# Is matrix algebra a special case of Grassmann (exterior) algebra, and if so what is more general case?

Just a little question. I only recently heard about Grassmann algebra while reading a book on the history of vector algebra and quaternions.

I still don't understand what does exterior product mean exactly, and what is its most general case (according to the book I've read, Grassmann was trying to describe every possible algebraic system).

But this thing about matrices is easy to understand. Since the product of $m \times n$ and $n \times m$ matrices is $m \times m$ matrix, I can see that.

$$[a_1, a_2, a_3] \begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix} = a_1 b_1+a_2 b_2+a_3 b_3$$

Which is $1 \times 1$ matrix, i.e. a number, a dot product of two vectors. So this operation lowers the rank of a tensor by one.

But in the other case

$$\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix} [a_1, a_2, a_3] = \begin{bmatrix}a_1 b_1 & a_2 b_1 & a_3 b_1 \\a_1 b_2 & a_2 b_2 & a_3 b_2 \\a_1 b_3 & a_2 b_3 & a_3 b_3 \end{bmatrix}$$

So this operation elevates the rank of a tensor by one, creates a square matrix from two vectors.

So is this a special case of exterior and interior products? And if it what is the more general case?

## 1 Answer

The second product you give is equivalent to a tensor product; if it were an exterior product instead, the resulting matrix would be antisymmetric. Also notice that the trace of that matrix is the inner product. So the inner product can be identified with the trace, and the exterior product could be identified with the antisymmetric part of this matrix, but there's still a symmetric, traceless piece.

That is, the tensor product captures both the inner product and the exterior product, but it also clearly has more stuff.

Exterior algebra can be described completely in terms of tensor algebra, but not vice versa.