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One of my review questions is:
True or false: if $E$ is a subspace of $V$, then $\dim(E)+\dim(E^\perp)=\dim(V)$? Justify.
My thought are that this is true, since E and E perp would sort of account for entire space of V, meaning there dimensions added would be the same as Dim V, or at least I am visualizing this to be true. I was wondering if anyone knew of a nice proof of this? I think I understand but I cannot really articulate it, so was hoping someone could help me.
The following is a sketch of the result that may help you out. I assume that you mean that $V$ is a vector space and the question is that if $E$ is a subspace of $V$ (not $V^4$) then the dimension of $E$ plus the dimension of its orthogonal complement equal the dimension of $V$. $V$ has an orthonormal basis by the Gram-Schmidt result. So does $E$. Call these $e_1, ..., e_{\dim{(E)}}$. We can take $k = \dim{V}-\dim{(E)}.$ We have at least $k$ vectors in the basis of $V$ that aren't spanned by the $e_i$ basis vectors of $E$. By the Gram-Schmidt procedure, we can construct these to be orthogonal to the orthonormal basis vectors of $E$. Call these new vectors $v_1, ..., v_k$. Given that the basis of $v$ are the $v_i$ and $e_i$ vectors, the dimension of $V$ is clearly $\dim{(E)} + K$.
It is pretty straightforward to show that the $v_i$ vectors space the orthogonal complement of $E$. Though let me know if it is not.
Edit: RE: $V$ versus $V^4$. If $V=\mathbb{R}$ and $E$ = $\mathbb{R}^2$, then $\dim{(E)} + \dim{(E^\perp)} = 4>\dim{(V)} = 1$.
Edit 2: I assumed, also, that we are dealing with finite-dimensional vector spaces here.
If $X$ is a vector space and if $Y,Z$ are subspaces of $X$ with $X=Y \oplus Z$, then we always have
$\dim X= \dim Y + \dim Z$
