What is the importance of the multiplicity of an eigenvalue of a matrix A?

Question

Here is the question, given that $A$ is an $m \times m$ matrix, is the statements in problem (3) and (4) true or false? I just can't figure out why the multiplicity of eigenvalues matters here.

Thanks in advance.

Answer 1

In the repeated eigenvalue case, you might want to consider looking at Jordan normal form. For the $m$ repeated eigenvalues case, we can write the entire matrix as just one Jordan block, or as $J=\lambda I + N$ for a nilpotent matrix $N$ with ones on the superdiagonal.

If we raise this to the $k$th power and apply it to a random vector $v\in \mathbf{R}^m$ and normalize to get $v_k := J^kv/\|J^k v\|$, this is the same as iterating the method above $k$ times (assuming we never get to the division by zero part). So, to figure out what happens we need to see if $J^k/\|J^k v\|$ converges. This does converge, but it depends on $N$. If $N$ is the zero matrix, we'll just get back the original vector. From now on we will assume $N$ is nonzero as well as $\lambda$.

If it is not the zero matrix, for large powers a few select entries will dominate. The reason why this happens can be seen from the binomial theorem: $$(\lambda I+N)^k = \sum_i \binom{k}{i}\lambda^{k-i} N^{i}.$$ In general you can't use this, but $\lambda I$ commutes with everything so it is fine here. Many of these terms vanish since $N$ is nilpotent, and for sufficiently large $k$ the number of terms becomes constant. The largest term as $k\to \infty$ will always be the one associated to the largest value of $i$, which we can see by dividing out by $\lambda^k$. This is because $\binom{k}{i}$ will grow on the order of $k^i$ for fixed $i$. To find this term, we look for the largest nonzero power of $N$. The nonzero entries in this matrix will indicate the dominant values in $J^k$.

Because we normalize the vector, we can scale the matrix by anything and get the same result at the end. Thus, for large $k$ we can treat $J^k$ as $N^i$ where $i$ is the largest nonzero power of $N$ because these are precisely the dominant entries. Thus, we get $N^i v/\|N^i v\|$ as our result in the limit.

For example, take the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ that you gave. The matrix $N$ will be $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, which has $N^2=0$. As a result, we will converge to $Nv/\|Nv\|$ which is either $\vec{0}$ or $(\pm1,0)$.

One point that is worth making is that while this does converge, it is quite slow. Look at the following SAGE code:

m=matrix(RR,2,2,[[1,1],[0,1]])

print m
M=m^10000
print M

v=vector((-1,-23))

print M*v/((M*v).norm())

m2=matrix(RR,2,2,[[2,1],[0,1]])

print m2
M2=m2^10000
print M2

print M2*v/((M2*v).norm())

The first gives us (-0.999999995000043, -0.0000999995647192882), while the second gives (-1.00000000000000, -4.80352388465618e-3011). Big difference!

Answer 2

Well, this is only a start! We could think more about how to rigorously show this ...

For now, I've done a few empirical experiments. If $A$ has eigenvalue of multiplicity $m$, then $A$ is proportional to the identity matrix, and if the eigenvalue is positive, the sequence converges to a vector parallel to the vector you start with. If the eigenvalue is negative, the sequence will not converge (the sign will flip with each iteration).

Similarly, if any of the eigenvalues are negative in the multiple eigenvalue case, we can see the same behavior.

If there are all positive eigenvalues, looks (only after a few experiments) to converge.