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We know $rank(A) = 1$ so I have $n-1$ eigenvalues which are $0$

So my characteristic polynomial is $(x-0)^{n-1}(x-a) = x^{n-1}(x-a)$ with $a$ the last eigenvalue to determine. Now, I found on some forum that the polynomial minimal is $x(x-a)$ with $a = Tr(A)$. Do you have a simple explanation about this without Jordan Blocks ?

(If it can help, in my case I have $Tr(A)=0$)

Marco Cantarini
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ReaperSala
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3 Answers3

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A rank $1$ matrix takes the form $uv^\top$, where $u, v$ are $n \times 1$ column vectors. We have, even for non-square matrices, $\operatorname{Tr}(AB) = \operatorname{Tr}(BA)$. Thus, $$\operatorname{Tr}(uv^\top) = \operatorname{Tr}(v^\top u) = v^\top u = u \cdot v,$$ since $v^\top u$ is $1 \times 1$, where $u \cdot v$ is the dot product of $u$ and $v$.

Let's plug this matrix into the polynomial: \begin{align} uv^\top(uv^\top - \operatorname{Tr}(uv^\top)I) &= uv^\top uv^\top - \operatorname{Tr}(uv^\top)uv^\top \\ &= (u \cdot v)uv^\top - (u \cdot v)uv^\top = 0. \end{align}

Is this the minimal polynomial? Let's look at the lesser factors of $x(x - \operatorname{Tr}(A))$. Certainly $x$ would not suffice, because then $A = 0$, which contradicts $A$ being rank $1$. On the other hand, $x - \operatorname{Tr}(A)$ will also not suffice if $n > 1$, because then $A = \operatorname{Tr}(A) I$, which is either full rank $n$ if $\operatorname{Tr}(A) \neq 0$, or the $0$ matrix if $\operatorname{Tr}(A) = 0$. (Note also that $x - \operatorname{Tr}A$ is the minimal polynomial in the $n = 1$ case.)

Theo Bendit
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The characteristic polynomial must be $p(t)=t^{n-1}(t-a)$. The eigenvalue $t=0$ has both the algebraic multiplicity and geometric multiplicity equal to $(n-1)$; because $rank(A-0I)=1$ implies the system $(A-0I)v=O$ has $n-1$ linearly independent solutions. This leads to a total of $n$ linearly independent eigenvectors (one corresponding to eigenvalue $t=a$ too!) thereby confirming that $A$ is diagonalizable. And for a diagonalizable matrix, the minimal polynomial (a factor of $p(t)$) must have distinct linear factors so that each eigenvalue must be its zero, so the only possibility is $m(t)=t(t-a)$ where $a=tr(A)$.

Nitin Uniyal
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Since $A$ has rank one, it can be written as $A=u v^T$. Using the Sylvester identify we get (for $\lambda \ne 0$), $\det(\lambda I -uv^T) = \det( \lambda (I -{1 \over \lambda} u v^T) = \lambda^n (1-{1 \over \lambda}v^T u) = \lambda^{n-1} (\lambda -v^Tu)$.

To see if $\lambda (\lambda -v^Tu)$ is the minimal polynomial, note that $A(A-v^Tu I)=u v^T ( u v^T - v^T u I) = v^T u u v^T -v^Tu u v^T = 0$.

copper.hat
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