Let be $T:\mathbb{R}^7\rightarrow \mathbb{R}^7$

Such that $(T-15I)^3=0$ and $\dim\text{Im}(T-15I)^2=2$ find the Jordan normal form of $T$

If $(T-15I)^3=0$ so the minimal polynomial can be $(T-15I)^3=0$,$(T-15I)^2$,$(T-15I)$ But because $\dim\text{Im}(T-15I)^2=2$ So it can not be $(T-15I)^2$ and $(T-15I)$ it had to be $\dim\text{Im}(T-15I)^2=7$

The Characteristic polynomial is $(T-15I)^7$.

So we have $B_1$ of size $3$ and we are left with 7-3=4 eigenvalues, how can I find the reminding jordan normal form blocks?

  • $\begingroup$ If you denote your other blocks by $B_i$ , what happens when you compute $(T-15I)^2$? What blocks allows you to have $\dim \text{Im}(T-15I)^2=2$? $\endgroup$ – Thibaut Dumont Sep 2 '15 at 12:18
  • $\begingroup$ @ThibautDumont because $dimIm(T−15I)^2=2$ it surly is not the minimal polynomial therefore no blocks can have $dimIm(T−15I)^2=2$? $\endgroup$ – gbox Sep 2 '15 at 14:27
  • $\begingroup$ The dimension of the image of $(B_i-15I)^2$ is $1$, where $B_i$ is a block 3x3. So you need two of these blocks to have $\dim \text{Im}(T-15I)^2=2$. $\endgroup$ – Thibaut Dumont Sep 3 '15 at 14:29

The largest Jordan block size is $3$ because $(T-15I)^{3}=0$, and there have to be two of those because the rank of $(T-15I)^{2}$ is $2$. Because the dimension of the space is $7$, then there is also one Jordan block of size $1$.

  • $\begingroup$ We there must be two Jordan block of size 3? $\endgroup$ – gbox Sep 3 '15 at 19:18
  • 1
    $\begingroup$ @gbox : Because there are two linearly independent vectors $x_1$, $x_2$ such that $x_1 = (T-15I)^{2}x_1'$ and $x_2=(T-15I)^{2}x_2'$. So $x_1'$ and $x_2'$ are linearly independent with $(T-15I)^{2}x_1' \ne 0$, $(T-15I)^{2}x_2'\ne 0$ and, of course, $(T-15I)^{3}x_1'=0$, $(T-15I)^{3}x_2'=0$. That gives you two independent subspaces with cyclic vectors of order $3$, which means two Jordan blocks of size 3 with diagonal $15$. That's what $\left[\begin{array}{ccc}15 & 1 & 0 \\ 0 & 15 & 1 \\ 0 & 0 & 15\end{array}\right]$ describes. $\endgroup$ – DisintegratingByParts Sep 3 '15 at 19:23
  • $\begingroup$ I got the points about the two linearly independent vectors but I did not understand why I must have two Jordan block of size 3, I know why I must have one. $\endgroup$ – gbox Sep 3 '15 at 19:37
  • $\begingroup$ @gbox : Ah! Good, so the explanation helped. $\endgroup$ – DisintegratingByParts Sep 3 '15 at 19:41
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    $\begingroup$ @gbox : There is exactly one eigenvector for each Jordan block. $v_n \mapsto v_{n-1}=(T-\lambda I)v_n\mapsto \cdots \mapsto v_1 = (T-\lambda I)v_2 \mapsto 0 = (T-\lambda I)v_1$. And, the set of vectors in one cycle is independent of the set of vectors in another cycle. $\endgroup$ – DisintegratingByParts Sep 3 '15 at 20:26

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