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Let $P$ be the operator on $R^2$ which projects each vector onto the $x$-axis, parallel to the $y$-axis: $P(x,y)=(x, 0)$. Show that $P$ is linear. What is the minimal polynomial for $P$?

My attempt: Let $(x,y),(r,s)\in \Bbb{R}^2$ and $c\in F$. Then $P(c(x,y)+(r,s))=P(cx+r,cy+s)=(cx+r,0)=c(x,0)+(r,0)=cP(x,y)+P(r,s)$. Thus $P$ is linear. Let $B=\{e_1,e_2\}$ be standard basis of $\Bbb{R}^2$. Then $P(e_1)=(1,0)$ and $P(e_2)=(0,0)$. So $[P]_B=\begin{bmatrix} 1&0\\ 0&0\\ \end{bmatrix}$. Thus $\exists B$ basis of $\Bbb{R^2}$ such that $[P]_B$ is diagonal. So $P$ is diagonalizable. Characteristic polynomial function of $P$ is $f:\Bbb{R}\to \Bbb{R}$ such that $f(x)=\det (xI_2-[P]_B)=(x-1)x$, $\forall x\in \Bbb{R}$. So $0,1$ are distinct eigenvalue of $P$. By this post, minimal polynomial of $P$ is $m=(x-1)x$. Is my proof correct?

Edit: User Anne Bauval comment made me realize, I unnecessary computed characteristic polynomial. Since $P$ is diagonalizable, I get eigenvalue directly from diagonal of matrix $[P]_B$. No need to compute characteristic polynomial. I would like to add one more point which I thought about recently. Let $V$ be $n$-dimensional vector space over $F$ and $T\in L(V,V)$. Suppose $T$ is diagonalizable. Then $\exists B$ basis of $V$ such that $[T]_B$ is diagonal, i.e. $[T]_B=\begin{bmatrix}c_1 & &\\ & \ddots & \\ & & c_n\end{bmatrix}$. We claim $c_1,…,c_n$ are all eigenvalues of $T$. Assume towards contradiction, $\exists c_{n+1}\in F$ such that $c_{n+1}$ is eigenvalue of $T$ and $c_{n+1}\neq c_i$, $\forall i\in J_n$. Characteristic polynomial of $T$ is $f(x)=\det (xI_n-[T]_B)=(x-c_1)\cdots (x-c_n)$, $\forall x\in F$. Since $c_{n+1}$ is eigenvalue of $T$, we have $f(c_{n+1})=(c_{n+1}-c_1)\cdots (c_{n+1}-c_n)=0$. Which implies $c_{n+1}-c_j=0$, for some $j\in J_n$. So $c_{n+1}=c_j$. Thus we reach contradiction. Hence $c_1,…,c_n$ are all eigenvalues of $T$.

user264745
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    Yes, it is correct. A less boaring way to prove simultaneously that $P$ is linear and that $[P]_B=\begin{pmatrix}1&0\0&0\end{pmatrix}$ is to notice that $\begin{pmatrix}x\0\end{pmatrix}=\begin{pmatrix}1&0\0&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}.$ – Anne Bauval Dec 29 '22 at 17:47
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    Your proof is perfectly fine. Although I find it odd for a linear algebra textbook to ask to ckeck the linearity of a projection (which is basic linear algebra) while talking about minimal polynomials (which is not as basic). – Zag Dec 29 '22 at 17:51
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    Also, the minimal polynomial of a non-trivial projection (i.e. projection distinct from the $0$ map and the identity map) is always $x(x-1)$ since more generally (and directly, whithout computing the characteristic polynomial and whithout your link to the other post), the minimal polynomial of a diagonal matrix $D$ is $\prod_{\lambda\in S}(x-\lambda)$ where $S$ is the set of all diagonal elements of $D.$ – Anne Bauval Dec 29 '22 at 18:00
  • @AnneBauval yeah. Noticing that equality proves $[P]B=\begin{bmatrix}1&0\ 0&0\ \end{bmatrix}$ (by uniqueness of matrix representation of a linear map) but that is not that “trivial”, IMO. Your comment made me realise, I unnecessary computed characteristic polynomial. Since $P$ is diagonalizable, I get eigenvalue directly from diagonal of matrix $[P]_B$. No need to compute characteristic polynomial. That link post is intermediate in proof of minimal polynomial of a diagonal matrix $D$ is $\prod{\lambda \in S}(x-\lambda)$. – user264745 Dec 29 '22 at 20:40
  • @Zag I guess, author wanted to remind readers that we define minimal polynomial notion on linear map, not on generic map. This is just a speculation. – user264745 Dec 29 '22 at 20:44
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    Not only does this equality prove that $[P]_B$ is that matrix: it first proves that $P$ is linear. Such a method is very efficient, to avoid wasting time and ink on "automatic writing" of usual boaring proofs of linearity ("$P(cu+v)=\dots=cP(u)+P(v)$"). And it is trivial when you get used to: if $f(x,y)=(ax+by,cx+dy),$ just write $\begin{pmatrix}ax+by\cx+dy\end{pmatrix}=\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}.$ – Anne Bauval Dec 29 '22 at 22:39
  • Btw if $c_1,…,c_n$ are distinct, then we’re done, since there exists at most $n$ roots of characteristic polynomial of $T$. – user264745 Dec 31 '22 at 12:12

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