# Math Help - Orthonormal basis and linear combinations

1. ## Orthonormal basis and linear combinations

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!

2. 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 $$ and $$, then... :

$=$

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

$+=c_1+c_2$

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

And $=0$ because it's an orthonormal basis.

Therefore, $=c_1 \cdot 1^2+c_2 \cdot 0=c_1$

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 )