Practicing for the GRE I found this question and I was wondering if anyone had any general tips to approach this type of questions or any literature I could review to approach them.

Which of the following sets are dense in the set of square n×nn \times n square matrices over C\mathbb{C}?

I) Invertible matrices

II) Unitary Matrices

III) Symmetric Matrices

IV) Diagonalizable Matrices.

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Just to encourage contemplation of such questions… these are useful issues to consider! 🙂

– paul garrett

Oct 20 at 23:24

Well, I) and IV) are equivalent (at least I think–unless being over C\mathbb{C} changes something about matrices that I’m not aware of).

– Jared

Oct 20 at 23:25

1

@Jared Consider the matrix (1101)\begin{pmatrix} 1 & 1\\ 0 &1\end{pmatrix}.

– Jacky Chong

Oct 20 at 23:29

1

Sorry @symplectomorphic, I will put better titles in the future.

– user110320

Oct 20 at 23:30

1

PS: I have taken all the official GRE practice tests and the real thing once (95th percentile), and based on that experience I don’t think you’d ever be asked a question that depends on the density of diagonalizable matrices. You must be taking some unofficial practice test.

– symplectomorphic

Oct 20 at 23:36

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2 Answers

2

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Symmetric matrices aren’t dense in Cn2\mathbb{C}^{n^2} since they form a non-trivial subspace (unless n=1n = 1). Unitary matrices aren’t dense in Cn2\mathbb{C}^{n^2} since they form a bounded set.

Diagonalizable matrices are indeed dense. To see this, note that any complex matrix is similar to an upper triangular matrix. Given A∈Cn2A \in \mathbb{C}^{n^2}, we can find an invertible PP and an upper triangular BB such that A=P−1BPA = P^{-1} B P and the diagonal elements of BB are precisely the eigenvalues of AA (counting algebraic multiplicity). By perturbing the diagonal elements of BB slightly (call the result B′B’), we can make all the eigenvalues of B′B’ distinct and so B′B’ will be diagonalizable and hence also P−1B′PP^{-1} B’ P which can be made arbitrary close to AA.

Invertible matrices are dense since if AA is a matrix, A+λIA + \lambda I will be invertible for all but finitely many λ∈C\lambda \in \mathbb{C} and so we can find an invertible matrix arbitrary close to AA.

For the last one, a proof I like better is that (−∞,0)∪(0,∞)(-\infty,0) \cup (0,\infty) is dense, and the invertible matrices are the preimage of this set under the determinant. Ironing out the details of this proof is a bit messy, though.

– Ian

Oct 20 at 23:36

Case 1: Invertible Matrices

For each A∈Mn×n(C)A \in M_{n\times n}(\mathbb{C}), then us consider the characteristic polynomial

pA(z)=det(zI−A)\begin{align}

p_A(z) = \det(zI-A)

\end{align}

then AA is not invertible if and only if zero is a root of of p_A(z)p_A(z). However, if we perturb AA by a diagonal matrix, i.e.

\begin{align}

B= A+\epsilon I

\end{align}\begin{align}

B= A+\epsilon I

\end{align}

then we see that BB is invertible since the characteristic polynomial of BB is p_B(z) = p_A(z-\epsilon)p_B(z) = p_A(z-\epsilon) no longer has zero as its root for any \epsilon \neq 0\epsilon \neq 0. Hence the class of invertible matrices are dense in M_n(\mathbb{C})M_n(\mathbb{C}). (In fact, the set of invertible matrices is actually open in M_n(\mathbb{C})M_n(\mathbb{C}). )

Case 2: Unitary Matrices

A quick way to see that this class not dense is to consider the following matrix

\begin{align}

A =

\begin{pmatrix}

2 & 0\\

0 & 2

\end{pmatrix}

\end{align}\begin{align}

A =

\begin{pmatrix}

2 & 0\\

0 & 2

\end{pmatrix}

\end{align}

which is not unitary. Hence we see that

\begin{align}

\operatorname{Tr}|A-U|\geq 1.

\end{align}\begin{align}

\operatorname{Tr}|A-U|\geq 1.

\end{align}

Since all norms are equivalent on finite dimensional vector spaces, we have that the class of unitary matrices can’t be dense in M_n(\mathbb{C})M_n(\mathbb{C}).

Case 3: Symmetric Matrices

A again a quick way to see that this class is not dense is to consider the matrix

\begin{align}

C =

\begin{pmatrix}

1 & 100\\

0 & 1

\end{pmatrix}

\end{align}\begin{align}

C =

\begin{pmatrix}

1 & 100\\

0 & 1

\end{pmatrix}

\end{align}

which is clearly not symmetric. Now, if for any symmetric matrix SS, we have

\begin{align}

C-S =

\begin{pmatrix}

1-a & 100-b\\

-b & 1-c

\end{pmatrix}.

\end{align}\begin{align}

C-S =

\begin{pmatrix}

1-a & 100-b\\

-b & 1-c

\end{pmatrix}.

\end{align}

If we take the Frobenius norm of C-SC-S, we get

\begin{align}

\|C-S\|_F = \sqrt{|1-a|^2+|1-c|^2+b^2+|100-b|^2}\geq \sqrt{b^2+|100-b|^2} \geq \frac{100}{\sqrt{2}}.

\end{align}\begin{align}

\|C-S\|_F = \sqrt{|1-a|^2+|1-c|^2+b^2+|100-b|^2}\geq \sqrt{b^2+|100-b|^2} \geq \frac{100}{\sqrt{2}}.

\end{align}

Hence the class of symmetric matrices is not dense in M_n(\mathbb{C})M_n(\mathbb{C}).

Case 4: Diagonalizable Matrices

Let AA be an arbitrary matrix in M_n(\mathbb{C})M_n(\mathbb{C}). Let us show that AA can be approximated by diagonal matrix.

Using the fact that a matrix is diagonalizable if it has nn distinct eigenvalues, we shall construct a BB matrix arbitrarily close to AA in norm and BB has distinct nn-eigenvalues.

Suppose p_A(z) =\det(zI-A) =(z-\lambda_1)(z-\lambda_2)\cdots (z-\lambda_n)p_A(z) =\det(zI-A) =(z-\lambda_1)(z-\lambda_2)\cdots (z-\lambda_n) is the characteristic polynomial of AA where |\lambda_1\leq |\lambda_2|\leq \ldots \leq |\lambda_n||\lambda_1\leq |\lambda_2|\leq \ldots \leq |\lambda_n|, then consider a diagonal matrix

\begin{align}

D=

\begin{pmatrix}

d_1 & 0 & \ldots & 0\\

0 & d_2 & \ldots & \vdots\\

\vdots & \ldots &\ddots & 0\\

0 & \ldots & 0 & d_n

\end{pmatrix}

\end{align}\begin{align}

D=

\begin{pmatrix}

d_1 & 0 & \ldots & 0\\

0 & d_2 & \ldots & \vdots\\

\vdots & \ldots &\ddots & 0\\

0 & \ldots & 0 & d_n

\end{pmatrix}

\end{align}

where the |d_1|<|d_2|<\ldots <|d_n||d_1|<|d_2|<\ldots <|d_n|. Then by considering the Jordan canocial form of AA, we have that
\begin{align}
A+\epsilon D
\end{align}\begin{align}
A+\epsilon D
\end{align}
has distinct eigenvalues, i.e. diagonaliable for any \epsilon\neq 0\epsilon\neq 0. Hence the set of diagonalizable matrices are dense in M_n(\mathbb{C})M_n(\mathbb{C}).