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Given that $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent vectors in $\ker(A-\lambda I)^k$ but not in $\ker(A - \lambda I)^{k-1}$ for some $\lambda$ and $k \geq 2$, how can it be proven that any possible linear combination of these vectors will also not be in $\ker(A - \lambda I)^{k-1}$? Therefore, more generally, that $$\left(\ker(A-\lambda I)^k\setminus\ker(A-\lambda I)^{k-1}\right)\cup\{\mathbf{0}\}$$is a subspace?

This question emerged when I was trying to prove that any set of jordan chains is linearly independent (if the top vectors also were linearly independent), once I got stuck, I found some posts that went through an analogous approach I did but at a certain point they concluded something like $$(A-\lambda I)^{k-1}(a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + a_3\mathbf{v}_3) = 0 \implies a_1,a_2,a_3=0$$ which wasn't very obvious to me.

ultralegend5385
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excitedGoose
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    The set difference of two subspaces is never a subspace because it does not contain the zero vector. – lhf Sep 03 '24 at 15:57

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It's not closed under addition. Let $A$ be the matrix $$\begin{bmatrix} 0 & 1\\0 & 0 \end{bmatrix}$$ $\lambda=0$ is an eigenvalue, and $A^2=0$. Take $v_1=[1,3]^T$ and $v_2=[1,-3]^T$ Then $v_1,v_2 \in\ker(A^2),\notin \ker A$, but $v_1+v_2 \in\ker(A)$.

ultralegend5385
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Paul
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  • Then, the final conclusion on posts like this, in the part it says that "$a_i=0$ for each $i$ is required" is not necessarily true? – excitedGoose Sep 03 '24 at 17:38
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    @excitedGoose As best I can tell, either you are mixing up / and , or you're mixing up set complements with complementary subspaces. Needless to say, this answer in no way applies to that one. – Brian Moehring Sep 03 '24 at 22:35