Let $\mathcal{S}$ be a set of all linear maps from $\mathbf{L\,}(\mathbb{C^3})$ such that no two maps are similar and the following equality holds $$\mathbf{A^{10}-3A^9+2A^8=0}$$Find the maximum number of elements in $\mathcal{S}$.

Attempt at solution : Two matrices $\mathbf{A}$ and $\mathbf{B}$ are similar if there exists matrix $\mathbf{P}$ such that $\mathbf{B}=\mathbf{P^{-1}\mathbf{A}}\mathbf{P}\,$. Also if two matrices are similar they share minimal polynomial, characteristic polynomial, rank, geometric multiplicity... Furthermore we know that for polynomial $f(x)= x^{10}-3x^9+2x^8 \Rightarrow$ $f(\mathbf{A})=0\,$ so minimal polynomial $\mu{_x}$ divides $f(x)$. I am not sure how to proceed from here.

  • $\begingroup$ Do you know about Jordan canonical form? $\endgroup$ Jan 16 '17 at 18:35

Hint: Note that $\mu_x \mid f(x) = x^8(x-2)(x-1)$. Thus, if $A$ is a matrix satisfying the equation, then the eigenvalues of $A$ come from $\{0,1,2\}$. The largest possible Jordan block associated with $1$ or $2$ must have size at most $1$, while the largest Jordan block associated with $0$ has size at most $8$.

It suffices to count the similarity classes, which is to say that we may assume without loss of generality that the elements of $\mathcal S$ are in Jordan form

$\mathcal S$ can contain any one of diagonal matrices that satisfy the above constraints (i.e. the matrices with diagonal entries in $\{0,1,2\}$), of which there are $3^3 = 27$. $\mathcal S$ can contain the non-diagonalizable matrices with a size-$2$ Jordan block for $\lambda = 0$, of which there are $3$. Finally, $\mathcal S$ can contain the size-$3$ Jordan block for $\lambda = 0$.

All together, $\mathcal S$ contains at most $27 + 3 + 1 = 31$ elements.

  • $\begingroup$ So given that we are dealing with vector space $\mathbf{L\,}(\mathbb{C^3})$ it must be that $\mu_x = x(x-2)(x-1)\,$. Since no matrix in $\mathcal{S}\,$ is similar to $\mathbf{A}\,$, Jordan form is unique up to permutation of blocks. But also equality holds for null matrix and identity matrix, so maximum number of elements is 3? I not sure if my logic is correct here. $\endgroup$
    – Tino
    Jan 16 '17 at 19:16
  • $\begingroup$ Not quite. We could have $\mu_x = (x-2)$, for example $\endgroup$ Jan 16 '17 at 19:22
  • $\begingroup$ Also, I don't know what you mean when you say "no matrix in $\mathcal S$ is similar to $\mathbf A$" $\endgroup$ Jan 16 '17 at 19:24
  • $\begingroup$ $\mu_x \,$ can be any of the following $\mu_x = x(x-2)(x-1)\,$ , $\mu_x = x\,$ , $\mu_x = (x-1)\,$ , $\mu_x = (x-2)\,$ , $\mu_x = x(x-2)\,$ , $\mu_x = x(x-1)\,$ , $\mu_x = (x-2)(x-1)\,$ , $\mu_x = x^2\,$ , $\mu_x = x^2(x-2)\,$ $\mu_x = x^2(x-1)\,$ , $\mu_x = x^3\,$ it cannot be $\mu_x = x^4\,$ because $deg(\mu_x)\le\,dim(V)=3$. And e.g. if $\mu_x = x-2\,$ then we have matrix looking like this $$\begin{bmatrix} 2 & 0 & 0 \\ 0& 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}$$ multiplicity of root gives the dimension of the biggest block. Is that correct ? $\endgroup$
    – Tino
    Jan 16 '17 at 22:55
  • 1
    $\begingroup$ See my latest edit $\endgroup$ Jan 16 '17 at 23:27

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