A doubt in solving a projection mapping and orthogonality question

Question

Let $l^2= \{ (x_1,x_2,x_3,\dots):x_n \in \mathbb{R} \text{ for all } n \in \mathbb{N} \text{ and } \sum_{n=1}^{\infty} x_n^2 < \infty \}.$

For a sequence $(x_1,x_2,x_3,\dots) \in l^2,$ define $$\lvert \lvert (x_1,x_2,x_3,\dots)\rvert\rvert_2= \bigg( \sum_{n=1}^{\infty} x_n^2\bigg)^{\frac{1}{2}}.$$

Consider the subspace $$M=\{ (x_1,x_2,x_3,\dots) \in l^2 :\sum_{n=1}^{\infty} \frac{x_n}{4^n}=0 \}.$$

Let $M^{\perp}$ denote the orthogonal complement of $M$ in the Hilbert space $(l^2, \lvert \lvert \cdot \rvert \rvert_2).$

Consider $\big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big) \in l^2.$

If the orthogonal projection of $\big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big)$ onto $M^{\perp}$ is given by $\alpha\big( \sum_{n=1}^{\infty} \frac{1}{n4^n}\big)\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big)$ for some $\alpha \in \mathbb{R},$ then what is the value of $\alpha$ ?

My attempt:

I understand that I have to first decompose the given vector $\big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big)$ as sum of a vector in $M$ and a vector in $M^{\perp}.$ But for that I need basis for both $M$ and $M^{\perp}.$ This is how we have been taught in finite dimensional cases. But I am confused here as to how to proceed.

I am able to see that $\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big) \in M^{\perp}.$

Is this true : No non-zero vector other than $\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big)$ is in $M^{\perp}.$ How to see this ?

The projection of $\big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big)$ onto $M^{\perp}$ is given by $\frac{\langle \big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big) ,\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big) \rangle}{\lvert \lvert \big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big) \rvert \rvert} \big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big).$

The coefficient of $\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big)$ in the above expression is $$\frac{\langle \big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big) ,\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big) \rangle}{\lvert \lvert \big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big) \rvert \rvert} = \frac{\big( \sum_{n=1}^{\infty} \frac{1}{n4^n}\big)}{\frac{4}{\sqrt{15}}}.$$ So we have that $\alpha = \frac{\sqrt{15}}{4 }$

Please help me understand. The value of $\alpha$ is $15.$

Answer 1

Note that $M=\{(x_1,x_2,\ldots):\langle (x_1,x_2,\ldots),(\frac{1}{4},\frac{1}{4^2},\ldots)\rangle=0\}=(span\{(\frac{1}{4},\frac{1}{4^2},\ldots)\})^{\perp}$. So: $$M^{\perp}=((span\{(\frac{1}{4},\frac{1}{4^2},\ldots)\})^{\perp})^{\perp}=span\{(\frac{1}{4},\frac{1}{4^2},\ldots)\}$$ where the last step is true as $span\{(\frac{1}{4},\frac{1}{4^2},\ldots)\}$ is a finite dimensional subspace.

The only thing to note is that: $$\|(\frac{1}{4},\frac{1}{4^2},\ldots)\|^2=\langle (\frac{1}{4},\frac{1}{4^2},\ldots),(\frac{1}{4},\frac{1}{4^2},\ldots)\rangle=\sum_{n=1}^{\infty}\frac{1}{4^{2n}}=\sum_{n=1}^\infty\frac{1}{16^n}=\frac{\frac{1}{16}}{1-\frac{1}{16}}=\frac{1}{15}$$ And hence $\alpha={15}$.

Answer 2

Taking motivation from the above answer:

Let $v=\big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4}, \dots \big).$ We shall decompose this $v$ as $v=m+m',$ where $m \in M$ and $m' \in M^\perp.$ Since, $$M^\perp=\text{span} \{x\},$$ where $x=\big(\frac{1}{4},\frac{1}{4^2},\frac{1}{4^3}, \dots \big).$

So, we have that $m'=ax,$ for some $a \in \mathbb{R}.$ Now, $v=m+m'$ implies that $m=v-ax.$ Also, $\langle m,m' \rangle=0.$ Solve for $a$ in the above equation, we get $\langle v-ax,ax \rangle=0.$

On simplifying, we get $a\langle v, x\rangle= a^2\langle x, x\rangle.$ As $a\ne0,$ we have $\langle v, x\rangle= a\langle x, x\rangle.$ This means that $a=\frac{\langle v, x\rangle}{\langle x, x\rangle}.$

So, $$a=\frac{\big( \sum_{n=1}^{\infty} \frac{1}{n4^n}\big)}{\frac{1}{15}}=15\bigg( \sum_{n=1}^{\infty} \frac{1}{n4^n}\bigg)$$ So, $\alpha=15.$

Is there anything wrong in this solution ? Please check.