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This question already has an answer here:

By inspecting its formula one can see easily that the matrix for projection onto a subspace is equal to its transpose.

But what is the underlying "geometric" reason for this equality? I have hard time figuring out why it has to be so?

EDIT: After reviewing the suggested links, it became more clear, but the reasoning was still escaping my intuition. However, I think that I can (based on the formal arguments in the links) put forward a rough argument. Basically, for any operator $A$ and two vectors $x$, $y$ $$<Ax,y>=<x,A^Ty>$$

For a vector $y$ in the orthogonal complement to the subspace S we're projecting onto using the $P^S$ projection operator, $$<P^Sx,y>=0=<x,(P^S)^Ty>$$ Because we took any vector $x$, it means that $(P^S)^Ty=0$. As $y$ is in the orthogonal complement, $P^Sy=0$ this means that $$(P^S)^Ty=0=P^Sy$$ (in plain English, for any vector in the orthogonal complement to $S$ the action of $P^S$ makes it 0, which is a symmetric action).

For $y$ in $S$ and any $x$, $<P^Sx,y>=<x,y>$ and $P^Sy=y$ (by the properties of the projection operation), but also $<P^Sx,y>=<x,(P^S)^Ty>$ (true for any operator). Putting these together that yields that:$$(P^S)^Ty=y=P^Sy$$.

(in plain English, for any vector in $S$ the action of $P^S$ is the identity, which is a symmetric action).

To sum it up, this shows that the operator $P^S$ acts symmetrically on both elements of S and on elements in its orthogonal complement. Being that it is a linear operator, then it acts symmetrically on all vectors $x$.

A big thing why this works (I think) is that the concept of projection implies (finite) dimension inner product space and that is much more structured than a simple (finite) dimensional vector space.

LAST EDIT: Even more clear after digesting the answer of Trial and Error. Given a subspace M to project onto, any vector $z_1$ can be written as the sum of two orthogonal components:

$$z_1= (z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,$$

But the key geometric thing is that these two components belong to orthogonal subspaces, M and $\perp M$ (any vector from one subspace is perpendicular on any vector from the other subspace).

The projection operator will leave the component in M unchanged and annihilate the component in $\perp M$.

Because of the fact that the subspaces themselves are orthogonal the product of ANY two arbitrary vectors $z_1, z_2$

$$<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,(z_2-P_{\mathcal{M}}z_2) + P_{\mathcal{M}}z_2>$$ becomes (by the orthogonality of M and $\perp M$: $$<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1),(z_2-P_{\mathcal{M}}z_2)>+<P_{\mathcal{M}}z_1,P_{\mathcal{M}}z_2>$$

But applying $P_M$ annihilates/zeroes the first bracket and leaves unchanged the second bracket. This holds whether we apply it to $z_1$ or to $z_2$.

Ergo the effect of $P_M$ is the same whether applied to the first of second term of the scalar product. So its matrix should be equal to its transpose.

The bolded text indicates key issues that I was not fully appreciating before. Thank you for your patience!

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marked as duplicate by Martin Sleziak, E. Joseph, C. Falcon, Namaste, Daniel W. Farlow Dec 3 '16 at 1:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You’re only considering orthogonal projections. Other types or projections do not have this property. See the linked question in GLay’s comment. $\endgroup$ – amd Dec 2 '16 at 16:47
  • $\begingroup$ So it is a feature particular to inner product spaces. How does it fail in non-inner product normed spaces? $\endgroup$ – Dacian Bonta Dec 2 '16 at 20:08
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As you learned in Calculus, the orthogonal projection $P$ of a vector $x$ onto a subspace $\mathcal{M}$ is obtained by finding the unique $m \in \mathcal{M}$ such that $$ (x-m)\perp \mathcal{M}. \tag{1} $$ So the orthogonal projection operator $P_{\mathcal{M}}$ has the defining property that $(x-P_{\mathcal{M}}x)\perp \mathcal{M}$. And $(1)$ also gives $$ (x-P_{\mathcal{M}}x) \perp P_{\mathcal{M}}y,\;\;\; \forall x,y. $$ Consequently, $$ \langle P_{\mathcal{M}}x,y\rangle=\langle P_{\mathcal{M}}x,(y-P_{\mathcal{M}}y)+P_{\mathcal{M}}y\rangle= \langle P_{\mathcal{M}}x,P_{\mathcal{M}}y\rangle $$ From this it follows that $$ \langle P_{\mathcal{M}}x,y\rangle=\langle P_{\mathcal{M}}x,P_{\mathcal{M}}y\rangle = \langle x,P_{\mathcal{M}}y\rangle. $$ That's why orthogonal projection is always symmetric, whether you're working in a real or a complex space.

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