# Does an eigenvalue that does NOT have multiplicity usually have a one-dimensional corresponding eigenspace?

I'm trying to understand this statement in my book:

In general, the multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace.

So then, if an eigenvalue does NOT occur as a multiple root, is that kind of like saying it has a multiplicity of one and therefore the dimension of its eigenspace is less than or equal to one?

I guess I'm just struggling to understand the concept of multiplicity and why it matters, and what having multiple roots of the characteristic polynomial even means or how it affects anything.

The eigenspace of a particular eigenvalue is guaranteed to have dimension at least $1$. This is because in finding the eigenvalues of a matrix $A$, we require $\det(A-\lambda I)=0$, which guarantees that the system $(A-\lambda I)\mathbf{x}=\mathbf{0}$ will have a nontrivial solution. So there will be at least one eigenvector which satisfies $A\mathbf{x}=\lambda\mathbf{x}$ for each eigenvalue $\lambda$.

The issue when an eigenvalue $\lambda$ has multiplicity greater than $1$ is that we cannot guarantee that the eigenspace will have the same dimension as the multiplicity. We can only guarantee that the eigenspace will have dimension at least $1$.

In particular, it tells us about diagonalizability of a matrix. If an $n\times n$ matrix $A$ has $n$ distinct eigenvalues (i.e. each eigenvalue has multiplicity $1$), then it is diagonalizable. It may still be diagonalizable if it doesn't have $n$ distinct eigenvalues, but there is no guarantee.

Let's look at it from the point of view of the dimension of the eigenspace. Being an eigenvalue means that there's a nontrivial corresponding eigenspace, i.e. the dimension has to be at least $1$. And on the other hand, this dimension cannot exceed the multiplicity of the eigenvalue. So we have the following double inequality:

$$1\le\dim(\text{eigenspace})\le\text{multiplicity of eigenvalue}.$$

If the multiplicity is $1$, then from this double inequality we immediately see that the dimension has to be $1$ too.

In fact, I find the following terminology more illuminating. There are two types of multiplicities defined for an eigenvalue $\lambda$: the algebraic multiplicity of $\lambda$ is its multiplicity as a root of the characteristic equation, while the geometric multiplicity of $\lambda$ is the dimension of the corresponding eigenspace. With this terminology, the double inequality above can be stated as

$$1\le\text{geom.mult.}(\lambda)\le\text{alg.mult.}(\lambda).$$

Even if we really care about the geometric multiplicity, i.e, about eigenspaces and their dimensions, it's not easy to find, as that requires much more work. That's one reason why we care about algebraic multiplicities: they are easier to find, and at least they tell us something, some bounds on those dimensions. And if we're lucky to discover that all eigenvalues are distinct, we know all dimensions (all equal to $1$) without doing too much work!

• Thanks, that makes sense. But I actually don't understand why the dimension of the eigenspace is less than the multiplicity of the eigenvalue in the first place? – dagny Feb 18 '17 at 4:38
• To be honest, I can't think of a short answer. A long answer is to look at the Jordan normal form of the matrix. Basically, in this form all eigenvalues are lined up on the main diagonal according to their algebraic multiplicities. Either all of them or only some of them have corresponding eigenvectors, hence the geometric multiplicity can't exceed algebraic. – zipirovich Feb 18 '17 at 5:27