The Unapologetic Mathematician

Mathematics for the interested outsider

The Multiplicity of an Eigenpair

As usual, let T:V\rightarrow V be a linear transformation on a real vector space V of dimension d. We know that T can be put into an almost upper-triangular form

\displaystyle\begin{pmatrix}A_1&&*\\&\ddots&\\{0}&&A_m\end{pmatrix}

where each block A_j is either a 1\times1 matrix or a 2\times2 matrix with no eigenvalues. Now I assert that

  • If \lambda is a real number, then exactly \dim\left(\mathrm{Ker}\left((T-\lambda I_V)^d\right)\right) of the blocks are the 1\times1 matrix whose single entry is \lambda.
  • If (\tau,\delta) is a pair of real numbers with \tau^2<4\delta, then exactly

    \displaystyle\frac{\dim\left(\mathrm{Ker}\left((T^2-\tau T+\delta)^d\right)\right)}{2}

    of the blocks are 2\times2 matrices with characteristic polynomial X^2-\tau X+\delta. Incidentally, this shows that the dimension of this generalized eigenspace is even.

This statement is parallel to the one about multiplicities of eigenvalues over algebraically closed fields. And we’ll use a similar proof. First, let’s define the polynomial p(X) to be X-\lambda if we’re trying to prove the first part, and to be X^2-\tau X+\delta if we’re trying to prove the second part, and let n be the degree of p. We’ll proceed by induction on the number m of blocks along the diagonal.

If m=1 then V is either one- or two-dimensional. Then the statement reduces to what we worked out by cases earlier. So from here we’ll assume that m>1, and that the statement holds for all matrices with m-1 blocks.

Let’s define U\subseteq V to be the subspace spanned by the first m-1 blocks, and \hat{U} be the subspace spanned by the last block. That is, U consists of all but the last one or two rows and columns of the matrix, depending on whether A_m is 1\times1 or 2\times2. Clearly U is invariant under T, and the restriction T|_U of T to U has matrix

\displaystyle\begin{pmatrix}A_1&&*\\&\ddots&\\{0}&&A_{m-1}\end{pmatrix}

with m-1 blocks. Thus the inductive hypothesis tells us that exactly

\displaystyle\frac{1}{n}\dim\left(\mathrm{Ker}\left(\left(p\left(T|_U\right)\right)^d\right)\right)

of the blocks from A_1 to A_{m-1} have characteristic polynomial p.

We’ll also define \hat{T}:\hat{U}\rightarrow\hat{U} to be the linear transformation acting by the block A_m. This is essentially the action of T on the quotient space V/U, but we’re viewing \hat{U} as giving representatives in V for vectors in the quotient space. This way, if v\in\hat{U} is a vector in this subspace of representatives we can write Tv=Sv+u_1 for some u_1\in U. Further, T^2v=S^2v+u_2 for some other vector u_2\in U. No matter which form of p we’re using, we can see that p(T)v=p(S)v+u_3 for some u_3\in U, and further that p(T)^dv=p(S)^dv+u_4 for some u_4\in U.

Now, either A_m has characteristic polynomial p or not. If not, then I say that \mathrm{Ker}\left(p(T)^d\right)\subseteq U. This implies that

\displaystyle\mathrm{Ker}\left(\left(p(T)\right)^d\right)=\mathrm{Ker}\left(\left(p\left(T|_U\right)\right)^d\right)

and thus that both the dimension of this kernel and the number of blocks with characteristic polynomial p are the same as for T|_U.

So let’s assume v\in\mathrm{Ker}\left(p(T)^d\right) and write v=u+\hat{u} with u\in U and \hat{u}\in\hat{U}. Then

\displaystyle0=p(T)^dv=p(T)^du+p(T)^d\hat{u}=p(T)^du+u_5+p(S)^d\hat{u}

for some u_5\in U. This implies that p(S)^d\hat{u}=0, but since p is not the characteristic polynomial of S, it is invertible on \hat{U}. Thus \hat{u}=0 and v\in\mathrm{Ker}\left(p\left(T|_U\right)^d\right).

On the other hand, if the characteristic polynomial of A_m is p, then we want to show that

\displaystyle\dim\left(\mathrm{Ker}\left(p(T)^d\right)\right)=\dim\left(\mathrm{Ker}\left(\left(p\left(T|_U\right)\right)^d\right)\right)+n

The inclusion-exclusion principle tells us that

\displaystyle\begin{aligned}\dim\left(\mathrm{Ker}\left(p(T)^d\right)\right)&=\dim\left(U\cap\mathrm{Ker}\left(p(T)^d\right)\right)+\dim\left(U+\mathrm{Ker}\left(p(T)^d\right)\right)-\dim(U)\\&=\dim\left(\mathrm{Ker}\left(p\left(T|_U\right)^d\right)\right))+\dim\left(U+\mathrm{Ker}\left(p(T)^d\right)\right)-(d-n)\end{aligned}

We’ll show that U+\mathrm{Ker}\left(p(T)^d\right)=V, and so its dimension is d, and we have the result we want.

So take \hat{u}\in\hat{U}. Because the characteristic polynomial of S is p, we know that p(S)=0. Thus p(T)\hat{u}\in U. Then

\displaystyle p(T)^d\hat{u}=p(T)^d\left(p(T)\hat{u}\right)\in\mathrm{Im}\left(p\left(T|_U\right)^{d-1}\right)=\mathrm{Im}\left(p\left(T|_U\right)^d\right)

where the last equality holds because the dimension of U is d-n, and so the image has stabilized by this point. Thus we can choose u\in U so that p(T)^d\hat{u}=p\left(T|_U\right)^du. And so

\displaystyle\begin{aligned}p(T)^d(\hat{u}-u)&=p(T)^d\hat{u}-p(T)^du\\&=p(T)^d\hat{u}-p\left(T|_U\right)^du\\&=0\end{aligned}

which shows that \hat{u}-u\in\mathrm{Ker}\left(p(T)^d\right). And thus \hat{u}=u+(\hat{u}-u) is in U+\mathrm{Ker}\left(p(T)^d\right). Since \hat{u} was arbitrary, the whole subspace \hat{U}\subseteq U+\mathrm{Ker}\left(p(T)^d\right), which shows that V=U+\hat{U}\subseteq U+\mathrm{Ker}\left(p(T)^d\right)\subseteq V, which completes our proof.

Now, all of that handled we turn to calculate the characteristic polynomial of T, only to find that it’s the product of the characteristic polynomials of all the blocks A_m. That is, we will have \dim\left(\mathrm{Ker}\left((T-\lambda I_V)^d\right)\right) factors of (X-\lambda) and \frac{1}{2}\dim\left(\mathrm{Ker}\left((T^2-\tau T+\delta)^d\right)\right) factors of (X^2-\tau X+\delta). We can thus define this half-dimension to be the multiplicity of the eigenpair. Like the multiplicity of an eigenvalue, it counts both the number of times the corresponding factor shows up in the characteristic polynomial of T, and the number of blocks on the diagonal of an almost upper-triangular matrix for T that have this characteristic polynomial.

April 8, 2009 Posted by | Algebra, Linear Algebra | Leave a Comment

   

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