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If $P$ is the matrix that projects $R^n$ onto a subspace $S$, explain why every vector in $S$ is an eigenvector, and so is every vector in $S^{\perp}$. What are the eigenvalues (Note the connection to $P^2 = P$, which means that $λ^2 = λ$ .)

I feel like I'm missing something from the question and therefore have no idea where to begin.

The only thing I can think of to get started is that if $S$ is a proper subspace of $R^n$ then, by the Invertible Matrix Theorem, 0 is an eigenvalue which also holds since $0^2=0$ from the given hint.

This is Review Question 5.6 Linear Algebra and Its Applications Fourth Edition by Gilbert Strang

DoubleV
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    Would this in-site reference Find eigenvalues, kernel and Image of an Orthogonal projection be helpful for you? – Hanno Nov 17 '22 at 17:06
  • If $\lambda^2=\lambda$, then $\lambda^2-\lambda=0$, so $\lambda(\lambda-1)=0$. That means the only possible eigenvalues of $P$ are $\lambda=0$ and $\lambda=1$. – Arturo Magidin Nov 17 '22 at 17:32
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    The problem is incomplete/incorrect as stated. There is no unique matrix that project "onto a subspace $S$". For example, both $P_1\left(\begin{array}{cc}1&0\0&0\end{array}\right)$ and $P_2\left(\begin{array}{cc}1&-1\0&0\end{array}\right)$ are matrices that project $\mathbb{R}^2$ onto $S={(x,0)\mid x\in\mathbb{R}}$. But while $S^{\perp}$ is the eigenspace of $P_1$ corresponding to $0$, for $P_2$ we have $P_2(0,1)^t = (-1,0)$, so $S^{\perp}$ does not consist of eigenvectors of $P_2$. You need to specify that $P$ is an orthogonal projection. – Arturo Magidin Nov 17 '22 at 17:37

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As someone else stated, this works for an orthogonal projection. Think about it geometrically. Let $\bf{v}\in S$. Then the projection of $\bf{v}$ onto $S$ is just $\bf{v}$ itself. Thus $P\bf{v}$ $=\bf{v}$ $=1\bf{v}$, so $\bf{v}$ is an eigenvector with corresponding eigenvalue $\lambda =1$. Similarly, if $\bf{u}\in S^\bot$, then the projection of $\bf{u}$ onto $S$ is $0$. So $P\bf{u}$ $=\bf{0}$ $=0\bf{u}$, so $\bf{u}$ is an eigenvector with corresponding eigenvalue $\lambda =0$.

mathimposter
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