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

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.

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?

• What metric are you using on matrices? The operator norm or something else? Sep 1, 2016 at 3:00
• The easiest way to show that a set $S$ isn't dense is to find an open set that contains no elements of $S$. Sep 1, 2016 at 3:04
• Its a finite dimensional vector space, so whichever norm is most convenient. I've been working with the operator norm and the 2 norm mostly, $|| \, \cdot \, ||_{op}$ or $d(A,B) = \sqrt{\sum_{i,j} |a_{ij} - b_{ij}|^2}.$ Sep 1, 2016 at 3:05
• @Omnomnomnom I assume that if one wants to go that route, using the determinant to define such an open set is the way to go? I can maybe see how I would do it for showing $SL(2,C)$ isn't dense, pick the open set of matrices with determinant within (2,3), then clearly no element of $SL(2)$ is in this. But I'm struggling to see how I would construct such an open set for this Jordan block problem. Kind of new to this type of linear algebra/analysis. Sep 1, 2016 at 3:14

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.

• Though that sounds plausible, but how does one show that the discriminant is continuous?
– user99914
Sep 1, 2016 at 3:29
• The discriminant of a polynomial is a polynomial in its coefficients, this could probably be safely assumed. Very slick, thanks for the insight! Sep 1, 2016 at 3:31

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.

• So the argument is that the sequence of Jordan blocks will always have its eigenvalue with algebraic multiplicity 3, and therefore can't converge to a matrix with distinct eigenvalues? Sep 1, 2016 at 3:09
• @Merkh Yes. ${}{}{}$
– user99914
Sep 1, 2016 at 3:10

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.

• Interesting, very concrete way to do this.. I like it, thanks! Sep 1, 2016 at 20:22

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.

• When you say $f(A) = 0$ in the line "if A has one Jordan block...", you mean F(A)? I am a bit confused about what is happening here. You define $F(B)$ = det$(B-\lambda I) - (\lambda - 1)^3.$ If we plug in $A,$ which has 1 Jordan block, then $F(A) = - (\lambda - 1)^3?$ Also, the exam tells us what "dense" means in terms of any matrix being the limit point of the set in question. They probably were looking for something less topological but I don't see any better ways. Sep 1, 2016 at 12:05
• Fixed a typo in the definition. My first answer included an example, hence why the 1 instead of tr(A)/3. The point is that the pre image of zero is a proper closed subset containing the set of matrices with one Jordan block. A proper closed set can't contain a dense set. Put another way you seem to be comfortable, F is continuous. If the set in question was dense, then F would be zero for every matrix. But it's certainly nonzero on any matrix with distinct eigenvalues. Sep 1, 2016 at 18:01