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Let a linear transformation $T:$ $\mathbb{R}^3$ → $\mathbb{R}^3$ rotate a vector around the z-axis by $45^{o}$ followed by an orthogonal projection onto the x-axis. Determine the standard matrix for this transformation from the images of the standard basis vectors.

How will I find $T(e_{i})$'s before the standard matrix?

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    $\begingroup$ What's wrong with just applying the given recipe to the vectors $e_1$, $e_2$, and $e_3$? $\endgroup$ – Marc van Leeuwen Dec 24 '14 at 16:28
  • $\begingroup$ Oh, now I see that was a silly question to ask. $\endgroup$ – Kerem Dec 24 '14 at 16:34
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    $\begingroup$ "there aren't silly questions but there are sillies that don't ask" $\endgroup$ – janmarqz Dec 24 '14 at 16:36
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The standard matrix for $T$ is defined as the matrix $[T(e_1),T(e_2), T(e_3)]$ - so we need to find where $T$ sends the standard basis. Let's follow through on the construction.

First, $T$ rotates the $xy$-plane (aka, $e_1e_2$-plane) by $45^\circ$; this sends $e_1$ to $\frac{\sqrt{2}}{2}e_1+\frac{\sqrt{2}}{2}e_2$; $e_2$ to $-\frac{\sqrt{2}}{2}e_1+\frac{\sqrt{2}}{2}e_2$, and it sends $e_3$ to itself.

Now projecting onto the $x$-axis means killing off the $e_2$- and $e_3$- parts of each vector; so combining with the previous step, \begin{align*} T(e_1)&=\frac{\sqrt{2}}{2}e_1\\ T(e_2)&=-\frac{\sqrt{2}}{2}e_1\\ T(e_3)&=0\end{align*}

Putting this all together we can write the standard matrix for $T$ as $$T=\left( \begin{array}{ccc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right).$$

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