Show $\operatorname{tr}(P) = 1$ where $P$ is an orthogonal projector onto the one-dimensional space spanned by vector b

Question

I am struggle-bussing over this proof right now. This is what I have so far: since $ P $ is a projection, $ P^2 = P$, and the $\operatorname{dim}U = 1$, so by a property, basis $B = b \in \mathbb{R}^{m \times 1}$. Then $P = bb^T, tr(bb^T)=b^tb.$ In the span, $b_i$, isn't technically orthonormal so $b^tb$ isn't necessarily the identity matrix, correct or is it?

I'm not sure where to go from here, I feel like I need to find the way to sum over $a_{ij}$ where $i = j$ of the matrix to be able to use another property where that sum is equal to $1$.

I thought of going the route of having $P = bb^t = \|b\|^2$ due to the span of $U$ having a dim of 1 and then finding the $\operatorname{tr}(\|b\|^2)$, but I'm not really sure where to even take the next step in that one.

Any hints would be appreciated. Thanks.

For $bb^\top$ to be an orthogonal projector, you'd need $b$ to be a unit vector (spanning the one-dimensional subspace onto which you are projecting). So, as you point out, $\operatorname{tr}(bb^\top) = b^\top b = |b|^2 = 1$. — Theo Bendit, Sep 12 '23 at 04:33
Does this answer your question? Proving: "The trace of an idempotent matrix equals the rank of the matrix", or https://math.stackexchange.com/questions/3675161/proving-the-trace-equals-the-dimension-of-an-image, or https://math.stackexchange.com/questions/985879/relation-between-trace-and-rank-for-projection-matrices — leslie townes, Sep 12 '23 at 04:57

Ryszard Szwarc · Answer 1 · 2023-09-18T15:56:46.200

1

There is no need that the projection is orthogonal. It suffices that $P^2=P.$ Any one dimensional operator is of the form $Px=(u^Tx)v,$ where $u,v\neq 0.$ The operator $P$ is a projection iff $u^Tv=1.$ Next $${\rm Tr}\,P=\sum_{j=1}^m e_j^TPe_j =\sum_{j=1}^m(e_j^Tv)(u^Te_j)=u^Tv$$

edited Sep 18 '23 at 15:56

answered Sep 12 '23 at 04:53

Ryszard Szwarc

44,675

score 1 · Accepted Answer · answered Sep 15 '23 at 12:34

Let $V$ be your vector space. If you have an orthonormal basis $\{v_1, \ldots, v_n\}$ then trace of any operator $T:V\to V$ is simply $\sum_{i=1}^{n} v_i'Tv_i$. Now take $T = P$ and $v_1 = b/|b|$. Since all other vectors in the basis are orthogonal to $b$ their projection on $b$ is going to be $0$ i.e. $Pv_i = 0$ for all $i \not= 1$. Thus trace of $P$ equals $v_1'Pv_1 = v_1'v_1 = |v_1|^2 = 1$.

Show $\operatorname{tr}(P) = 1$ where $P$ is an orthogonal projector onto the one-dimensional space spanned by vector b

2 Answers2