I would like to solve the following problem.
Let $n \in \mathbb{N}$ such that $n \geq 2$. Let $\mathrm{M} \in \mathfrak{M}_{n}(\mathbb{C})$. Under what condition(s) do there exist two diagonalizable matrices $\mathrm{A}$ and $\mathrm{B}$, both with rank $1$, such that $\mathrm{M} = \mathrm{A} + \mathrm{B}$?
I have already done that.
Important Note. --- Let us start by noting that it is sufficient to do this for a matrix similar to matrix $\mathrm{M}$. Indeed, if there exist $\mathrm{P}\in\mathrm{GL}_{n}(\mathbb{C})$, and if there exist $\mathrm{A}'$ and $\mathrm{B}'$, both diagonalizable of rank $1$, such that $\mathrm{P}^{-1}\mathrm{M}\mathrm{P}=\mathrm{A}'+\mathrm{B}'$, then $\mathrm{M}=\mathrm{P}\mathrm{A}'\mathrm{P}^{-1}+\mathrm{P}\mathrm{B}'\mathrm{P}^{-1}$. Moreover, $\mathrm{P}\mathrm{A}'\mathrm{P}^{-1}$ and $\mathrm{P}\mathrm{B}'\mathrm{P}^{-1}$ are of rank $1$ (multiplication by an invertible matrix does not change rank) and diagonalizable (since they are similar to diagonalizable matrices).
This will allow us to work with a reduced matrix.
Necessary Conditions. --- Let $\mathrm{M}\in\mathfrak{M}_{n}(\mathbb{C})$. Suppose there exist two diagonalizable matrices $\mathrm{A}$ and $\mathrm{B}$ of rank $1$, such that $\mathrm{M}=\mathrm{A}+\mathrm{B}$. We have $\mathrm{im}(\mathrm{M})=\mathrm{im}(\mathrm{A}+\mathrm{B})\subset\mathrm{im}(\mathrm{A})+\mathrm{im}(\mathrm{B})$. The Grassmann formula then ensures that $\mathrm{M}$ has rank $\leq 2$.
Sufficient Condition. --- Suppose $\mathrm{rank}(\mathrm{M})\leq 2$. Let's consider different cases.
$\triangleright$ If $\mathrm{rank}(\mathrm{M})=0$, then $\mathrm{M}=0$ and we have $\mathrm{M}=\mathrm{E}_{1,1}-\mathrm{E}_{1,1}$. The matrices $\mathrm{E}_{1,1}$ and $-\mathrm{E}_{1,1}$ are clearly of rank $1$ and diagonalizable since they are projector matrices.
$\triangleright$ If $\mathrm{rank}(\mathrm{M})=1$, let $v$ be a generator of $\mathrm{im}(\mathrm{M})$. As $v\neq 0$, the singleton $(v)$ is linearly independent, so we can complete it to a basis $\mathcal{B}$ of $\mathbb{C}^n$. The matrix of the endomorphism canonically associated with $\mathrm{M}$ in $\mathcal{B}$ is
$$ \mathrm{M}' = \begin{bmatrix} a_{1}&a_{2}&\cdots&a_{n}\\ 0&0&\cdots&0\\ \vdots&\vdots&&\vdots\\ 0&0&\cdots&0\\ \end{bmatrix}. $$
According to the observation, it is sufficient to decompose $\mathrm{M}'$ to conclude.
$\bullet$ If $a_{1}\neq 0$, then the decomposition $\mathrm{M}'=\frac{1}{2},\mathrm{M'}+\frac{1}{2},\mathrm{M'}$ is suitable.
$\bullet$ If $a_{1}=0$, then the decomposition $$ \mathrm{M}' = \begin{bmatrix} 1&a_{2}&\cdots&a_{n}\\ 0&0&\cdots&0\\ \vdots&\vdots&&\vdots\\ 0&0&\cdots&0\\ \end{bmatrix} + \begin{bmatrix} -1&0&\cdots&0\\ \hphantom{-}0&0&\cdots&0\\ \hphantom{-}\vdots&\vdots&&\vdots\\ \hphantom{-}0&0&\cdots&0\\ \end{bmatrix} $$ works since both matrices in the sum are of rank and diagonalizable due to their non-zero traces.
$\triangleright$ If $\mathrm{rank}(\mathrm{M})=2$.
