Rank of Operators in a Two-Dimensional Operator Subspace

Question

Let $L$ be a two-dimensional subspace in the space of all linear operators from $\mathbb{E}$ to $\mathbb{F}$ ($\mathbb{E}$ and $\mathbb{F}$ are linear spaces, possibly finite-dimensional). Let $A,B,C$ be three non-proportional (i. e. every two of them form a basis for $L$) linear operators from $L$. It is easy to prove that if each of $A,B,C$ has rank $1$ then every non-zero operator from $L$ has rank $1$.

I'm interested in a generalization of this result to the case of operators with higher rank. More precisely,

Question. Let each of $A,B,C$ has rank $k$. Is it true that every operator from $L$ has rank $\leq k$?

If it is necessary, one can consider existence of more than three operators with the required property.

I will be grateful for any help.

To make it clear I give a proof for the original statement (i. e. when $A,B,C$ are operators of rank $1$).

Proof. It is convenient to assume that $C=A+B$ (otherwise, $C=a A + b B$, since $A$ and $B$ form a basis, and we put $A' = a A$, $B'=b B$). Let $\operatorname{Ran} D$ denote the range (the image) of operator $D$. There are two possibilities

Case 1: $\operatorname{Ran}A=\operatorname{Ran}B$. In this case the statement is obvious since every $D \in L$ is $D=a A + b B$ for some $a,b$, and, therefore, $\operatorname{Ran}D=\operatorname{Ran}A=\operatorname{Ran}B$ if $D$ is non-zero, i. e. $a^2+b^2\not=0$.

Case 2: $\operatorname{Ran}A \cap \operatorname{Ran}B = 0$. There is a vector $v_{0} \in \mathbb{E}$ such that $Av_{0}=e_{A}$ and $Bv_{0}=e_{B}$ is a basis vectors for $\operatorname{Ran}A$ and $\operatorname{Ran}B$ respectively. Indeed, otherwise, for every $v \in \mathbb{E}$ we have $Av=0$ or $Bv=0$ and, therefore, $\operatorname{Ran}C \supset (\operatorname{Ran}A \cup \operatorname{Ran}B)$ and $\operatorname{Ran}C$ is at least two-dimensional that is impossible. Vector $e_{C}=Cv_{0}=e_{A}+e_{B}$ is a basis for $\operatorname{Ran}C$. Now for every $v \in \mathbb{E}$ we have $Av=\alpha e_{A}$, $Bv=\beta e_{B}$ and $\alpha e_{A}+\beta e_{B}= Cv = \gamma e_{C} = \gamma e_{A} + \gamma e_{B}$. So, since $e_{A}$ and $e_{B}$ are linearly independent, $\alpha=\beta=\gamma$. Consequently, for $D \in L$, $D=a A+ b B$, we have $$Dv=(aA+bB)v = a \gamma e_{A} + b \gamma e_{B} = \gamma (a e_{A}+b e_{B}).$$ If $D$ is non-zero, i. e. $a^2 + b^2 \not=0$, then $\operatorname{Ran}D$ is one-dimensional or, equivalently, $\operatorname{rank}D=1$.

Answer 1

Nope. Take $A = {\rm diag}(1, 1,0)$, $B = {\rm diag}(-1,0,1)$, and $C = {\rm diag}(0,1,1)$. Then $A + B = C$ so they are non-proportional, and each of them has rank $2$, but $2A + B = {\rm diag}(1, 2, 1)$ has rank $3$.

Answer 2

A generalization should be like this.

Proposition. Let $L$ be a two-dimensional subspace in the space of all linear operators from $\mathbb{E}$ to $\mathbb{F}$ ($\mathbb{E}$ and $\mathbb{F}$ are finite-dimensional linear spaces). Suppose there are $k+2$ pairwise non-proportional operators $A_{1},\ldots,A_{k+2}$ with rank $\leq k$; then every operator from $L$ has rank $\leq k$.

Proof. We start with an observation. Let $A, B$ be two $p \times q$-matrices ($p,q \geq k$) and consider the matrix $B-xA$, $x \in \mathbb{R}$. Since every minor of order $k$ is a polynomial of degree $\leq k$ there are only two possibilities

1. $\operatorname{rank}(B-xA) \leq k$ for every $x$;

2. $\operatorname{rank}(B-xA) \leq k$ only for at most $k$ numbers $x$.

Now suppose that there is an operator $B \in L$ with $\operatorname{rank}B \geq k+1$. Then $\operatorname{rank}(B-xA_{1}) \leq k$ only for at most $k$ numbers $x$. But since $A_{1}$ and $B$ forms a basis for $L$, for $i=2,\ldots,k+2$ $$A_{i} = \alpha_{i} A_{1} + \beta_{i} B = \beta_{i} \left(B+\frac{\alpha_{i}}{\beta_{i}}A_{1}\right)$$ and numbers $\frac{\alpha_{i}}{\beta_{i}}, i=2,\ldots,k+2$ are distinct due to non-proportionality of $A_{i}$'s. So, we have $k+1$ numbers $x$ such that $\operatorname{rank}(B-xA_{1}) \leq k$. This is a contradiction.