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Let $V = \text{span}\{(1,1,0,-1), (0, 1, 1, -2) (1,2,2,1)\}$

  1. Find an orthonormal basis for V.
  2. Find the projection of $x = (2, 4, 3, -2)$ onto V. Interpret your answer geometrically.

For the most part, I have no problems here. Using Gram-Schmidt, I found a normalised orthonormal basis;

$B = \frac{1}{\sqrt3}\begin{pmatrix}1\\1\\0\\-1\end{pmatrix}, \frac{1}{\sqrt3}\begin{pmatrix}-1\\0\\1\\-1\end{pmatrix}, \frac{1}{\sqrt78}\begin{pmatrix}1\\4\\6\\5\end{pmatrix} $

These are all unit vectors, say $v_1, v_2, v_3$, therefore $\text{Proj}_{V}x = (x\cdot v_1)v_1 + (x\cdot v_2)v_2 + (x\cdot v_3)v_3$

So I find a projection; $\text{Proj}_{V}x = \begin{pmatrix}\frac23\\\frac53\\3\\2\end{pmatrix} $

But what does it mean by "Interpret your answer geometrically"?

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Your subspace $V$ has dimension 3 and you are in a space of dimension 4. To simplify things and make it visual, imagine we were in a space of dimension 2, and your subspace $V$ was a line (dimension 1). An orthogonal projection of a vector $u$ on $V$ would look like the shade that $u$ would make on $V$ if the sun was radiating in a direction totally perpendicular to $V$. So $V$ would be the "ground", $u$ a vector, the projection of $u$ in $V$ would be the length of the shade that $u$ makes on $V$. If you draw it, you will be able to explain it in a more scientific way, I hope.

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    I understand how to visualise a projection, and that V can encompass an entire 3 dimensional space. But I don't see to get there from $\mathbb{R}^4$. – 2017-02-08
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    I don't think anybody can "see" in $\mathbb{R}^4$ , so I think you can just explain how it would look like in a lower dimension space. And refer to the triangle in which the original vector is the hypothenuse and the projection is one of the sides, something like this? – 2017-02-08