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In an exercise stated as the following.

Consider the real inner product space $\mathbb{R}^3$ (using scalar-product as the inner product), and let $W=\text{Span}(v_1,v_2)$ where

$$v_1 = \begin{bmatrix}1\\2\\1\end{bmatrix}, v_2 = \begin{bmatrix}2\\1\\2\end{bmatrix}$$.

The orthogonal projection $P_W : \mathbb{R}^3 \rightarrow W$ is given by

$$P_W(\begin{bmatrix}a\\b\\c\end{bmatrix})=\begin{bmatrix}\alpha\\\beta\\\alpha\end{bmatrix}$$.

Find $\alpha,\beta$.

I was able to use the hints to determine that

$$\alpha=\frac{1}{2}(a+c),\quad \beta=b$$

but I don't understand why, and I don't know how to solve similar problems without the hints. Can anyone explain why $\alpha, \beta$ are given as they are.

I think one should be able to determine them from the orthogonal projection formula


Given a inner product space $V$ with $W=\text{Span}(v_1,\dots, v_n)$, then any vector $v\in V$ can be written as

$$v=p+h$$

where $p\in W,h\in W^\perp$ and where we can write $p$ as

$$p=\sum_{i=1}^n \frac{\langle v, v_i \rangle}{\langle v_i, v_i \rangle} v_i$$.


However, I'm unable to relate the two.

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  • $\begingroup$ That last formula is only valid when the $v_i$ are orthogonal. $\endgroup$ – amd Apr 15 '20 at 7:18
  • $\begingroup$ Ah yes @amd. The exercise specifically states that $v_1,v_2$ are orthogonal. $\endgroup$ – lunalux Apr 15 '20 at 7:20
  • $\begingroup$ However, $v_1$ and $v_2$ in the first part of your question are not. $\endgroup$ – amd Apr 15 '20 at 7:22
  • $\begingroup$ You're right @amd. I'm just confused. $\endgroup$ – lunalux Apr 15 '20 at 8:03
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The formula at the end of your question is only applicable when $v_1$ and $v_2$ are orthogonal, which they obviously aren’t. However, if the projection of an arbitrary vector is of the form $[\alpha,\beta,\alpha]^T$, this means that another basis for $W$ is $w_1=[1,0,1]^T$ and $w_2=[0,1,0]^T$. These vectors are orthogonal, so the formula can be used with them. We then have $$\alpha = {v\cdot w_1\over w_1\cdot w_1} = {a+c\over2} \\ \beta = {v\cdot w_2\over w_2\cdot w_2} = b.$$

These values can of course be derived without knowing the projection formula. If $[\alpha,\beta,\alpha]^T$ is the projection of $[a,b,c]^T$, then $[a,b,c]^T-[\alpha,\beta,\alpha]^T$ must be orthogonal to both $v_1$ and $v_2$. Setting the inner products equal to zero gives you a system of two linear equations that you can solve for $\alpha$ and $\beta$, giving the same result as above.

Later on you will learn how to compute the projection directly from $v_1$ and $v_2$, without first finding an orthogonal basis or solving a set of equations: setting $A=[v_1\;v_2]$, $P_W(v)=A(A^TA)^{-1}A^Tv$. This is basically a generalization of the projection formula near the end of your question to an arbitrary basis of $W$.

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If you apply the Gram-Schmidt process to $\{v_1,v_2\}$, you will get $\{e_1,e_2\}$, with$$e_1=\frac1{\sqrt6}\begin{bmatrix}1\\2\\1\end{bmatrix}\text{ and }e_2=\frac1{\sqrt3}\begin{bmatrix}1\\-1\\1\end{bmatrix}.$$So,\begin{align}P_W\left(\begin{bmatrix}a\\b\\c\end{bmatrix}\right)&=\left(\begin{bmatrix}a\\b\\c\end{bmatrix}.e_1\right)e_1+\left(\begin{bmatrix}a\\b\\c\end{bmatrix}.e_2\right)e_2\\&=\begin{bmatrix}\frac{a+c}2\\b\\\frac{a+c}2\end{bmatrix}.\end{align}

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  • $\begingroup$ This answer is correct, but the book stated this example specifically before introducing the Gram-Schmidt process. If possible, I'd like to see if one can reach the same result using just orthogonal projection. $\endgroup$ – lunalux Apr 15 '20 at 7:15

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