Similar matrices of rank 1

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 ?

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

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.

• I am looking for a proof that does not use Jordan reduction. Commented Feb 16, 2016 at 13:54
• @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. Commented Feb 16, 2016 at 13:59
• The matrix is not always diagonalizable. But let me see your answer! Commented Feb 16, 2016 at 14:14
• 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. Commented Feb 16, 2016 at 14:15
• In dimension 2, think of $\begin{pmatrix} 0 & 1 \\ 0&0\end{pmatrix}$ Commented Feb 16, 2016 at 14:16

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

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