# What's the “geometry” in “geometric multiplicity”?

The geometric multiplicity of an eigenvalue is defined as the dimension of the associated eigenspace, i.e. number of linearly independent eigenvectors with that eigenvalue.

Here are my questions:

• Where is the name "geometric multiplicity" from in math history?
• Is there any thing in geometry related to this concept? Does it only mean the number of linearly independent eigenvectors with that eigenvalue one can "draw" as the definition says?
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Well, an eigenspace is a vector subspace, so if the geometric multiplicity is 1 then it is a line through the origin, if it is 2 then it is a plane through the origin, etc. This is just a wild guess, but it seems plausible that the name might come from this interpretation. – lentic catachresis Aug 4 '11 at 21:01
The term seems self-explanatory. If you're talking about the dimension of a space, you're talking about geometry. – Michael Hardy Aug 4 '11 at 22:49
Would be cool to find the historic roots of this term, that is: when was it first used? Great question. From OED: "3. Math. The number of times, for a given root a, that (x − a) occurs as a factor of f(x) (or its Taylor expansion about the point x = a). 1841 J. J. Sylvester in London, Edinb., & Dublin Philos. Mag. 18 136, I use the word multiplicity to denote a number, and distinguish between the total and partial multiplicities of the roots of an algebraic equation." But when was "geometric multiplicity" first used? It's not in the OED. – neuronet Jan 4 at 21:12

@Bruno is essentially correct. It's important to see that geometric multiplicity is meant to be distinguished from algebraic multiplicity of eigenvalues, the latter being the total number of times an eigenvalue occurs as a root of the characteristic equation. An analogy can be made with roots of any polynomial. For example, $x^2 + 2x + 1$ has a single root $-1$ of multiplicity 2. Algebraically, there are always two roots for a quadratic (at least over $\mathbb{C}$), and in this case, those roots are $-1$ and $-1$. But (geometrically) there is only one $x$-intercept for the function $y = x^2 + 2x + 1$.
Given a linear map $L:V \to V$; $V$, $W$ vector spaces, the eigenspace $E$ associated with an eigenvalue $\lambda$ is that subspace of $V$ where $L$ acts as a scalar, i.e., $L$ acts by stretching vectors. The geometric multiplicity of $\lambda$ is the dimension of the subspace of $V$ (in the domain) where $T$ acts as a scalar, with scaling coefficient $\lambda$, i.e., $T$ when restricted to the eigenspace is a map given by $T(v)=\lambda v$.
Maybe the clearest example is that given by the identity map $I:V \to V$, $I(v)=v$, where $V$ is $n$-dimensional ($n < \infty$), with associated matrix the identity matrix. Here, the only eigenvalue is $1$, and you can easily see, e.g, by looking at the associated matrix ($M-\lambda I:=(I-1I)=0$ that the geometric multiplicity is $n$; this means that the identity acts on the whole of $V$ by scaling by $1$. Substituting the identity matrix for a scalar matrix (i.e. $a_{ii}=c$; $a_{ij}=0$; $i\neq j$) illustrates the same point, e.g., if our matrix is the matrix ($a_{ij}:a_{ii}=2$; $a_{ij}=0$ if $i \neq j$ ) then $M$ acts like a scalar for every vector.
At the other extreme, if our matrix represents a linear transformation of a rotation by an angle $\theta \neq 0, \neq 2n\pi$, then the eigenspace is $0$-dimensional, since each point will be sent to another point with the same radius (because it will send a point $p$ in a circle $C$ to another point $p'$ in the same circle), and no point will be rotated into itself. The eigenspace would then be $0$-dimensional, since only the dimension-zero subspace will be sent to itself.