Let $W \subset V$ with $\dim V= n$. Suppose $w_1,\ldots,w_m$ is an orthogonal basis for $W$ and $w_{m+1},\ldots,w_n$ is an orthogonal basis for $W^\perp$.

a.) Prove that the combination $w_1,\ldots,w_n$ form an orthogonal basis of $V$.

b.) Show that if $v=c_1w_1+\cdots+c_nw_n$ is any vector in $V$, then its orthogonal decomposition $v=w+z$ is given by $w=c_1w_1 + \cdots+c_mw_m \in W$ and $z=c_{m+1}w_{m+1}+\cdots+c_nw_n\in W^\perp$

How will I be able to prove this?

I know that if $\dim W=m$ and $\dim V=n$, then $\dim W^\perp = n-m$ and since $W\subset V$ then its orthogonal basis $w = w_1,\ldots,w_m$ is an orthogonal complement of $V$ iff $\langle w_i,v_i \rangle = 0$, but how will I be able to prove that using the conditions given in the question?

  • $\begingroup$ I think that either you're lacking some basic understanding of the very definitions and properties of things, or else you're not paying due attention to this exercise, which is as close as being trivial as one can expect. I tell you because this is not the first question about these theme that you ask today... $\endgroup$ – DonAntonio Nov 6 '12 at 4:24
  • $\begingroup$ @DonAntonio you are correct I am. My book only had three pages explaining this topic. It was very vague. $\endgroup$ – diimension Nov 6 '12 at 4:28
  • 2
    $\begingroup$ Perhaps its time to find a better book. $\endgroup$ – EuYu Nov 6 '12 at 4:32

For the first part, let us call the first basis $\mathcal{B}$ and the second $\mathcal{C}$. Then we're interested in showing $\mathcal{B}\cup\mathcal{C}$ is an orthogonal basis. $\mathcal{B}$ itself is orthogonal and so is $\mathcal{C}$. Now because these are basis of orthogonal complements you can show that $w_i \perp w_j$ for $i\le m$ and $j > m$. This will show that the vectors are pairwise orthogonal. Orthogonal (non-zero) vectors always form a linearly independent set.

For the second part, first show that the given decomposition is a valid decomposition. Then suppose that $\mathbf{v}$ can be decomposed in two ways $$\mathbf{v} = \mathbf{w}+\mathbf{z} = \mathbf{w'}+\mathbf{z'}$$ with the $\bf{w}$s in $W$ and the $\bf{z}$s in $W^\perp$. Then what can you say about $$\mathbf{w}-\mathbf{w'} = \mathbf{z}' - \mathbf{z}$$

I suggest that you follow the above and fill in all the details with care. I've skipped over statements which may require proof.

  • $\begingroup$ So $\mathbf{w}-\mathbf{w'} = \mathbf{z}' - \mathbf{z}$ must be equal hence they are 0? Thank you very much EuYu. $\endgroup$ – diimension Nov 6 '12 at 4:30
  • $\begingroup$ Have you proven that orthogonal complements have trivial intersection? If not then that would be a good exercise. $\endgroup$ – EuYu Nov 6 '12 at 4:33
  • $\begingroup$ No, I didnt .. Thanks for the suggestion, I am going to try it. $\endgroup$ – diimension Nov 6 '12 at 4:35
  • $\begingroup$ Sry to bother you on this question again Euyu, but i am having trouble understanding part a (I kind of understand b). I know in order to be an orthogonal complement than every vector in $v\in V$ must be orthogonal to every member of $w\in W$. so $z = v -w $ but what exactly are they asking for in the question? $\endgroup$ – diimension Nov 17 '12 at 9:09
  • $\begingroup$ Well, you know the fact that $\dim W + \dim W^\perp = \dim V$. So in a sense the two orthogonal complements really are complements in the sense that together, they form the whole vector space. Part a is essentially asking you to show this in a more precise manner. If you have a basis for $W$ and a basis for $W^\perp$ then you can put them together and get a basis for the entire vector space $V$. $\endgroup$ – EuYu Nov 17 '12 at 19:34

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