Visual representation of matrices

Question

I am used to seeing most basic mathematical objects being visually represented (for instance, a curve in the plane divided by the xy axis; the same goes for complex numbers, vectors, and so on....), However, I never saw a visual representation of a matrix. I do not mean the disposition in rows and columns, of course. I would like to know whether they can be graphically represented in some intuitive way. If so, how? Could you illustrate it, say, with a 2x2 quadratic matrix?

Thanks a lot in advance

Answer 1

There is no one way to "visualize" matrices. In fact, visualizing abstract quantities can be useful but also lead to loss of information through the visualization.

As far as 2x2 matrices go, one way of visualizing them is to note that they have the same mathematical structure as complex numbers when it comes to addition and multiplication (See http://en.wikipedia.org/wiki/2_%C3%97_2_real_matrices#2.E2.80.89.C3.97.E2.80.892_real_matrices_as_complex_numbers)

Of course, this is highly specific as far as matrices go. A more useful approach to "visualizing" a matrix is to view it a linear transformation on a vector -- and observe its action on a standard set of vectors. Linear transformations and their properties in terms of dilating and rotating vectors are useful ways to understand matrices, although they are not the only useful information contained inside a matrix.

I would recommend Artin's text as a useful first resource to highlight some of these connections. Depending on what you want to do with matrices, you can expand in many different directions.

Answer 2

You can represent a $2 \times 2$ matrices $A = \left[\begin{smallmatrix}a & b\\c & d\end{smallmatrix}\right]$ as a parallelogram $Q_A \subset \mathbb{R}^2$ with vertices $(0,0), (a,c), (b,d)$ and $(a+b,c+d)$. If one identify the plane $\mathbb{R}^2$ with $M^{2\times 1}(\mathbb{R})$, the space of $2 \times 1$ column matrices, then $Q_A$ is the image of the unit square $[0,1] \times [0,1]$ under linear transform

$$[0,1] \times [0,1] \ni \begin{bmatrix}x \\ y\end{bmatrix} \quad \mapsto \quad \begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix}a & b\\c & d\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} \in Q_A $$

Since a linear transformation is uniquely determined by its action on a basis, this provides a faithful represent of the $2 \times 2$ matrices. Under this representation, some geometry related operations now correspond to familiar geometric shapes. e.g.

• The matrix $\left[\begin{smallmatrix}s & 0\\0 & s\end{smallmatrix}\right]$ represents a scaling of geometric objects. It corresponds to a square of side length $s$, axis aligned with the standard $x$- and $y$-axis.

• The matrix $\left[\begin{smallmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \;\cos\theta\end{smallmatrix}\right]$ represents a counterclockwise rotation of angle $\theta$. It corresponds to the unit square rotated counterclockwisely for angle $\theta$.

• The matrices $\left[\begin{smallmatrix}1 & m\\ 0& 1\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}1 & 0\\ m& 1\end{smallmatrix}\right]$ represent sheer mappings in horizontal and vertical directions. They can visualized as parallelogram with one pair of its sides staying horizontal and vertical respectively.

This sort of shapes provides a useful visual mnemonics for what are the effects of those matrices (when viewed as a transformation of the plane).

Finally, one can also use this to introduce the concept of determinant to students.

• What is the the determinant of a matrix $A$? It is just the area of $Q_A$.
• What does it mean $\det A < 0$? It just mean the $Q_A$ has been flipped.

I'm sure there are other ways to use this visualization as a teaching tool.

Answer 3

You probably want a visual or mental picture of a matrix. Now that's truly hard. Let me explain why with a simple example:

Imagine you have a door and you open it. You can prove to me that the door is in two different states because you did a certain transformation to it and you actually rotated the door around the hinge. You did that transformation of a 3D object in TIME. So a transformation is nothing else but a structure in 4D space-time for the door, and you're lucky here because you transform just 2 coordinates, x, y(you can't rise up the door when you open it). Math lovers tend to generalize stuff and say...ok...but I'm a math guru, so I make the time dimension a spatial one...and I add another one...n actually.

Going back now to 4D a 2x2 matrix could actually be seen in 4D space. But that would also mean that for you opening a door is actually meaningless in 4D space(it's important this, when it's time it has meaning), because you see a geometrical object that has all the transitory states between the two states(door opened in state 1 or state 0). In time, it would mean that you see events in space-time like single stationary objects. For example a human speaking to you wouldn't actually be speaking to you. You would see him before he spoke and as it spokes and for you is a single object, even though for mere mortals he passed thru an infinite number of transition.

Pretty nice stuff though...helps alot in physics.

Answer 4

I understand your frustration with wanting to "visualize" matrices (even simple $2 \times 2$s)!

So, let's start first with the complex $1 \times 1$ matrices $[z]$ with $z \in \mathbb{C}$. Let's also constrain this set of matrices by imposing a rule that each of these matrices must be unitary; namely, $z$'s conjugate transpose is also its inverse. So $\bar{z} = z^{-1}$ implies $|z| = 1$, which means the set of all such $1 \times 1$ matrices is in fact the 1-dimensional unit circle. If we replace $\mathbb{C}$ with $\mathbb{R}$, we get two points: $\lbrace -1, 1 \rbrace$ (the 0-dimensional "circle"). If we lose the unitary requirement, then the real $1 \times 1$ matrices just form an infinite line: $\mathbb{R}$.

Ok, now let's move up to $2 \times 2$ real matrices. We know such matrices belong in $\mathbb{R}^{4}$ (in fact, all together they are $\mathbb{R}^4$), but how do we make sense of them?

Let's consider all such matrices (real and $2 \times 2$) which are also invertible. They form an open subspace of $\mathbb{R}^4$ that is disconnected: one component represents the matrices of positive determinant; the other negative. Ok, so this isn't very helpful, but if you think of these matrices as what they really are (linear transformations), then it will begin to make sense.

Note that a linear transformation does two things to a vector: (1) it scales it, and (2) it rotates it. Then we get a new vector. So we are sending $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$. Scaling can be parameterized by a single real number $\lambda$; similarly, rotation can be parameterized by a single real angle $\phi \in \mathbb{R} / \pi\mathbb{Z}$ (in fact, the set of all real $2 \times 2$ rotation matrices is the unit circle). These two parameters can be thought of as your extra $2$ dimensions; namely, to visualize an invertible real $2 \times 2$ matrix $A$, you could view it as a $4$-tuple $(x, y, \lambda, \phi)$ ($\in \mathbb{R}^4$), where $v = [x, y]^\top$ is a vector in $\mathbb{R}^2$, and $\lambda$ and $\phi$ are the scaling and rotating components that will send $v$ to $Av$.

Another kind of hand-wavy way is to say that a $2 \times 2$ real invertible matrix can be decomposed into it's scaling part and it's rotating part (each independent). The rotating part, as claimed above, is really just the circle $S^1$. The scaling part is really just the real line $\mathbb{R}$. So together they form sort of an infinite cylinder $\mathbb{R} \times S^1$. So the "total space" is $\mathbb{R}^2 \times \mathbb{R} \times S^1 = \mathbb{R}^3 \times \mathbb{R}/\mathbb{Z}$ which is "in" (hand-wavy) $\mathbb{R}^4$.

In line with the door comment, this is a space in which the vector $v$ under transformation by $A$ is "preserved" through all states until it reaches its final state $Av$.