Show the non-zero vectors $v_1,v_2,⋯,v_n$ are linearly independent if $(A−λI)^{j+1}v_j=0$, where $j=0,1,⋯,n$.

Question

My attempt at a solution uses the linear dependence lemma. Assume on the contrary that the list of vectors is linearly dependent. Then it is possible to find some $v_j$ that can be written as a linear combination of the remaining vectors: that is, $$v_j = a_1v_1 + a_2v_2 + \cdots + a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + \cdots + a_nv_n.$$ Applying the operator $(A - \lambda I)^{j+1}$ to both sides, we have $$ 0 = (A - \lambda I)^{j+1}(a_1v_1 + a_2v_2 + \cdots + a_{j-1}v_{j-1}) + (A - \lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + \cdots + a_nv_n).$$ The first sum is zero, since the vectors there require the operator $A - \lambda I$ applied at most $j$ times to be $0$. Hence, $$0 = (A - \lambda I)^{j+1}(a_{j-1}v_{j-1} + a_{j+1}v_{j+1} + \cdots + a_nv_n).$$ And this is where I'm stuck. How do I use this to show a contradiction and conclude they are linearly independent?

Answer 1

I use the assumption that $$ (A-\lambda I)^jv_j\ne 0\quad\text{and}\quad (A-\lambda I)^{j+1}v_j=0,$$ otherwise it is not true. For example $$ \lambda=0, \quad A=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right), \quad v_j=\binom{0}{1}, \quad j\in\mathbb N. $$ In order to prove that $v_1,\ldots,v_n$ are linearly independent, it suffices to show that, for all $k$, the vector $v_k$ can not be expressed as a linear combination of $v_1,\ldots,v_{k-1}$.

Clearly, for $k=1$, this holds, since $(A-\lambda)v_1\ne 0$, and hence $v_1\ne 0$.

Assume that $v_k=c_1v_1+\cdots+c_{k-1}v_{k-1}$. Then $$ 0\ne (A-\lambda I)^kv_k =(A-\lambda I)^k\big(c_1v_1+\cdots+c_{k-1}v_{k-1}\big)=0 $$