If $V$ is an orthogonal matrix, and $V^T e=0$, then is it true that the sum of any subset of the columns also $= 0$?

I am struggling to prove this, but I'm unable to come up with a counterexample....


closed as unclear what you're asking by Brian Borchers, Lord Shark the Unknown, max_zorn, mechanodroid, user91500 Sep 16 '18 at 9:18

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    $\begingroup$ It's not clear what you're asking. Could you please provide an example? $\endgroup$ – Adrian Keister Sep 16 '18 at 0:27

If $V$ is an orthogonal matrix, then by definition

$V^TV = VV^T = I, \tag 1$

the identity matrix. Thus

$V^T = V^{-1}, \tag 2$

and with

$V^Te = 0, \tag 3$

we have

$e = Ie = (VV^T)e = V(V^Te) = V(0) = 0; \tag 4$

thus any vector $e$ satisfying (3) vanishes, which comes as no surprise, since the columns of $V^T$ are linearly independent: if we write $V^T$ in columnar form as

$V^T = [W_1 \; W_2 \; \ldots \; W_n], \tag 5$

the $W_i$ being the columns of $V^T$, then taking any vector

$f = \begin{pmatrix} f_1 \\ f_2 \\ \vdots \\ f_n \end{pmatrix}, \tag 6$

we have

$V^T f = [W_1 \; W_2 \; \ldots \; W_n]\begin{pmatrix} f_1 \\ f_2 \\ \vdots \\ f_n \end{pmatrix} = \displaystyle \sum_1^n f_i W_i; \tag 7$

that is, the entries of the vector $f$ form the coefficients of the linear expansion

$\displaystyle \sum_1^n f_i W_i; \tag 8$

of $V^Tf$ in terms of the columns of $V^T$; since the columns of an orthogonal matrix are, in fact, orthonormal as vectors (which follows from (1)), and hence linearly independent, the expresion (8) may only vanish if each of the $f_i$ does; that is, only if $f = 0$. (4) is nothing more than a condensed, abstract presentation of these remarks in more algebraic terms.

Since there are no non-zero vectors such as $e$ or $f$ satisfying (3), we may conclude from (5)-(8) that no sum of the $W_i$ may vanish.

Note Added in Edit, Sunday 16 September 2018 8:27 AM PST: To affirm further that the $f_i = 0$, $1 \le i \le n$, note that the $W_i$ orthonormal means

$\langle W_i, W_j \rangle = \delta_{ij}, \; 1 \le i, j \le n; \tag 9$

thus taking $\langle W_j, \cdot \rangle$ with each side of

$\displaystyle \sum_1^n f_i W_i = 0 \tag{10}$

we obtain

$f_j = \displaystyle \sum_1^n f_i \delta_{ij} = \sum_1^n f_i \langle W_j, W_i \rangle = \langle W_j, \sum_1^n f_i W_i \rangle = \langle W_j, 0 \rangle = 0 \tag{11}$

for all $f_j$, $1 \le j \le n$. A simple sum of the $W_i$ occurs if we only take each $f_i = 0, 1$; but we see here this is impossible unless all the $f_i = 0$.

Furthermore, I should probably point out that for $V$, and hence $V^T$, orthogonal matrices, and $e = (1, 1, \ldots, 1)^T$, there is no way that

$V^T e = 0 \tag{12}$

can possible bind; so in effect, the hypothesis of the problem are self-contradictory; vacuous.

I mean, for sure the implication

$V^Te = 0 \Longrightarrow V_m^T e = 0, \tag{13}$

is true; but the hypothesis being false renders indeterminate the truth value of any conclusion. End of Note.

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    $\begingroup$ and +1 for clear explanation $\endgroup$ – Ahmad Bazzi Sep 16 '18 at 1:47
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    $\begingroup$ @AhmadBazzi: and I thank you sir! $\endgroup$ – Robert Lewis Sep 16 '18 at 1:51
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    $\begingroup$ You seem to be saying that it is actually impossible for a sum of those columns to be $0$? $\endgroup$ – nundo Sep 16 '18 at 14:36
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    $\begingroup$ @nundo: yes, that is correct. If $\sum f_i W_i = 0$ for non-zero, orthogonal vectors $W_i$, then all the $f_i = 0$. Think of two or three orthonormal basis vectors in $\Bbb R^2$ or $\Bbb R^3$, respectively. Cheers! $\endgroup$ – Robert Lewis Sep 16 '18 at 15:25
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    $\begingroup$ @nundo: I added a few notes to my answer. Hope they clarify! $\endgroup$ – Robert Lewis Sep 16 '18 at 16:06

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