-1
$\begingroup$

If P is a projection operator, show that I-P is a projection operator. Determine the range and nullspace of I-P.

How can I solve this? What should be my approach?

$\endgroup$
0
$\begingroup$

A linear operator $P:V\to V$ on a vector space $V$ is a projector iff $P^2=P$. For such a projector it follows that $(1-P)^2=1^2+P^2-2P=1+P-2P=1-P$ and therefore $1-P$ is a projector too.

Consider $P\circ (1-P) = P - P^2 = 0$. This implies that the $\mathrm{range}(1-P)\subset\ker(P)$. In the same way $(1-P)\circ P = 0$, implying $\mathrm{range}(P)\subset\ker(1-P).$

With the dimension theorem between kernel and range of $P$ and $1-P$, the subsets turn out to be improper: $$\ker(1-P)=\mathrm{range}(P)$$ $$\mathrm{range}(1-P)=\ker(P)$$


Some more details on the eigenstructure of projectors, as it is mentioned in Nir Regev's answer:

If $\lambda$ is an eigenvalue of $P$ then $Pv=\lambda v$ for a corresponding eigenvector $v$. Letting $P$ act twice gives $P(Pv)=\lambda^2 v$. Because $P^2=P$ it follows that $\lambda^2=\lambda$. The only possible eigenvalues of a projector are therefore $0$ and $1$. With those eigenvalues, the eigenvector equation for $v\neq 0$ becomes $Pv=v$ and $Pv=0$, which define the range and the kernel of $P$ respectively.

The eigenstructure of projectors can be used for a more constructive proof of what you ask for: $v \in \ker(P) \Leftrightarrow Pv=0 \Leftrightarrow (1-P)v=(v-Pv)=v \Leftrightarrow v \in \mathrm{range}(1-P)$ and similarly for the nullspace of $1-P$.

$\endgroup$
0
$\begingroup$
  1. since $P$ is a projection matrix $P^2=P$. Using this it is easy to show that (I-P)(I-P) = I-P. Hence I-P is also Projection (symmetry property also preserves).
  2. Since the eigenvalues of a projection matrix are either ones or zeros you can use eigendecomposition: $P = U \begin{pmatrix} I_M & \bf{0}\\ \bf{0} &\bf{0} \end{pmatrix} U^T$ where $M$ is the dimension of the subspace to which the projectios is carried it's easy to show that $I - U \begin{pmatrix} I_M & \bf{0}\\ \bf{0} &\bf{0} \end{pmatrix} U^T = U \begin{pmatrix} \bf{0} & \bf{0}\\ \bf{0} & I_{N-M} \end{pmatrix} U^T$.

From here it is easy to extract the 4 sub-spaces of each projection operator.

The above analysis is correct for orthogonal projectors. For oblique projectors a similar calculation can be done using singular value decomposition.

$\endgroup$
  • $\begingroup$ Oblique projectors don't diagonalise like this $\endgroup$ – Jazzmaniac Mar 27 '16 at 12:09
  • $\begingroup$ Right. Obviously the problem consider orthogonal projectors. $\endgroup$ – Nir Regev Mar 27 '16 at 12:13
  • $\begingroup$ I don't see any reference to orthogonal projection in the question $\endgroup$ – Jazzmaniac Mar 27 '16 at 12:14
  • $\begingroup$ @Jazzmaniac you're right. I updated the answer $\endgroup$ – Nir Regev Mar 27 '16 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.