- #1

RJLiberator

Gold Member

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## Homework Statement

Which of the following sets of vectors in ℂ^3 is an orthogonal set? Which is an orthonormal set? Which is an orthonormal basis?

[tex] \begin{pmatrix}

1/\sqrt{2}\\

0\\

1/\sqrt{2}

\end{pmatrix},

\begin{pmatrix}

-1/\sqrt{2}\\

0\\

1/\sqrt{2}

\end{pmatrix},

\begin{pmatrix}

0\\

(1+i)/\sqrt{2}\\

0

\end{pmatrix}[/tex]

## Homework Equations

## The Attempt at a Solution

Question 1: [/B] If I can conclude that the set is not orthogonal, then I can automatically conclude that the set is not orthonormal.

**My answer: YES.**

Question 2:If I can conclude that the set is orthogonal, but not orthonormal, then I can conclude that the set is not an orthonormal basis.

Question 2:

**My Answer: YES**

I want to get these two understandings out of the way first. These are by definition, is this correct?Next, My work for this particular set:

If we let the first vector = v, second vecotr = w, third vector = z. By simple dot product algebra

v*w=0

v*z=0

w*z = 0

So we

**conclude**that this set is orthogonal.

Now, we check for orthonormality by taking each vector and inner product it with itself. The interesting note here is that when you take the inner produce of z and itself you must use the complex conjugate of z and z itself. so <z*|z> which equals 1, instead of <z|z> which equals -i.

Here, in using that special element of the inner product we

**conclude**that all 3 vectors inner producted with themselves is indeed 1 and so we see orthonormality.

Now, for orthonormal basis, we

**conclude**that is IS an orthonormal basis considering there is a theorem that states that any orthonormal set of N vectors in V is an orthonormal basis for V. Since this set has 3 vectors and it is in ℂ ^ 3 we conclude that this is an orthonormal basis as well.

Is my analysis spot on? I feel like I am starting to finally get the hang of it.