$\bullet$ If $\mathrm{M}|_{\mathrm{im}(\mathrm{M})}$ induces an automorphism of $\mathrm{im}(\mathrm{M})$, the matrix $\mathrm{M}$ is similar to a matrix of the form $$ \mathrm{M}'= \begin{bmatrix} \mathrm{M}_{1}&\mathrm{O}_{2,n-2}\\ \mathrm{O}_{n-2,2}&\mathrm{O}_{n-2,n-2} \end{bmatrix} $$ where $\mathrm{M}_{1}\in\mathrm{GL}_{2}(\mathbb{C})$ can be deduced from the form $$ \mathrm{M}_{1} = \begin{bmatrix} a&b\\ 0&c \end{bmatrix}\qquad \text{o\`u $ac\neq 0$}. $$ It is enough to decompose $\mathrm{M}_{1}$ to conclude. Now, the decomposition $$ \mathrm{M}_{1} = \underbrace{\begin{bmatrix} a&0\\ 0&0 \end{bmatrix}}_{\mathrm{U}} + \underbrace{\begin{bmatrix} 0&b\\ 0&c \end{bmatrix}}_{\mathrm{V}} $$ allows us to write $$ \mathrm{M}' = \begin{bmatrix} \mathrm{U}&\mathrm{O}_{2,n-2}\\ \mathrm{O}_{n-2,2}&\mathrm{O}_{n-2,n-2} \end{bmatrix} + \begin{bmatrix} \mathrm{V}&\mathrm{O}_{2,n-2}\\ \mathrm{O}_{n-2,2}&\mathrm{O}_{n-2,n-2} \end{bmatrix}. $$
Both matrices in the sum are of rank $1$ and diagonalizable due to their non-zero traces.
$\bullet$ If $\mathrm{im}(\mathrm{M})\cap\ker(\mathrm{M})\neq\{0\}$ I'm not making progress !!!!
I feel that the case where $\mathrm{M}^2=0$ is really problematic.
When $\mathrm{im}(\mathrm{M})\cap\ker(\mathrm{M})$ has dimension 1, decompositions are possible. For example, the matrix decomposition below works. $$ \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0&1/2&1/2\\ 0&1/2&1/2\\ 0&0&0 \end{bmatrix} + \begin{bmatrix} 0&\hphantom{-}1/2&-1/2\\ 0&-1/2&\hphantom{-}1/2\\ 0&\hphantom{-}0&\hphantom{-}0 \end{bmatrix} $$
Can you help me finish the proof?
Thank you very much.
I would use the fact that you can trigonalize $M$, thus consider $M$ is $\begin{pmatrix} D & B\\O & T\end{pmatrix}$ where $D=\begin{pmatrix} a_1 & a\\0& a_2\end{pmatrix}$ with $|a_1|\geqslant |a_2|$, $B$ is a $(n-2)\times 2$ matrix, $O$ is the $(n-2)\times 2$ matrix with all entries set to zero and $T$ is a $(n-2)\times (n-2)$ upper-triangular matrix with zeros on its diagonal. Then you retrieve your different cases depending on $a_1$ and/or $a_2$ null and the most difficult ones correspond to $a_2=0$ as I develop thereafter.
$\bullet$ if $a_1$ and $a_2\neq 0$ then as rank$(M)\leqslant 2$ each column of $\begin{pmatrix}B\\T\end{pmatrix}$ is a combination of $C_1=\begin{pmatrix}a_1\\0 \\ \vdots\\0\end{pmatrix}$ and $C_2=\begin{pmatrix}a\\a_2\\0 \\ \vdots\\0\end{pmatrix}$ hence all entries in $T$ are zeroes. Then you have $M=M_1+M_2$ where the first row of $M_1$ is that of $M$ and the other entries are zeros and the second row of $M_2$ is that of $M$ with its other entries being zeros, $M_1$ and $M_2$ are obviously rank 1 diagonalizable matrices.
