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If $U$ is set of all matrices that commute with matrix:

$A=\begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 1 \\ 3 & 0 & 4\end{bmatrix} $ prove that $U$ is subspace of $\mathbb{M}_{3X3}$ determine if it contains $span(I, A, A^2, ...)$ and then determine dimensions and one basis of given subspaces.

I know that square matrices commute with $I, A, A^2, ...$ but i don't know if there are more matrices that commute with this one, so i don't know how to determine elements of the set $U$, which is subspace of $\mathbb{M}_{3X3}$ as it's stated, but i don't know how to prove that since i don't know how to determine $U$ which means that i don't know if it contains given span either. Any ideas?

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  • $\begingroup$ Not all square matrices commute with $A$. Consider $S=\pmatrix{0&0&1\\0&0&0\\0&0&0}$ where $SA= \pmatrix{3&0&4\\0&0&0\\0&0&0}$ while $AS=\pmatrix{0&0&2\\0&0&0\\0&0&3}$ $\endgroup$ – Mark Fischler Jun 6 '16 at 14:53
  • $\begingroup$ @MarkFischler Ok,thank you, i'll keep that in mind, but that fact won't help me to solve this. Do you know something more that could help me solve this particular problem? $\endgroup$ – cdummie Jun 6 '16 at 15:00
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Let $B=(b_{ij})$ be an arbitrary $3\times 3$-matrix satisfying $AB=BA$. Then this matrix equation is equivalent to $9$ equations in the variables $b_{ij}$, which have a very easy solution: $$ B=\begin{pmatrix} b_1 & 0 & b_1-b_5 \cr 3(b_1 - b_5 - b_8) & b_5 & b_8 \cr 3(b_1 - b_5) & 0 & 3b_1 - 2b_5 \end{pmatrix}. $$ Here I rewrote the matrix with the $9$ entries as $b_1,\ldots ,b_9$. This is the general solution. Of course, powers of $A$ commute with $A$ by definition.

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  • $\begingroup$ Ok, i see, and it is relatively easy to prove that this is the subspace of $\mathbb{M}_{3X3}$ but how can i determine if it contains $span(I, A, A^2, ...)$? $\endgroup$ – cdummie Jun 6 '16 at 15:04
  • $\begingroup$ @cdummie Is $A^k\cdot A = A\cdot A^k$? Is then $A^k$ in $U$? (Remember that $A^k\cdot A = \underbrace{A\cdot A\cdots A}_{k~\text{times}}\cdot A$) $\endgroup$ – JMoravitz Jun 6 '16 at 15:05
  • $\begingroup$ You can also compute explicitly what $A^k$ is. Note that the Jordan form of $A$ is given by $\begin{pmatrix} 1 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 5\end{pmatrix}$. $\endgroup$ – Dietrich Burde Jun 6 '16 at 15:09
  • $\begingroup$ @JMoravitz Obviously $A^k$ is in $U$, but, span is set of all linear combinations of it's elements. How can i know that every linear combination of $I, A, A^2, ...$ is in $U$? $\endgroup$ – cdummie Jun 6 '16 at 15:13
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    $\begingroup$ @cdummie if each individually is in $U$, then any linear combination of them is too... Suppose you have two things you know are in $U$. If you have $B$ and $C$ which are in $U$, then $(bB+cC)A = bBA+cCA = bAB + cAC = A(bB)+A(cC)=A(bB+cC)$ $\endgroup$ – JMoravitz Jun 6 '16 at 15:16

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