Franklin's Notes


A projector is a square matrix $P=\mathbb F^{m\times m}$ that satisfies the equation $P^2 = P$. These are sometimes called idempotent matrices. Trefethen and Bau describe projections as transformations for which $Pv$ can be interpreted as the shadow that the point $v$ casts on the subspace $\text{range}(P)$, if the light shines from a certain angle.

If $P$ is a projector, then the matrix $I-P$ is also a projector, for The projector $I-P$ is called the complementary projector of $P$. It's easy to show via algebra that the range of a projector equals the null space of its complementary projector, and the null space of a projector is the range of its complementary projector. That is, We can also prove algebraically that the only vector in the null space of both $P$ and $I-P$ is the zero vector. This means that and the projector $P$ divides the vector space into two complementary subspaces. In fact, every vector can be uniquely decomposed as the sum of a vector from $\text{null}(P)$ and a vector from $\text{range}(P)$. An orthogonal projector is one for which these two subspaces are orthogonal (i.e. any pair of vectors taken from them is orthogonal).

Proposition 1. Over $\mathbb C$, projector matrices can only have eigenvalues of $0$ and $1$, and they have a full basis of eigenvectors (i.e. they are nondefective).

Proof. Suppose that $v$ is an eigenvector of a projector matrix $P$ so that $Pv=\lambda v$. Then $P^2 v = \lambda^2 v$, but since $P^2=P$, we have that $\lambda^2 v = \lambda v$, or $\lambda^2 = \lambda$. This implies that either $\lambda=0$ or $\lambda=1$, as claimed.

Now, since $P^2=P$, we have that $P^n=P$ for all $n\in\mathbb N$. This means that all generalized null spaces of $P$ are simply equal to the null space of $P$, the dimension of which is equal to the multiplicity of the linear factor $t$ in the characteristic polynomial $p_P(t)$. On the other hand, since $I-P$ is also a projector, we have that the dimension of the eigenspace of $P$ corresponding to $\lambda=1$ is equal to the multiplicity of the linear factor $t-1$ in $p_P(t)$. Since the degree of $p_P(t)$ always equals the dimension of the overall space, and $p_P(t)$ can be factored into powers of the linear factors $t$ and $t-1$, we have that the dimensions of the eigenspace of $\lambda=0$ and the eigenspace of $\lambda=1$ must sum to the dimension of the overall space, meaning that we have a full basis of eigenvectors. $\blacksquare$

We will use this result to derive a simple criterion for determining whether a projector is orthogonal.

Proposition 2. Over $\mathbb{C}$, a projector $P$ is an orthogonal projector iff it is Hermitian , that is, if $P^\ast = P$.

Proof. If $P^\ast=P$, then we have that for any vector $v$, meaning that the range of $P$ is orthogonal to the range of $I-P$, making $P$ an orthogonal projector.

To prove the reverse direction, we use the previous proposition. Suppose $P$ is an orthogonal projector. Since we have a full basis of eigenvectors according to Proposition 1, we may let ${q_1,...,q_r}$ be the set of eigenvectors corresponding to $\lambda=0$, and ${q_{r+1},...,q_m}$ be the set of eigenvectors corresponding to $\lambda=1$. Note that these two sets space $\text{null}(P)$ and $\text{range}(P)$ respectively. Thus, since $P$ is an orthogonal projector, these subspaces are orthogonal, and ${q_1,...,q_m}$ is an orthogonal basis for $\mathbb C^m$. Thus, if $Q$ is the matrix whose columns are the vectors $q_1,...,q_m$ in order, we have $P$ has the eigenvalue decomposition $P=Q\Lambda Q^\ast$, where $\Lambda$ is a diagonal matrix consisting of $r$ ones and $m-r$ zeroes along the diagonal. From this decomposition, we have that $P^\ast = Q\Lambda^\ast Q^\ast = Q\Lambda Q^\ast = P$ as claimed. $\blacksquare$



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