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Let $$B = \left\{ \begin{bmatrix} 3\\ -3\\ 0\end{bmatrix},\begin{bmatrix} 2\\ 2\\ -1\end{bmatrix},\begin{bmatrix} 1\\ 1\\ 4\end{bmatrix}\right\},\qquad v =\begin{bmatrix} 5\\ -3\\ 1\end{bmatrix}.$$ a) Verify that $B$ is an orthogonal basis of $\mathbb{R}^3$.
b) Transform $B$ into an orthonormal basis.
c) Write $v$ as linear combination of $B$.

I am really lost in class. I don't even know where to start. Please show steps and answers for the exercise problem so that I can learn. Thank you

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Why don't you write down the definitions of the things you don't understand? – user66345 Apr 8 '13 at 3:24
I am assuming the same thing as user66345, and I have made an edit to that effect. – Zev Chonoles Apr 8 '13 at 3:25
@JoMo: Note that you need to end a line with two spaces in order for the line break to appear. – Zev Chonoles Apr 8 '13 at 3:26
My family is a rather huge set so I think they must be linear dependent... – DonAntonio Apr 8 '13 at 3:29
@DonAntonio Ha ha! Yes, most likely. It is good to know you can depend on your relatives. – 1015 Apr 8 '13 at 3:32


a) Take the dot product of the vectors and it should be zero for any two different vectors.

b) Divide each vector by its length.

c) $v = \alpha b_1 + \beta b_2 + \gamma b_3 $

To determine the $\alpha,\beta$, and $\gamma$, take the dot product of $v$ with $b_1,b_2$, and $b_3$ and note that $b_i.b_i=1,\, \forall i=1..3$. Note that, $\alpha,\beta$, and $\gamma$ are known as the Fourier coefficients.

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For a) But that just shows that they are orthogonal set. But how do we show that B is an orthogonal basis of R^3? – Jo Mo Apr 10 '13 at 2:03

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