I am trying to prove this theorem in my book. Because it is provided without proof, please let me know what you think!

$\mathbf{Theorem:}$ Let V be a finite , $n$ -dimensional vector space and let U be a subspace of V.

Define, $T: V \rightarrow V$ as $$ T(v)=proj_U(v)$$ ( ie, the orthogonal projection)

Then the following are true;

i) T is a linear transformation

ii) $Im(T)=U$ and $kerT=U^{\perp} $ where $U^{\perp}$ represents the orthogonal complement.

iii) $Dim(U)+Dim(U^{\perp})= n$


(i) Let $v , w \in V$ , $k \in \mathbb R$, and let $\mathbb B=\{u_1,…u_n\}$ be an orthogonal basis for V.

then we have $$T(v)= \frac{\langle v , u_1 \rangle}{\langle u_1, u_1 \rangle}u_1 +…+ \frac{\langle v, u_n \rangle}{\langle u_n, u_n \rangle}u_n$$

and $$T(w)=\frac{\langle v , w_1 \rangle}{\langle w_1, w_1 \rangle}w_1 +…+ \frac{\langle v, w_n \rangle}{\langle w_n, w_n \rangle}w_n$$

then the rest follows by invoking the inner product properties of linearity. I have it written but it will take my a long time to type that out , so for that I would just be wondering if that is the correct approach?


Must show that $Im(T) \subseteq U$ and that $U \subseteq Im(T)$

Let $w \in Im(T)$ then there exists a$ v \in V$ such that $T(v)=w$

and we have seen that this would imply w is just the projection i.e. a linear combination of vectors $u_i$ and for $U \subseteq Im(T)$ we can simply take the projection of $u$ onto itself. Thus, showing $Im(T)=U$

Now suppose $w \in U^{\perp}$ then $$T(w)=\frac{\langle w,u_1 \rangle}{\langle u_1, u_1 \rangle }u_1+…+\frac{\langle w , u_n \rangle}{\langle u_n, u_n \rangle}u_n=0$$ ie $w \in Ker(T)$ and now now suppose $w \ in Ker(T)$ then $T(w)=0$ , using the fact that for any $w \in V$ we have $w-T(w) \in U^{\perp}$ then this gives $w-0 \in U^{\perp}$ that is $w \in U^{\perp}$ i.e. $Ker(T) \subseteq U^{\perp}$

(iii) This follows directly from the rank nullity theorem, which states $dim(ImT)+dim(KerT)=dim(V)$

Let me know what you all think, any suggestions etc, thanks!

  • $\begingroup$ What's $\phantom{x}^\perp$? $\endgroup$
    – GFauxPas
    Commented Apr 20, 2015 at 20:57
  • $\begingroup$ What is $proj_U$? And how can you assume $V$ has an orthogonal basis? $\endgroup$ Commented Apr 20, 2015 at 21:03
  • $\begingroup$ @GFauxPas the orthogonal complement $\endgroup$ Commented Apr 20, 2015 at 21:04
  • $\begingroup$ @HagenvonEitzen the (orthogonal) projection onto $U$ $\endgroup$ Commented Apr 20, 2015 at 21:04
  • $\begingroup$ @HagenvonEitzen It's usually given as a theorem that every inner product space has an orthogonal basis. The proof is essentially given by the Gram Schmidt algorithm. $\endgroup$ Commented Apr 20, 2015 at 21:05

1 Answer 1


I think you meant $\mathbb{B}=\{u_1,…,u_r\}$ is an orthogonal basis for $U$. You can extend it, by adding vectors in the corresponding orthogonal complements, to an orthogonal basis $\{u_1,…,u_r,v_{r+1},…,v_n\}$ for $V$. There is no need to normalize the vectors. Now, if $v=\sum <v,u_i>u_i+\sum <v,v_i>v_i$, then $T(v)=\sum<v,u_i>u_i$. From that explicit formula everything is simpler, for example, $v\in Ker(T) \Leftrightarrow \;<v,u_i>=0,$ for all $i=1,…,r$, which is equivalent to $v \in U^\bot$.


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