The question, which is in my Linear Algebra textbook (no answer is provided), goes as follows:
Let $W=\text{span}\{v_1= [2,2,-2,0]^T, \hspace{.2cm}v_2=[0, 1, 1, -1]^T, \hspace{.2cm} v_3=[1,1,2,-4]^T\}$ be a subspace of $\mathbb{R}^4$ endowed with the usual Euclidean inner product.
- Find the angle between the vectors $v_1, v_2$ and $v_1, v_3$. What do you observe on where $v_1$ belongs?
I calculate: For $v_1, v_2: \hspace{.2cm} \cos{\theta}=\displaystyle\frac{v_1^Tv_2}{||v_1|| \cdot ||v_2||}=0\Rightarrow \theta=\pi/2$ and the same for $v_1, v_3.$ So, does $v_1$ lie on the same line as the other two vectors?
- Find an orthonormal basis for $W$.
I use the Gram-Schmidt process and find the basis $B=\left\{\begin{bmatrix} \displaystyle\frac{\sqrt{3}}{3}\\ \displaystyle\frac{\sqrt{3}}{3} \\ \displaystyle - \frac{\sqrt{3}}{3} \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ \displaystyle\frac{\sqrt{3}}{3} \\ \displaystyle\frac{\sqrt{3}}{3} \\ \displaystyle - \frac{\sqrt{3}}{3} \end{bmatrix}, \begin{bmatrix} \displaystyle\frac{\sqrt{195}}{65} \\ \displaystyle\frac{4\sqrt{195}}{195} \\ \displaystyle\frac{7\sqrt{195}}{195} \\ \displaystyle\frac{11\sqrt{195}}{195} \end{bmatrix} \right\}$. Is there a better basis?
- Find an orthonormal basis for the orthogonal complement $W^{\perp}$ of $W$.
I just calculated the basis of the null space which I found to be $B_{W^{\perp}}=\left\{\begin{bmatrix} \displaystyle\frac{5}{3} \\ \displaystyle - \frac{1}{3} \\ \displaystyle\frac{4}{3} \\ 1 \\ \end{bmatrix}\right\}$
- Find the orthogonal projection of the vector $u=(1, -1, 1, 1)$ on $W$ and $W^{\perp}$
Is this not the sum of the orthogonal projections of $u$ onto all of the vectors of $W$ and $W^{\perp}$?
I am so sorry for writing such a long exercise, but I would appreciate help with these, by either pointing out my mistakes or providing your own solution (particularly on that last one). It is just that it's the first exam I'm taking and I'm nervous that I need to know what I am doing.
The vector $v_1$ is orthogonal to both $v_2$ and $v_3$. So, $v_1$ belongs to the plane orthogonal to $v_2$ and to $v_3$. On the other hand, $v_2$ and $v_3$ are not orthogonal.
When I apply the Gramm-Schmidt process to $\{v_1,v_2,v_3\}$, what I get is the basis $\{w_1,w_2,w_3\}$ equal to$$\left\{\left[\frac1{\sqrt3},\frac1{\sqrt3},-\frac1{\sqrt3},0\right]^T,\left[0,\frac1{\sqrt3},\frac1{\sqrt3},-\frac1{\sqrt3}\right]^T,\left[\frac3{\sqrt{51}},-\frac4{\sqrt{51}},-\frac1{\sqrt{51}},-\frac5{\sqrt{51}}\right]^T\right\}.$$On the other hand, the second and third vectors of the basis that you got are not orthogonal.
The orthogonal projection of $u$ on $W$ is$$(u.w_1)w_1+(u.w_2)w_2+(u.w_3)w_3=\left[-\frac{14}{51},-\frac{38}{51},-\frac{1}{51},\frac{12}{51}\right]^T.$$And the orthogonal projection of $u$ on $W^\perp$ is$$u-\left[-\frac{14}{51},-\frac{38}{51},-\frac{1}{51},\frac{12}{51}\right]^T=\left[\frac{65}{51},-\frac{13}{51},\frac{52}{51},\frac{39}{51}\right]^T.$$