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I am able to prove that two matrices of rank 1 are similar if and only if they have the same trace.

But, my proof is long and complicated. In particular, I use the fact that the union of non-trivial subspaces can't be equal to the whole space. My proof goes like this

  1. I prove the equivalent result for endormorphims of $E$, let's call them $f$ and $g$.

  2. I write $f = a\cdot e$ and $g = b\cdot f$, with $a, b\in E^*$ and $e,f\in E$.

  3. We can assume $a, b \neq 0$

  4. Then, I write $E = k.e_a\oplus \ker(a)$ and $E = k.e_b\oplus \ker(b)$. With $a(e_a)=1$ and $b(e_b)=1$.

  5. Then, I decompose $e$ in $k.e_a\oplus \ker(a)$. It is $\lambda_a \cdot e_a + x_a$. And same for $b$. We can prove that $\text{tr}\ u = \lambda_a$.

  6. Etc, etc. I already feel at this point that the proof is too complicated. Then, I build the automorphim $\phi$ by doing $e_a \to e_b$ and $\ker a \to \ker b$. But, we need to have also $\phi(x_a) = x_b$.

  7. So, we need $x_a \neq 0$ and $x_b \neq 0$. We do this by taking $x_a$ and $x_b$ not being eigen vectors of $u$ and $v$. This where I use the result about the union...

It all seems too complicated. Am I missing something ?

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    $\begingroup$ Does equivalence of matrices mean $A = P \cdot B \cdot Q^{-1}$ to you or $A = P \cdot B \cdot P^{-1}$? $\endgroup$
    – j4GGy
    Commented Feb 16, 2016 at 12:06
  • $\begingroup$ I mean similarity, hence $PBP^{-1}$. $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 12:11

2 Answers 2

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The only possible Jordan Normal forms for a rank $1$ matrix are

$$J_1= \left(\begin{array}{c|ccc} c & 0 & \cdots & 0 \\ \hline 0 & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{array}\right) J_2= \left(\begin{array}{cc|ccc} 0 & 1 & 0&\cdots & 0 \\ 0 & 0 & 0 &\cdots &0\\ \hline 0 & 0 & 0 & \ldots &0\\ {\vdots} & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{array}\right)$$

This makes the claim almost trivial: if $A$, $B$ are two rank one matrices, then $$\text{tr}(A)=c_{A},\qquad \text{tr}(B)=c_{B} $$ where $c_A$, $c_B$ are either $0$ or the only non-zero eigenvalues of the two matrices. If $c_A=c_B$, the two matrices have the same normal form and hence are similar, otherwise they are not.


Without using the Jordan decomposition, one can notice that the kernel of a rank one operator $A$ has dimension $n-1$. Consider a basis of $\ker A=\langle v_2,\ldots, v_n \rangle$ and a linearly independent vector $v_1\not \in \ker A$. We can visualize the kernel of a matrix as the eigenspace of eigenvalue $0$. If there exists one non-zero eigenvalue, algebraic and geometric multeplicity of $0$ coincide. Let $c$ be $A$'s only non-zero eigenvalue. For dimensionality restrictions, we know that $c$'s eigenspace $V_c$ has dimension one. Since eigenspaces relative to different eigenvalues are in direct sum, $$V=\ker A \oplus V_c$$ implying that $A$'s matrix is exactly $J_1$. This shows that if $\text{tr}(A)=c_A,~\text{tr}(B)=c_B$, with $c_A,c_B\neq 0$ then $A$ is similar to $B$ if and only if they have the same trace.

Now, if $A$ has only $0$ as an eigenvalue, then $A$'s characteristic polynomial is $x^n$. This shows that if $A,B$ are two rank one matrices such that $\text{tr}(A)=0,~ \text{tr}(B) \neq 0$ they are not similar, because their characteristic polynomials are different.

If $\text{tr}(A)=\text{tr}(B)=0$, the matrices are similar to two of the following kind:

$$A'= \left(\begin{array}{c|ccc} 0 & 0 & \cdots & 0 \\ \hline a_1 & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ a_n & 0 & \cdots & 0 \end{array}\right) B'= \left(\begin{array}{c|ccc} 0 & 0 & \cdots & 0 \\ \hline b_1 & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ b_n & 0 & \cdots & 0 \end{array}\right) $$

I cannot think of an elementary method to show that they are similar at the moment, I'll edit the post if something comes to mind.

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  • $\begingroup$ I am looking for a proof that does not use Jordan reduction. $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 13:54
  • $\begingroup$ @Colas I'll edit my post accordingly. Jordan theory is not indispensable in this case, but provides a quick tool to reduce your claim to triviality. There is no actual use of generalized eigenvectors, but only of simple ones because the matrix is diagonalizable. $\endgroup$
    – Lonidard
    Commented Feb 16, 2016 at 13:59
  • $\begingroup$ The matrix is not always diagonalizable. But let me see your answer! $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 14:14
  • $\begingroup$ Your proof works if the matrix has a non-zero eigen value, but it is not always the case. Think of nilpotent matrix of rank 1. $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 14:15
  • $\begingroup$ In dimension 2, think of $\begin{pmatrix} 0 & 1 \\ 0&0\end{pmatrix}$ $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 14:16
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Note that two similar matrix have the same trace. Indeeed if$A$ and $B$ are similar then exist an invertible matrix $Q$ such that $$A=Q^{-1}BQ$$ and taking trace in both sides we obtain $$tr A=tr(Q^{-1}QB)=tr A=tr B$$ If a matrix has rank $1$ for the theorem of nullity+rank the $\ker$ associated to two matrix has dimension $n-1$. Therefore you can choose a new basis $(v_1,.....,v_{n-1},v_n)$ with the first $n-1$ vectors that belongs to $\ker$ of two matrix while $v_n$ belongs to $im f$. In this new basis the two matrix are similar to: $$P= \left(\begin{array}{c|ccc} c & 0 & \cdots & 0 \\ \hline * & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ * & 0 & \cdots & 0 \end{array}\right)$$ if the trace is equal then are similar.

Edit

Note that a matrix of rank $1$ and trace$\neq 0$ is diagonalizable and similar to $$J= \left(\begin{array}{c|ccc} c & 0 & \cdots & 0 \\ \hline 0 & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{array}\right)$$ Therefore if two matrix have the same trace are similar to same diagonal matrix and they are similar

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  • $\begingroup$ It is false that you can find a basis $v_i$ where the two matrices are or the given form. $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 13:53
  • $\begingroup$ No it's true... $\endgroup$ Commented Feb 16, 2016 at 13:55
  • $\begingroup$ Note that a matrix of rank 1 and trace$\neq 0$ is diagonalizable and similar $$J= \left(\begin{array}{c|ccc} c & 0 & \cdots & 0 \\ \hline 0 & 0 &\cdots &0\\ {\vdots} & \vdots& \ddots & \vdots\\ 0 & 0 & \cdots & 0 \end{array}\right)$$ $\endgroup$ Commented Feb 16, 2016 at 13:59
  • $\begingroup$ False : In dimension 2, think of $\begin{pmatrix} 0 & 1 \\ 0&0\end{pmatrix}$ $\endgroup$
    – Colas
    Commented Feb 16, 2016 at 14:17
  • $\begingroup$ But $tr =0$.... $\endgroup$ Commented Feb 16, 2016 at 14:19

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