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?


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 $1-P$.

  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.

  • $\begingroup$ 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

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