$A$ is an $n$ by $n$ matrix over $\mathbb C$ such that $Rank(A)=1$ and $Tr(A)=0$, prove that $A^2=0$

Question

This is what I did: Because $\mbox{rank}(A)=1$, then from rank nullity theorem $$\dim\ker A + \mbox{rank} A = n \implies \dim \ker A = n-1$$

and $gm(0)=dimE(0)=dimKerA=n-1$ where $E(0)$ is the eigenspace of the eigenvalue $0$.

So the characteristic polynomial of $A$ is: $C_A(x)=x^{n-1}(x-\lambda)$ where $\lambda$ could be $0$.

It is known that the sum of all eigenvalues of A equals the trace of A (Shown this using the fact that all matrices under $\mathbb C$ are triangularizable), so from that we'll conclude that $\lambda=0$ and $C_A(x)=x^n$ From here the minimal polynomial of A is $M_A(x)=x^k$ where $k\leq n$

From Cayley–Hamilton $M_A(A)=0 \implies A^k=0$

I'm not sure how to progress from here, I don't really know what I can say about the index of a nilpotent matrix when I know its rank. And I don't think I can say anything about the minimal polynomial given that I know the geometric multiplicity of the only eigenvalue

Answer 1

As noted in @Lost in Space answer, we can write $A = zw^T$, as a result of $\operatorname{rank}(A) = 1$. Now $0 = \operatorname{Tr}(A) = \operatorname{Tr}(zw^T) = \operatorname{Tr}(w^Tz) = w^Tz$ implies $w^Tz = 0$.

Therefore, $A^2 = zw^Tzw^T = z0w^T = 0$.

Answer 2

Let $A$ be a rank $1$ matrix . Then $A=zw^T$

We know that $0$ is an eigenvalue of multiplicity at least $n-1$ i.e it can have atmost one non zero eigenvalue of multiplicity $1$.

Since $\text{Trace}(A) =0$ , $0$ is the only eigenvalue of $A$ i.e $0$ has full multiplicity.

Since $A$ is defined over Algebraic closed field, $0$ is an only eigenvalue iff $A$ is nilpotent.

Claim: The degree of nilpotency of $A$ is $2$.

Suppose $A^n=0$ but $A^{n-1}\neq 0$ . Then $\exists x(\neq 0) $ such that $A^{n-1}x\neq 0$

Then $\{Ax,A^2x,\ldots ,A^{n-1}x\}$ is L.I (check! or see here)

Hence $\text{rank}(A) \ge n-1$ and $A(A^{n-1}x) =0$ implies $\text{rank}(A) = n-1$

Given $\text{rank}(A)=1$ , hence $n=2$.

Thus the degree of nilpotency of $A$ is $2$.

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Note : Matrix over complex field is very important to conclude $\text{spec}(A) =\{0\} $ implies $A$ is nilpotent.

For an example choose $p(x) =x(x^2+1) \in \Bbb{R}[x]$ monic polynomial and the companion matrix $C(p) $ ,

$$C(p)=\begin{pmatrix} 0&0&0\\1&0&-1\\0&1&0\end{pmatrix}\in M_3(\Bbb{R})$$

Then $\text{spec}(A) =\{0\} $ but $A$ is not nilpotent.

Answer 3

By the dimension formula, the kernel of $A$ has dimension $n-1$, say $\ker A$ has basis $\{v_1, \cdots, v_{n-1}\}$. Extend the basis to be $\{v_1, \cdots, v_{n-1}, v\}$, a basis of $\mathbb C^n$. Now the image of $A$ is spanned by $Av$, and hence $A(Av)=\lambda Av$. Since $Av\not=0$, $\lambda$ must be an eigenvalue of $A$, and with your argument, $\lambda = 0$.

Also it's clear that $A^2v_i=0$, hence $A^2$ vanishes over a base, hence must be $0$.

This doesn't depend on $A$ being over $\mathbb C$ at all. In fact, it works for any field, and we don't even need eigenvalue theory to finish the argument.

Say $Av=a_1v_1+\cdots, a_nv_n + av$. Since the trace of $A$ is $0$, we must have $a=0$, hence $Av=\sum_{i=1}^{n-1}a_iv_i\in \ker(A)$, $A(Av)=0$.

Answer 4

Another solution - if $A$ is matrix over the complex field, it has a Jordan form. Since $Rank(A)=1$, we have that $\dim\ker(A)=n-1$, meaning that $\lambda=0$ is an eigenvalue of $A$, and moreover it's of geometric multiplicity $n-1$. Since $Tr(A)=0$ and we know that $Tr(A)=Tr(J_A)$ with $J_A$ being the Jordan canonical form, we know that $$\sum_{\lambda \text{ is an eigenvalue of A}}\lambda=0$$ So the only eigenvalue of $A$ is $0$. By the geometric multiplicity, we know that $J_A$ has $n-1$ blocks corresponding the the eigenvalue $0$, meaning $n-2$ of them are of size $1\times 1$ and the last one is of size $2\times 2$. Since the algebraic multiplicity of the eigenvalue $0$ in th minimal polynomial is the size of the greatest block corresponding to $0$ in the Jordan canonical form, we have that $m_A(x)=x^2$. We know that $m_A(x)$ vanishes on $A$, so $A^2=0$.

Answer 5

Yet another flavor: if $E$ is any finite-dimensional vector space (over an arbitrary field) and $\varphi:E\rightarrow E$ is a linear map with rank $1$ and trace $0$, we can prove $\varphi^2=0$ using the composition algebra $E^*\otimes E$, where $E^*$ is dual to $E$. Since $\varphi$ has rank $1$, we can write $\varphi=x^*\otimes x$ for some $x^*\in E^*$ and $x\in E$. By assumption, $$\langle x^*,x\rangle=\mathrm{tr}(x^*\otimes x)=\mathrm{tr}(\varphi)=0$$ so $$\varphi^2=(x^*\otimes x)^2=\langle x^*,x\rangle(x^*\otimes x)=0$$