Sums of cubes of two complex matrices are closed under matrix multiplication

Question

Show that the set of matrices $\{M^3+N^3:\ M,N\in M_n(\mathbb C)\}$ is closed under matrix multiplication.

Thanks for your help.

Answer 1

I think you may be able to produce a proof by fleshing out the following sketch:

1. Every diagonalizable matrix is a cube of a matrix.

2. Every matrix is a sum of two diagonalizable matrices.

3. It follows that every matrix is a sum of two cubes.

Fleshing this out:

Let $A$ be a diagonalizable matrix. That means there's a diagonal matrix $D$, and an invertible matrix $P$, such that $A=P^{-1}DP$. There's a diagonal matrix $E$ such that $E^3=D$; just let each entry of $E$ be a cube root of the corresponding entry of $D$. Let $B=P^{-1}EP$. Then $$B^3=P^{-1}EPP^{-1}EPP^{-1}EP=P^{-1}E^3P=P^{-1}DP=A$$ Thus, every diagonalizable matrix is a cube.

Now every matrix $C$ is similar to a matrix in Jordan form, that is, there is a matrix $J$ in Jordan form and an invertible matrix $P$ such that $C=P^{-1}JP$. A matrix in Jordan form is composed of Jordan blocks $J_i$, $i=1,\dots,m$, $$J=\pmatrix{J_1&0&\dots&0\cr0&J_2&\dots&0\cr\vdots&{}&\ddots&\vdots\cr0&0&\dots&J_m\cr}$$ where each block $J_i$ is of the form $$J_i=\pmatrix{a_i&1&0&\dots&0\cr0&a_i&1&\dots&0\cr\vdots&{}&\ddots&\ddots&\vdots\cr\vdots&{}&{}&a_i&1\cr0&\dots&\dots&0&a_i\cr}$$ Each $J_i$ can be written as a sum of two diagonlizable matrices, $J_i=R_i+S_i$, $$R_i=\pmatrix{1&1&0&\dots&0\cr0&2&1&\dots&0\cr\vdots&{}&\ddots&\ddots&\vdots\cr\vdots&{}&{}&m-1&1\cr0&\dots&\dots&0&m\cr}$$ and $$S_i=\pmatrix{a_i-1&0&0&\dots&0\cr0&a_i-2&0&\dots&0\cr\vdots&{}&\ddots&\ddots&\vdots\cr\vdots&{}&{}&a_i-(m-1)&0\cr0&\dots&\dots&0&a_i-m\cr}$$ $R_i$ is diagonalizable since it has distinct eigenvalues, and $S_i$ is diagonal. Then $J=R+S$, where $R$ and $S$ are composed of the blocks $R_i$ and $S_i$, respectively, and are diagonalizable, say $R=Q^{-1}FQ$ with $F$ diagonal ($S$ is already diagonal). So $$C=P^{-1}JP=P^{-1}(R+S)P=P^{-1}RP+P^{-1}SP=P^{-1}Q^{-1}FQP+P^{-1}SP=(QP)^{-1}F(QP)+P^{-1}SP$$ is a sum of two diagonal matrices.

So, every matrix is a sum of two diagonalizable matrices, and every diagonalizable matrix is a cube, hence, every matrix is a sum of two cubes, hence, the set of all sums of two cubes, being the same as the set of all matrices, is closed under multiplication.