$\bullet$ if $a_1\neq 0$ and $a_2= 0$ then if rank$(M)=1$ the dimension of Ker$(M)$ is $n-1$ hence $M$ is diagonalizable and similar to the diagonal matrice $M'$ with $a_1$ as its only diagonal non-zero entry, as you noticed we can then use $M'=1/2 M'+1/2 M'$ as a solution. If rank$(M)=2$ using that the characteristic polynomial of $M$ is $(X-a_1)X^{n-1}$ we have $\text{Ker}(M-a_1I)\oplus\text{Ker}(M^{n-1})=\mathbb C^n$ and using a basis adaptated to this decomposition we get $M$ similar to $$\left ( \begin{array}{cccc} a_1 & 0&\dots & 0\\ 0& & &\\ \vdots& &M' &\\ 0 & & & \\ \end{array}\right )$$ with $M'$ a $(n-1)\times (n-1)$ rank 1 nilpotent matrice. Then it is classical to prove that $M'$ is similar to $M''=\begin{pmatrix}0&0&\dots&0\\1&\vdots&\dots&\vdots\\ 0 &\vdots&\dots&\vdots\\\vdots&&&\vdots\\ 0& 0&\dots&0\end{pmatrix}$ and hence using your last idea of decomposition we get that $M$ is similar to $M''_1+M''_2$ with $M''_1=\begin{pmatrix}\frac{1}{2}a_1&-\frac{1}{2}a_1&0&\dots&\dots&0\\0&0&0&\dots&\dots&0\\-\frac{1}{2}&\frac{1}{2}&0&\dots&\dots&0\\0&0&0&\dots&\dots&0\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&&\dots&\dots&&0\end{pmatrix}$ and $M''_2=\begin{pmatrix}\frac{1}{2}a_1&\frac{1}{2}a_1&0&\dots&\dots&0\\0&0&0&\dots&\dots&0\\\frac{1}{2}&\frac{1}{2}&0&\dots&\dots&0\\0&0&0&\dots&\dots&0\\\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&&\dots&\dots&&0\end{pmatrix}$ both rank 1 diagonalizable matrices.
$\bullet$ if $a_1=a_2= 0$ $M$ is nilpotent. If rank$(M)=1$ we can conclude by using the previous case with $M$ instead of $M'$
Edit : that's wrong but we can decompose $\begin{pmatrix}0&0&\dots&0\\1&\vdots&\dots&\vdots\\ 0 &\vdots&\dots&\vdots\\\vdots&&&\vdots\\ 0& 0&\dots&0\end{pmatrix}$, similar to $M$, as $\begin{pmatrix}1&0&\dots&0\\1&\vdots&\dots&\vdots\\ 0 &\vdots&\dots&\vdots\\\vdots&&&\vdots\\ 0& 0&\dots&0\end{pmatrix}+\begin{pmatrix}-1&0&\dots&0\\0&\vdots&\dots&\vdots\\ 0 &\vdots&\dots&\vdots\\\vdots&&&\vdots\\ 0& 0&\dots&0\end{pmatrix}$, which is a sum of rank 1 diagonalizable matrices.
Note that in this case we have $\text{Im}(M)\subset\text{Ker}(M)$ but it can also happen in the remaining case which is $M$ nilpotent with rank 2. In that case $M^3$ must be null because otherwise $\text{rank}(M)>\text{rank}(M^2)>\text{rank}(M^3)$ would lead to the contradiction $\text{rank}(M)>2$. We may have $M^2\neq 0$ and hence letting any $X$ such that $M^2X\neq 0$ we get that $X$, $MX$ and $M^2X$ are linearly independant so they can be completed into a basis $\cal B$ of $\mathbb C^n$ by adding $n-3$ independant vectors from $\text{Ker}(M)$. In that basis $M$ is represented by the matrix $M'=\begin{pmatrix} 0&0&0&&\\1&0&0&&\\0&1&0&&\\0&0&0&&Z\\ \vdots&\vdots&\vdots&\\0&0&0&\end{pmatrix}$ where $Z$ is a block of zeros, then I think using your decomposition idea as above on the upper $3\times 3$ block works to finish the proof.
Finally we may also have $M^2=0$ i.e. $\text{Im}(M)\subset\text{Ker}(M)$. Let vectors $X_1$, $X_2$ such that $(MX_1,MX_2)$ is a basis of $\text{Im}(M)$, then complete this basis into a basis $(MX_1,MX_2,X_5,\dots )$ of $\text{Ker}(M)$ to get a basis ${\cal B}=(X_1,X_2,MX_1,MX_2,X_5,....)$ in which $M$ is represented by $M''=\begin{pmatrix} 0&0&\\0&0\\1&0\\0&1&&Z\\ \vdots&\vdots\\0&0&\end{pmatrix}$ where $Z$ is a block of zeros. This latter matrice cannot be decomposed by the same idea of yours we already used as a sum of two rank 1 diagonalizable matrices, but maybe it is possible to make it another way, at least the problem is easier to handle with this matrice.