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A projection operator $P$ is defined as $P^2$=$P$. Use this definition to find the eigenvalues of this operator.

In this question is it necessary to define what the projection operator is? And won't the eigenvalue just be zero?

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4 Answers 4

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Let $\lambda$ be an eigenvalue of $P$ for the eigenvector $v$. You have $\lambda^2 v = P^2 v = P v = \lambda v$. Because $v \neq 0$ it must be $\lambda^2 = \lambda$. The solutions of the last equation are $\lambda_1 = 0$ and $\lambda_2 = 1$. Those are the only possible eigenvalues the projection might have...

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The eigenvalues are $0$ and $1$. Indeed, we know that for a projector $P$ defined on a vector space $E$, we have $$ E= \ker P\oplus \operatorname{im}P=\ker P\oplus \ker(I-P) $$ $\ker P$ is the eigenspace associated with the eigenvalue $0$, $\ker(I- P) $ the eigenspace associated with the eigenvalue $1$. As $E$ is the direct sum of these eigenspaces, we have all eigenspaces, and all eigenvalues.

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If $\mathbf x$ is an eigenvector of $P$ with eigenvalue $\lambda$, then $P(\mathbf x) = \lambda \mathbf x$. If $P^2 = P(P) = P$, then $P(P(\mathbf x)) = P(\lambda \mathbf x) = \lambda^2 \mathbf x$, but this must also equal $\lambda \mathbf x$. Therefore the eigenvalues of $P$ can only be members of your base field such that $\lambda^2 = \lambda$.

Assuming your base field is $\Bbb R$, there's one more value that satisfies this equation besides $\lambda =0$. What is it?

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Geometrically, eigenvalues are the scaling factors by which particular vectors are scaled when multiplied by the respective matrix. Since the projection matrix projects a vector to its column space (or you can think of some nonzero subspace $W$), one such type of vector is the vector that is already living in that subspace. (Projecting a vector to its plane/subspace does not change it right?) However, when doing so, the length doesn't change, so one eigenvalue must be $1$.

What about the vector that is perpendicular to $W$? Where will it go after the projection? Yes, it goes to the origin and hence the second eigenvalue must be $0$.

Although this approach does not prove anything, it gives a nice neat geometric intuition.

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