Let $P :R^3 \rightarrow R^3$ be an orthogonal projection on the $x+2y+z=0$ plane.

Without calculating the projection matrix, determine its 3 eigenvalues and 3 linearly independent eigenvectors.


Let $V$ be the subspace of $\Bbb R^3$ defined by $$ V=\{(x,y,z)\in\Bbb R^3:x+2y+z=0\} $$ Now, note that every vector in $V$ is of the form $$ (-2y-z,y,z)=y(-2,1,0)+z(-1,0,1). $$ Putting $v_2=(-2,1,0)$ and $v_3=(-1,0,1)$ then gives $V=\operatorname{Span}\{v_2,v_3\}$.

The map $P:\Bbb R^3\to \Bbb R^3$ is defined by fixing $V$ and orthogonally projecting the vectors not in $V$ onto $V$.

The vector $v_1=(1,2,1)$ is normal to the given plane and $P$ sends this vector to the origin. Hence $P(v_1)=0\cdot v_1$ so that $v_1$ is an eigenvector of $P$ with eigenvalue $\lambda_1=0$.

Next, since $P$ fixes $V=\operatorname{Span}\{v_2,v_3\}$, we have $P(v_2)=v_2$ and $P(v_3)=v_3$. Hence $v_2$ and $v_3$ are eigenvectors of $P$ with eigenvalue $1$.

  • $\begingroup$ I'm sorry to bother you, but could you please explain how you concluded/calculated these two vectors from the subspace? $\endgroup$ – Eutherpy May 6 '14 at 19:37
  • $\begingroup$ @Eutherpy I added more to the answer. $\endgroup$ – Brian Fitzpatrick May 6 '14 at 20:03

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