The Problem. If $A$ is a complex $n \times n$ matrix and $b$ is a complex column vector and if $\text{rank }\begin{bmatrix} A – \lambda I & b \end{bmatrix} = n$ for every eigenvalue $\lambda$ of $A$, then $b$ is cyclic.
My Attempt. I know that $\text{rank}\begin{bmatrix} A – \lambda I \end{bmatrix} < n$ because $\text{nullity}(A – \lambda I)$ is nontrivial since $\lambda$ is an eigenvalue. I know that $\text{rank}\begin{bmatrix} A – \lambda I & b \end{bmatrix} = n$ implies that
- $b$ is not in $\text{range}(A – \lambda I)$,
- $\text{rank}\begin{bmatrix} A – \lambda I \end{bmatrix} = n – 1$
I know that point 2 implies that the geometric multiplicity of each eigenvalue is 1, so the minimal polynomial of $A$ equals the characteristic polynomial. I know that equality of these two polynomials occurs if and only if $A$ has a cyclic vector. What I don't know is: Why is $b$ specifically a cyclic vector?
My Research. I have posted about this already here where I have posted the problem from which this statement comes and where a kind person has provided a solution to the problem. But I still don't understand. I have reread chapters 6 and 7 of Hoffman and Kunze, which deal with eigendecomposition (chapter 6) and cyclic decomposition (chapter 7) and I have consulted the internet. So, please, can you provide some intuition about all of this? Specifically, how do I know that a cyclic vector not only exists, but that $b$ is that cyclic vector?
Best Answer
As you observed, ${\rm rank}(A-\lambda I)=n-1\Rightarrow {\rm ker}(A-\lambda I)=1,$ which implies that there is only one Jordan block for each eigenvalue and the characteristic polynomial agrees with the minimal polynomial.
(0) Basic setup: Let the characteristic polynomial be $$f(x)=\prod_{i=1}^r(x-\lambda_i)^{d_i},\sum d_i=n.$$ Let $W_i={\rm ker}(A-\lambda_i I)^{d_i}.$ Then $$V={\mathbb C}^n=\bigoplus_{i=1}^r W_i,\dim W_i=d_i.$$ Note that each $W_i$ is invariant under $A$.
(1) A cyclic vector exists: For each $i$, choose $b_i\in {\ker}(A-\lambda_i I)^{d_i}\setminus {\rm ker}(A-\lambda_i)^{d_i-1}$ (such $b_i$ exists, otherwise the minimal polynomial will be a proper factor of $f(x)$). Necessarily $b_i\notin {\rm Im}(A-\lambda_i I).$ Using definition, it is easy to show that $$b_i,(A-\lambda_i)b_i,\cdots,(A-\lambda_i I)^{d_i-1}$$ form a basis for $W_i,$ hence $W_i$ is spanned by $b_i,Ab_i,\cdots,A^{d_i-1}b_i.$ Now let $b=b_1+\cdots+b_r.$ Then one needs to show that $$<b,Ab,\cdots,A^{n-1}b>_{\mathbb C}=V.\qquad (1)$$ To see this, it suffices to show that $W_i\subset V$ for each $i$. Without loss of generality, one proves the case $i=1$. Note that $$\prod_{i\geq 2}(A-\lambda_i I)^{d_i}b=\prod_{i\geq 2}(A-\lambda_i I)^{d_i}b_1=[q(A)(A-\lambda_1 I)+\beta I]b_1,$$ where $q$ is some polynomial and $\beta=\prod_{i\geq 2}(\lambda_1-\lambda_i)^{d_i}\neq 0.$ Now it’s easy to check that $$b_1’:=[q(A)(A-\lambda_1 I)+\beta I]b_1\in {\rm ker}(A-\lambda_1 I)^{d_1}\setminus {\rm ker}(A-\lambda_1I)^{d_1-1}.$$ By exactly the same argument above for $b_i$, one sees that $$W_1=<b_1’,Ab_1’,\cdots,A^{d_1-1}b_1’>_{\mathbb C}\subset V.$$ Similarly, $W_i\subset V,i=2,\cdots,r,$ hence $(1)$ is proven and thus $b$ is a cyclic vector.
(2) The given $b$ in the question is a cyclic vector: To see this, write $b=b_1+\cdots+b_r$ in the decomposition $W_1\oplus\cdots\oplus W_r.$ One can show that each $$b_i\in {\rm ker}(A-\lambda_i I)^{d_i}\setminus{\rm ker}(A-\lambda_i I)^{d_i-1},$$ hence by (1), $b$ is a cyclic vector. Without loss of generality, one proves the case $i=1$, the other cases being similar. By the given assumptions $$B:=[A-\lambda_1I\qquad b_1+\cdots+b_r]$$ has rank $n$. Clearly $${\rm Im}(A-\lambda_1 I)\subset {\rm ker}(A-\lambda_1 I)^{d_1-1}\oplus W_2\oplus+\cdots\oplus W_r,\qquad (2)$$ since $$(A-\lambda_1I)^{d_1-1}((A-\lambda_1I)(v_1+\cdots+v_r))=0+\sum_{i\geq 2}(A-\lambda_1I)^{d_1}v_i$$ for any $v=v_1+\cdots+v_r$ in the decomposition $W_1\oplus\cdots\oplus W_r$ and $W_i$’s are invariant spaces. Now aiming for contradiction, assume that $b_1\in {\rm ker}(A-\lambda_1I)^{d_1-1}.$ Then $$b_1+\cdots+b_r\in {\rm ker}(A-\lambda_1I)^{d_1-1}\oplus W_2\oplus\cdots\oplus W_r.\qquad (3)$$ Combining $(2)$ and $(3)$, by dimension count, the rank of $B$ would be at most $n-1$. This is a contradiction. QED
Acknowledgement: I would like to thank Chris Sanders for pointing out problem in my earlier post which naively assumed that $A$ is diagonalizable.