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This question was asked in a masters exam for which I am preparing and I was unable to solve it.

Let $A$ be an $n\times n$ complex matrix that is not the scalar multiple of $I_n$. Then show that $A$ is similar to a matrix $B$ such that $B_{1,1}$( ie the top left entry of $B$) is $0$.

Well, I don't even have a intuition for this question's solution: I think in the case when $A$ is not a scalar multiple of $I_n$ but $\operatorname{rank} A = n$ then I don't think $B_{1,1}$ will be $0$.

I have studied theory from Hoffman and Kunze but I was unable to solve exercises due to my illness and I need help.

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    $\begingroup$ Just try to use a similarity transformation which changes this entry. Start with $n=2$ for an explicit calculation (to have some concrete understanding) and then generalize. $\endgroup$ Sep 9 at 8:00
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    $\begingroup$ The case of $2\times2$-matrices is a very good start. You can use the matrix $\begin{pmatrix}1&x\\0&1\end{pmatrix}$ as a similarity matrix in the first step. $\endgroup$
    – kabenyuk
    Sep 9 at 8:15
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    $\begingroup$ the minimal polynomial is degree at least 2 and $A$ is similar to its Rational Canonical Form $\endgroup$ Sep 9 at 19:35
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There are many ways to solve this question. Assume $n \geq 2$ (otherwise there is nothing to prove) and let $T = T_A \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ be the associated linear map given by $T_A(x) = Ax$. Assume that we can find $0 \neq v \in \mathbb{C}^n$ that is not an eigenvector of $T$. This means that $v \neq 0$ and $T(v)$ is not a scalar multiple of $v$ so $\{ v, T(v) \}$ is linearly independent. Complete $\{ v, T(v) \}$ to a ordered basis $\mathcal{B} = \left( v, T(v), v_3, \dots, v_n \right)$ of $\mathbb{C}^n$. Then the matrix representing $T$ with respect to $\mathcal{B}$ is similar to $A$ and has $$ \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} $$ as first column.

This leaves you with proving that if $A$ is not a scalar multiple of the identity then one can find at least one non-zero vector which is not an eigenvector of $A$. I'll leave this as an exercise (whose solution you can find on this website).

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  • $\begingroup$ I have some questions: Why "Then the matrix representing T with respect to B is similar to A" holds? and why it has$ \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$ as first column ? Please explain? $\endgroup$
    – James
    Sep 18 at 11:35
  • $\begingroup$ Can you please reply to my above comment? $\endgroup$
    – James
    Sep 21 at 15:33
  • $\begingroup$ @James: Whenever you represent a linear map with respect to two different bases, you get similar matrices. In this case, $T$ is represented by $A$ with respect to the standard basis. Regarding the second question, this follows immediately from the definition of the representing matrix... You have $T(v) = 0 \cdot v + 1 \cdot T(v) + 0 \cdot v_3 + \dots + 0 \cdot v_n$ which means that the coefficients, rearranged as a column, form the first column of the representing matrix. $\endgroup$
    – levap
    Sep 21 at 22:56

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