It's helpful to think like a computer scientist here...

a **tensor** is a type of **multidimensional array** with certain **transformation properties**

So if we wrote the spec for `array`

we just have to specify the collection of numbers.

```
array(N1, N2, N3)
```

If we specify tensor, we need to specify how this array "transforms" under certain matrix operations.

A spinor is going to be a tensor with a **group action** in addition to the other transformation rules. In particular an $SU(2)$ group action.

This already happens in linear algebra. Take the function $T: (1,0) \mapsto (1,1), (0,1) \mapsto (1,-1)$. We can write this as a $2 \times 2$ matrix:

$$\left[ \begin{array}{cr} 1 & 1 \\ 1 & -1 \end{array}\right]$$

These matrices have eigenvalues and eigenvectors. We have to solve the equation:

$$ \lambda^2 + 2 = 0$$

We get eigenvectors of $\lambda = \pm \sqrt{2}$. The eigenvectors are $(1 \pm \sqrt{2}, 1) $ and the diagonal form is:

$$\left[ \begin{array}{cr} \sqrt{2} & 0 \\ 0 & -\sqrt{2} \end{array}\right]$$

The same geometric object a "linear transformation" can be one of two different matrices depending on the basis we choose!

In differential geometry we can consider the **metric tensor**:

$$ dz^2 = dx^2 + dy^2 $$

If we switch to polar coordinates: $x = r \cos \theta, y = r \sin \theta$. Then $dx = -r \sin \theta \, d\theta + \cos \theta \, dr$ and $dy = r \cos \theta \, d\theta + \sin \theta \, dr$. Then

\begin{eqnarray*} dz^2 &=& (-r \sin \theta \, d\theta + \cos \theta \, dr)^2
+ ( r \cos \theta \, d\theta + \sin \theta \, dr)^2 \\
&=& (r d\theta)^2 + dr^2
\end{eqnarray*}

It should not matter if we compute the arc length in polar coordinates or cartesian, it should give the same answer.