## Related questions with answers

Let V be an inner product space with a subspace W having $B=\left\{\mathbf{w}_{1}, \mathbf{w}_{2}, \dots, \mathbf{w}_{n}\right\}$ as an orthonormal basis. Show that the function $T: V \rightarrow W$ represented by $T(\mathbf{v})=\left\langle\mathbf{v}, \mathbf{w}_{1}\right\rangle \mathbf{w}_{1}+\left\langle\mathbf{v}, \mathbf{w}_{2}\right\rangle \mathbf{w}_{2}+\cdots+\left\langle\mathbf{v}, \mathbf{w}_{n}\right\rangle \mathbf{w}_{n}$ is a linear transformation. T is called the orthogonal projection of V onto W.

Solutions

VerifiedWe have the linear transformation that projects the vectors $\bold{v}$ of $V$ onto the vectors $\bold{w}$ of $W$.

*How do we know that a function is a linear transformation?*

Take any vectors $u$ and $v$ in $V$. Using the axioms of inner product, we can get

$\begin{align*} T(u+v)&=\langle (u+v),w_1\rangle w_1+\langle (u+v),w_2\rangle w_2+\ldots+\langle (u+v),w_n\rangle w_n\\ &=\langle u,w_1\rangle w_1+\langle v,w_1\rangle w_1+\langle u,w_2\rangle w_2+\langle v,w_2\rangle w_2+\ldots+\langle u,w_n\rangle w_n+\langle v,w_n\rangle w_n\\ &=T(u)+T(v). \end{align*}$

That is, $T$ is closed under addition.

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