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Let $V$ be a finite-dimensional inner product space and $W$ be a subspace of $V$. Recall that any $v \in V$ can be written as $x+y$ with $x\in W$ and $y \in W^\perp$; and that the projection $\mathrm{proj}_{W}v$ is equal to $x$ by definition.

Prove that the function $T:V\rightarrow V$ defined by $T(v)=\mathrm{proj}_Wv$ is a linear transformation.

The algebra is supposed to be really easy, but I just can't seem to get it.
I did:
Let $v = v_1 + v_2$ with $v_1 \in W, v_2 \in W^\perp$;
$w = w_1 + w_2$ with $w_1\in W, w_2 \in W^\perp$.

$T(v+w)=T(v_1 + v_2 + w_1 + w_2) = T(\mathrm{proj}_Wv + v_2 + \mathrm{proj}_Ww + w_2) $ and I'm stuck..

Twenty-six colours
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4 Answers4

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If $\langle\cdot,\cdot\rangle$ is your inner product and if $(e_1,\ldots,e_k)$ is an orthonormal basis of $W$, then$$(\forall v\in V):T(v)=\sum_{j=1}^k\langle v,e_j\rangle e_j.$$This expresses $T$ as a sum of linear maps. Therefore, $T$ is linear.

José Carlos Santos
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If you have that $V=W\oplus W^{\perp}$ you can always write down $v=w_1+w_2, \forall v \in V, w_1 \in W, w_2 \in W^{\perp}$.

Now, for definition, your projection is given by the direct sum you have.

Definition: $\forall v \in V, \text{proj}_W(v)=w_1$ $\forall v \in V, \text{proj}_{W^{\perp}}(v)=w_2$.

So now you can easily verifiy that.

Alberto Andrenucci
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By definition of the projection onto $W$ we have $$ \DeclareMathOperator{proj}{proj} \proj_W v = \proj_W (x + y) = x $$ For two arbitrary scalars $\lambda_i$ and two vectors $v_i \in V$ we have individual splits $v_i = x_i + y_i$ with $x_i \in W$ and $y_i \in W^\perp$. We can combine them into $$ \lambda_1 v_1 + \lambda_2 v_2 = \lambda_1 (x_1 + y_1) + \lambda_2(x_2 + y_2) = (\lambda_1 x_1 + \lambda_2 x_2) + (\lambda_1 y_1 + \lambda_2 y_2) $$ due to the calculation rules in $V$.

Because of the closure of $W$ and $W^\perp$ each (a vector space property) we have $$ \lambda_1 x_1 + \lambda_2 x_2 \in W \\ \lambda_1 y_1 + \lambda_2 y_2 \in W^\perp $$ which implies $$ \proj_W(\lambda_1 v_1 + \lambda_2 v_2) = \lambda_1 x_1 + \lambda_2 x_2 $$ On the other hand we have $$ \lambda_i \proj_W(v_i) = \lambda_i x_i $$ so we get $$ \proj_W(\lambda_1 v_1 + \lambda_2 v_2) = \lambda_1 x_1 + \lambda_2 x_2 = \lambda_1 \proj_W(v_1) + \lambda_2 \proj_W(v_2) $$ This means $\proj_W$ is linear on $V$.

mvw
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$$T(v+w)=T(v_1+v_2+w_1+w_2)=T(v_1+w_1+v_2+w_2)=v_1+w_1=T(v)+T(w)$$

Edit: Let $v \in V$. Then $v=x+y$ where $x \in W,y \in W^\perp$. Then $T(v) = x$.

Now to show it is linear. $T(v)=T(v_1+v_2)=v_1$. Similarly, $T(w)=w_1$.

Also, $T(v+w)=T(v_1+v_2+w_1+w_2)$. Let $x=v_1+w_1,y=v_2+w_2$. Then $x \in W$ and $y \in W^\perp$. Therefore, $$T(v+w)=T(x+y)=x=v_1+w_1=T(v)+T(w)$$

Sahiba Arora
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