I am currently proving that an implicit Runge-Kutta Scheme for a vector-valued function with $\nu \geq 1$ stages is unconditionally stable. I actually managed to reduce the problem to "Show that a particular matrix has full rank".
Consider a collection of diagonalizable matrices $A_{j}$ for $j \in \lbrace 1, \ldots , \nu \rbrace$. We know that $A_{j} \in \mathbb{C}^{3\times3}$. Consider another collection of invertible matrices $K_{j}$ for $j \in \lbrace 1, \ldots , \nu \rbrace$. We know that $K_{j} \in \mathbb{C}^{3\times3}$.
Now let $T \in \mathbb{C}^{3\nu \times3 \nu}$ be a invertible matrix.
Now define $A = ( A_{1} \ldots A_{\nu}) \in \mathbb{C}^{3 \times 3\nu}$ and $K = ( K_1 \ldots K_{\nu})^{T} \in \mathbb{C}^{3 \nu \times 3}$.
My goal is to show that $A T K \in \mathbb{C}^{3 \times 3}$ has full rank for all possible values of $\nu$. I have proceded as follow:
We have that Rank$(ATK) \leq 3$. Hence it is sufficient to show Rank($ATK$)$\geq 3$.
Using Frobenius Inequality I can state Rank($ATK$) $\geq Rank(AT)+Rank(TK) - Rank(T)$. Since T is invertible we have that $Rank(T) = 3 \nu$, hence:
Rank($ATK$) $\geq Rank(AT)+Rank(TK) - 3 \nu$.
Moreover since $T$ is invertible and its dimensions are compatible with $A$,$K$ then $Rank(AT)= Rank(A)$ and $Rank(TK) = Rank(K)$. We can then write
Rank($ATK$) $\geq Rank(A)+Rank(K) - 3 \nu$.
Now each $A_j$ is diagonalizable this means it is invertible so it must have rank equal to $3$. This means that $A$ must also have rank equal to $3$ since the columns of some $A_j$ spans the whole space. Same thing should happen with $K$. This means we have:
Rank($ATK$) $\geq 6 - 3 \nu = 3(2- \nu)$ which is $\geq3$ only for $\nu = 1$.
Does anyone have any idea how to fix this problem? Maybe I should use other inequalities?