# A question related to similarity of a Complex matrix that is not scalar multiple of $I_n$

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.

• 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. Sep 9 at 8:00
• 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. Sep 9 at 8:15
• the minimal polynomial is degree at least 2 and $A$ is similar to its Rational Canonical Form Sep 9 at 19:35

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).
• 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? Sep 18 at 11:35
• @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. Sep 21 at 22:56