Prove the set of matrices with one Jordan block is not dense in $M_n(\mathbb{C}).$

Question

I would like to show that the set of matrices with one Jordan block is not dense in $M_n(\mathbb{C}),$ the set of all $3$ x $3$ matrices with complex entries.

I have done proofs showing that invertible matrices are dense, and diagonal matrices are dense, but I've been struggling with proofs showing that a given subset is not dense. Another one I've had trouble with was showing the $SL(2,\mathbb{C})$ is not dense too.

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The only reasonable approach seems to begin with assuming that the subset is dense, and reaching a contradiction. Intuitively, one should be able to state that $I$ is not the limit point of a sequence of Jordan blocks, but I have had trouble stating this rigorously. Any tips/suggestions/tricks?

Answer 1

You can prove both fairly elegantly with the following observation:

A continuous function which is constant on a dense set is constant on the entire space.

To show SL$(2, \mathbb{C})$ isn't dense, what function could you use to get a contradiction? (Hint: You use it in the definition of the special linear group.)

Matrices with only one Jordan block is maybe a little harder, but the function $M_3(\mathbb{C}) \to \mathbb{C}$ taking a matrix to the discriminant of its characteristic polynomial works.

Answer 2

Assume that $A , A_k \in M_n(\mathbb C)$, $A_k \to A$ (that is, $(A_k)_{ij} \to A_{ij}$ for all $i, j$) and $A_k$ has one Jordan block $J_{\lambda_k}$ (with eigenvalue $\lambda_k$) Then $p(\lambda) := \det (A-\lambda I)$ satisfies

$$p(\lambda)=\lim_{k\to \infty} p_k(\lambda).$$

But $$\det (A_k - \lambda I) = \det (J_{\lambda_k} - \lambda I) = (\lambda_k -\lambda)^n$$

This implies that $p(\lambda)$ also has $n$-identical eigenvalue. Thus some elements in $M_n(\mathbb C)$, for example those with distinct eigenvalue, does not lie in the closure of the set of matrix with one Jordan block.

Answer 3

It seems that a more strong version holds: the set of matrices with one eigenvalue is not dense in $M_n(\mathbb C)$.

Proof.

We show that the matrices in the set $\{\mathbf A\in\mathbb C^{n\times n}: trace(\mathbf A)=0,\ \det(\mathbf A)\ne 0\}$ are not limits of matrices $\mathbf B_k$ whose spectrum consists of one point $\lambda_k$.

Indeed, if $\|\mathbf A-\mathbf B_k\|\rightarrow 0$, then $n\lambda_k=trace(\mathbf B_k)\rightarrow trace(\mathbf A)=0$, i.e., $\lambda_k\rightarrow 0$. On the other hand, we should also have $\lambda_k^n=\det(\mathbf B_k)\rightarrow \det\mathbf A\ne 0$, a contradiction.

Answer 4

The answers here are, of course, correct. If you're taking the UCLA basic exam, I'd refrain from asserting that the roots of a polynomial are a continuous function of its coefficients. This fact, which would probably require you to write down the discriminant, would require some proof.

It's possible to do this only making the one assumption about the determinant. Again, if you're taking the basic exam, you can use freely that the map from $M_n(\mathbb{C})$ to $P_{\mathbb{C}}^n[\lambda]$, the set of polynomials of degree at most $n$, given by $A\mapsto \det(A-\lambda I)$ is continuous, since the coefficients are a polynomial in the entries.

Now define $F:M_n(\mathbb{C})\to P_{\mathbb{C}}^n[\lambda]$ by $$F(A)=\det(\lambda I-A)-(\lambda-\text{tr}(A)/3)^3.$$ If $A$ has one Jordan block, then $F( A)=0$. So $$\{A: A\text{ has one Jordan block}\}\subseteq F^{-1}(\{0\})\subsetneq M_n(\mathbb{C}).$$ Since $F^{-1}(\{0\})$ is closed, being the presage of a closed set under a continuous function, we're done.

Edit: Typo in definition of F, removed flawed summative ending.