\bf M)$ for a matrix $\bf{M},$ where matrices which crush any vectors satisfy $\text{det}(\bf M)=0.$

*It is also possible to solve this problem without using eigenvalues or determinants - can you find a way?*

### Why do this problem?

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.

Students might like to use this Matrix Multiplication calculator to test out their ideas.

### Possible approach

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.

### Key questions

*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 ${\bf Z}$.

There are more matrix problems in this feature.