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\noindent
{\bf Solutions to HW 10.}

\medskip
  
\begin{enumerate}
\item[1.]
What is the minimal polynomial of the
$3\times 3$ matrix of all $1$'s?
Does your answer suggest that the
matrix is diagonalizable or nondiagonalizable?
\end{enumerate}

\medskip

\color{blue}
      {\bf Solution.} (Assume that $3\neq 0$ in the scalar field.
      This happens, for example, when the scalar field is
      $\mathbb F = \mathbb Q, \mathbb R, \mathbb C$, or the $2$-element
      field $\mathbb F_2$.)
      
$\chi_A(\lambda)
%=\lambda^3-3\lambda^2
= \lambda^2(\lambda-3)$.
The minimal polynomial is a monic polynomial
that divides $\chi_A$ and has the same roots, so it is either
$\lambda^2(\lambda-3)$ or $\lambda(\lambda-3)$.
We can check that
\[A(A-3I)=
\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}
\begin{bmatrix}-2\hphantom{-}&1&1\\1&-2\hphantom{-}&1\\1&1&-2\hphantom{-}\end{bmatrix}
  =0,
  \]
so, since $A$ satisfies the lesser degree possibility $\lambda(\lambda-3)$,
the minimal polynomial
must be $\lambda(\lambda-3)$.

The fact that the minimal polynomial factors into distinct
linear factors over the scalar field
implies that $A$ is diagonalizable.
\bigskip

(Assume now that $3=0$ in the scalar field.
This happens, for example, when the scalar field is
$\mathbb F_3$.)
\medskip

Everything works the same way as above, but now
$\chi_A(\lambda)= \lambda^2(\lambda-3)=\lambda^3$,
and the minimal polynomial could be
$\lambda^3$,  $\lambda^2$, or $\lambda$.
We can check that $A$ is not a root of the polynomial $\lambda$,
since $A\neq 0$, but it is a root of $\lambda^2$,
since $A^2=3A=0$. It follows that 
the minimal polynomial of $A$ is $\lambda^2$. Since
the minimal polynomial does not factor into distinct
linear factors over this scalar field, $A$ is not diagonalizable.
\color{black}

\medskip

\begin{enumerate}
\item[2.]
How is the minimal polynomial of A
related to the minimal polynomial of $A-I$?
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}
For any matrix $A$ and any monic polynomial $p(x)$,
it is the case that $p(A)=0$ iff $q(A-I)=0$
for $q(x) = p(x+1)$. That is, $A$ satisfies a monic
polynomial $p(x)$ iff $A-I$ satisfies
the same-degree monic polynomial $p(x+1)$.
Similarly $A-I$ satisfies a monic polynomial $q(x)$
iff $A$ satisfies
the same-degree monic polynomial $q(x-1)$.
Since the minimal polynomial
of a matrix is the monic polynomial of least degree
that is satisfied by the matrix, we get that
$p(x)$ is the minimal polynomial of $A$ iff
$p(x+1)$ is the minimal polynomial of $A-I$.
\color{black}
  \medskip

\begin{enumerate}
\item[3.]
Show that if $A$
is diagonalizable, then $A-I$ is diagonalizable. 
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}
If $C^{-1}AC=\Lambda$ is diagonal,
then $C^{-1}(A-I)C=C^{-1}AC-C^{-1}IC=\Lambda-I$ is diagonal.
\color{black}

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