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The question is :

Let $X$ be a non-null vector.Then there exists an eigenvector $Y$ of $A$ belonging to the span of $\{X, AX, A^{2} X, ... \}$.

I have tried to the best of my ability to solve it. But I don't find any right way to proceed.Please help me.

I want to add a solution of my own question which I find just now in a pdf. Here's this :

Let $k$ be the least positive integer such that $X, AX, A^{2} X, ... , A^{k} X$ are linearly dependent. Now let us consider a relation $\sum_{i=0}^{k} c_{i} A^{i} X = 0$. Then we must have $c_k \neq 0$. Now let us consider a polynomial $g(t) = \sum_{i=0}^{k} c_{i} t^{i}$. Let $\beta_{1}, \beta_{2}, ... , \beta_{k}$ be the roots of the polynomial $g(t)$. Then $g(t) = c_{k} \prod_{i=1}^{k} (t - \beta_{i})$.

Hence, $\sum_{i=0}^{k} c_{i} A^{i} = g(A) = c_{k} \prod_{i=1}^{k} (A - \beta_{i} I)$.Taking $Y = (\prod_{i=2}^{k} (A - \beta_{i} I)) X$ , it is easy to see that $Y \neq 0$ by the minimality of $k$ and $(A - \beta_{1} I) Y = 0$. Hence the result follows.

But at last it is not clear to me why is $Y$ in the span of $\{X, A X, A^{2} X, ... \}$? Please anyone suggest me what is the trick behind it.

Thank you in advance.

Jonathan Hahn
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Arnab Chattopadhyay.
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2 Answers2

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Why $Y\ne 0$: Let $h(t)=\prod_{i=2}^k(t-\beta_i).$ Then $\deg (h)<k$ and $h(t)$ is not identically $0$ so by the minimality of $k$ we have $0\ne h(A)(X).$ But $h(A)(X)=Y.$

DanielWainfleet
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  • Ok.I have understood.But how can I show that $Y$ is in the span of ${X, AX, A^{2} X, ... }$? – Arnab Chattopadhyay. Dec 15 '16 at 18:19
  • The linear span of ${X,AX,...}$ is every vector of the form $\sum_{j=0}^na_jA^jX,$ which is every $p(A)(X)$ over all polynomials $p$ .And $h$ is a polynomial and $h(A)(X)=Y$..... BTW your proof should include a separate part for $k=1.$ If $k=1$ then $ c_1AX+c_0X=0$ with $c_1\ne 0,$ so $X$ is an eigenvector of $A.$ – DanielWainfleet Dec 15 '16 at 22:55
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What you have looks correct to me. To answer your question, $Y=(\prod_{i=2}^k(A−\beta_i I))X$ is in the span of $\{X, AX, A^2X, ...\}$ because $\prod_{i=2}^k(A−\beta_i I)$ is a polynomial in A: $$\prod_{i=2}^k(A−\beta_i I) =A^{k-1} + d_1 A^{k-1} + ... d_{k-2} A + d_{k-1} I$$ for some constants $d_1,...,d_{k-1}$.

For example, for $k=3$, $(A - \beta_2)(A-\beta_3) = A^2 - (\beta_2 + \beta_3)A + \beta_2\beta_3 I$. Then $(\prod_{i=2}^3(A−\beta_i I))X = A^2X - (\beta_2 + \beta_3)AX + \beta_2\beta_3 X$, which is a vector in the span of $\{X, AX, A^2X\}$.

Jonathan Hahn
  • 354