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I'm reading a book on numerical recipes and I'm having a bit of trouble trying to prove a statement made by the authors: given $B \in \mathbb{R}^{n \times r}$ with orthonormal columns (forming an orthogonal base in $\operatorname{Im} B = S$), then $P = BB'$ is an orthogonal projection on $S$ and its also unique.

Now the first part about $P$ being an orthogonal projection is clear, but its uniqueness is not so clear to me. Could anyone help me understand why $P$ is unique?

Rodrigo de Azevedo
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Paul Irofti
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2 Answers2

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The thing we need to prove is as follows: if $S$ is a fixed subspace, and if $P,P'$ are two projections onto $S$, then $P = P'$.

An easy way to prove this is as follows: we note that $P,P'$ must be identical if there is some basis $e_1,\dots,e_n$ for $\Bbb R^n$ such that $P e_i = P' e_i$ for all $1 \leq i \leq n$. We construct such a basis as follows:

We may select an orthonormal basis $e_1,\dots,e_r$ of $S$. We may extend this to an orthogonal basis of $\Bbb R^n$. We then note that we must have $$ P(e_i) = P'(e_i) = \begin{cases} e_i & 1 \leq i \leq r\\ 0 & r < i \leq n\end{cases} $$ The conclusion follows.

Ben Grossmann
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  • Is there something missing in your answer after "We construct the basis as follows:"? If not, I don't understand how extending the orthonormal basis holds for the equality you've written down. Could you expand a bit? – Paul Irofti Oct 01 '14 at 16:24
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    Would you agree that $P(x) = x$ for any $x \in S$? Would you agree that $P(x) = 0$ for any $x \in S^\perp$? – Ben Grossmann Oct 01 '14 at 16:32
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    This answer is missing the word "orthogonal" in front of "projections". I suppose its clear from context since the title of the question is "orthogonal projections." But a cavalier reader might think you are claiming that all projections onto a fixed subspace are identical. Of course that is false, for example $ \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} $ and $ \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} $ are distinct projections onto the $ x $ axis. As you noted above, its crucial that $ P,P' $ are orthogonal projections and so $ ker(P)=S^\perp=ker(P') $. – Ian Gershon Teixeira Mar 04 '23 at 22:05
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Say $P$ is an orthogonal projection onto $S$, then $span(P) = S$. Let $(u_1, \cdots, u_m)$ be an orthonormal basis of $S$, we can always extend it to an orthonormal basis $(u_1, \cdots, u_n)$ of $\mathbb{R}^n$. Suppose $Q$ is any orthogonal projection onto $S$, we only need to prove $P u_i = Q u_i, \text{ for } i = 1, \cdots, n$.

  • $\forall v\in S$, we can alway find an $x\in\mathbb{R}^n$ s.t. $Px = v$, it follows that $Pv = PPx = Px = v$; similarly $Qv = v$. Hence for $i=1, \cdots, m$, $P u_i = Q u_i = u_i$.
  • For $j = m+1, \cdots, n$, $u_j$ is orthogonal to $u_1, \cdots, u_m$ and thus every vector in $S = span(Q)$, hence $(Q u_j, u_j) = 0$. $(Q u_j, Q u_j) = (Q^*Qu_j, u_j) = (Q^2 u_j, u_j) = (Q u_j, u_j) = 0$, so $Q u_j = 0$. Similarly $P u_j = 0$, then $Q u_j = P u_j = 0.$
HM Fu
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