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Without using the Jordan forms, What are some interesting properties of $A - 2I$ when the matrix $A= \begin{pmatrix} 1&*&* \\ 0&2&* \\ 0&0&3\\ \end{pmatrix}$.

Attempt:

$A - 2I = \begin{pmatrix} -1&*&* \\ 0&0&* \\ 0&0&1\\ \end{pmatrix}$.

I am searching for answers to questions like :

$1$. When can we say for sure ( or for what matrix $A$) that $(A - 2I)^k = 0$ and for what values of $k$.

$2$. What can we say with surety about the dimension of $(A - 2I)^k$

I know the following results :

for some exponent $m$ :

\begin{multline} \text{null} (T - λ I )^0 \subset \text{null} (T - λ I )^1 \subset \text{null} (T - λ I )^2 \subset \cdots \subset \text{null} (T - λ I )^m \\ \phantom{(2)} \qquad =\text{null} (T - λ I )^{m+1} = \cdots = \text{null}(T - λ I )^ { \dim V }. \qquad (2) \end{multline}

Hence,

\begin{multline}\dim \text{null} (T - λ I )^0 = 0 < \dim \text{null} (T - λ I )^1 < \dim \text{null} (T - λ I )^2 < \cdots < \dim \text{null} (T - λ I )^m \\ \phantom{(3)} \qquad = \dim \text{null} (T - λ I )^{m+1} = \cdots = \dim \text{null} (T - λ I )^ { \dim V } \qquad (3)\end{multline}

But, I am not able to apply these results to the posed questions.

Thank you very much for your help in this regard.

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  • $\begingroup$ Have you tried computing $(A-2I)^k$ for small values of $k$? It will never be zero... What do you mean by "the dimension of $(A-2I)^k$? Its rank? $\endgroup$ – Najib Idrissi Apr 14 '15 at 8:32
  • $\begingroup$ yes, I did try computing it's higher powers. It was never $0$. So, I wanted to ask when is such an upper triangular matrix equal to $0$. And by dimension, yes, I mean rank. Thanks $\endgroup$ – MathMan Apr 14 '15 at 8:33
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We have

$$A= \begin{pmatrix} d_1 & * & \cdots & * \\ & d_2 & \cdots & * \\ & & \ddots & \vdots \\ & & & d_n \end{pmatrix} \implies A^k= \begin{pmatrix} d_1^k & * & \cdots & * \\ & d_2^k & \cdots & * \\ & & \ddots & \vdots \\ & & & d_n^k \end{pmatrix},$$

so in your case $(A - 2I)^k$ will never be zero.

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  • $\begingroup$ Thanks. What kind of a matrix should $A$ be for $(A - \lambda I)^k = 0?$ thanks.again $\endgroup$ – MathMan Apr 14 '15 at 8:42
  • $\begingroup$ $A-\lambda I$ should be nilpotent. $\endgroup$ – user207868 Apr 14 '15 at 8:45
  • $\begingroup$ Yes, but how do I identify a nilpotent matrix in a general case? $\endgroup$ – MathMan Apr 14 '15 at 8:46
  • $\begingroup$ Any triangular matrix with 0s along the main diagonal is nilpotent. $\endgroup$ – user207868 Apr 14 '15 at 8:46
  • $\begingroup$ The only eigenvalue for $A-\lambda I$ is 0 is another sufficient condition. $\endgroup$ – user207868 Apr 14 '15 at 8:47
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$A-2I$ has two nonzero eigenvalues ($-1$ and $1$); therefore its powers are never zero.

However, you may try to prove that $A-2I=(A-2I)^3$ and hence $A-2I=(A-2I)^{2k+1}$ for every natural number $k$.

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  • $\begingroup$ In the general case, how can I investigate about the rank of each higher power of $A-\lambda I$? $\endgroup$ – MathMan Apr 14 '15 at 8:50

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