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Let $A$ be an $n \times n$ matrix over $\mathbb{K}$ such as $\text{rank}(A)=1.$ Show that it's minimal polynomial is $m_A(x) = λ(λ-a) , a \in \mathbb{K}.$

I tried to prove it using induction. My work so far:

Let $n =2$ and $A = \begin{bmatrix}a&b\\c&d \end{bmatrix}$:

  • If $c=0_{\mathbb{K}}:$

    Since $\text{rank}(A)=1$, $d=0_{\mathbb{K}}$. So, $A=\begin{bmatrix}a&b\\0&0 \end{bmatrix} \implies \det(A-λI) = λ(λ-α) = m_A(λ)$

  • If $c\ne 0_{\mathbb{K}}$

    $A = \begin{bmatrix}a&b\\c&d \end{bmatrix} \to \begin{bmatrix}a&b\\0&-ad/c +b \end{bmatrix} \implies b = ad/c \implies A = \begin{bmatrix}a&ad/c\\c&d \end{bmatrix} \implies \det(A-λI) = λ(λ-(α+d)) , (a+d) \in \mathbb{K}$

On that point, I tried letting $n=3$ and writing the matrix in the form of $A = \begin{bmatrix}A_1&0\\0&A_2 \end{bmatrix}$,after Gauss elimination, where $A_1$ is a $2 \times 2 $ matix and $A_2$ is a $1 \times 1$ matrix and working my way from $n=k$ to $ n =k+1$, but it seems like I'll be doing way too much work. I there a better/shorter way of doing it?

Rodrigo de Azevedo
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Alex Matt
  • 429

1 Answers1

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There is a general proposition about full rank decomposition:

If $A$ is an $m\times n$ matrix of rank $r$, then there exists full column rank($=r$) matrix $B$ ($m\times r$) and full row rank($=r$) matrix $C$ ($r\times n$), such that $A=BC$.

Here, since $\mathrm{rank}(A)=1$, there exist nonzero column vectors $v$ and $w$, such that $A=vw^T$. Note that $A^2=v(w^Tv)w^T=w^Tv(vw^T)=aA$, where $a=w^Tv$ is a number. Therefore the minimal polynomial must divide $\lambda^2-a\lambda$, which annihilate the matrix $A$, $A^2-aA=0$. However, it cannot be $\lambda$ or $\lambda-a$ since this means $A$ is a scalar matrix, but scalar matrix with $n>1$ cannot have rank 1.

Mathis
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