Orthonormal basis and linear combinations

**pakman** · May 29th 2008, 04:09 PM

Let $\theta$ be a fixed real number and let $\textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right]$ and $\textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]$ .

Given a vector $\textbf{y}$ in $R^2$ , write it as a linear combination $c_1\textbf{x}_1+c_2\textbf{x}_2$ .

It also asked me to prove that $(\textbf{x}_1,\textbf{x}_2)$ was an orthonormal basis for $R^2$ so I did that.

However,
$\textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2$ . $<\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right]$ and $<\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]$

Can someone assist me on where to go from here? Thank you!

**Moo** · May 30th 2008, 12:31 AM

Hello,

Originally Posted by pakman

Let $\theta$ be a fixed real number and let $\textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right]$ and $\textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]$ .

Given a vector $\textbf{y}$ in $R^2$ , write it as a linear combination $c_1\textbf{x}_1+c_2\textbf{x}_2$ .

It also asked me to prove that $(\textbf{x}_1,\textbf{x}_2)$ was an orthonormal basis for $R^2$ so I did that.

However,
$\textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2$ . $<\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet a\end{array}\right]$ and $<\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array }\right]$

Can someone assist me on where to go from here? Thank you!

I'm not sure what you're trying to get ?

Did you prove that it was an orthonormal basis ?

If you're looking for $<y,x_1>$ and $<y,x_2>$ , then... :

$<y,x_1>=<c_1x_1+c_2x_2,x_1>$

Because the scalar product is bilinear, this is equal to :

$<c_1x_1,x_1>+<c_2x_2,x_1>=c_1<x_1,x_1>+c_2<x_2,x_1 >$

But $<x,x>=||x||^2$ . Here, $||x_1||=1$ because it's an orthonormal basis.

And $<x_2,x_1>=0$ because it's an orthonormal basis.

Therefore, $<y,x_1>=c_1 \cdot 1^2+c_2 \cdot 0=c_1$

If you prefer, the scalar product of y with the n-th vector of an orthonormal basis represents the n-th coordinate of y in this basis. (this is extra info and I'm not sure I explained it well

)

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