Question

Let [v1,v2,v3] be a set of nonzero vectors in r^m such that the (transpose of vi)*vj = 0 when i is not equal to j. show that the set is linearly independent.

Answer 1

Let $\mathbf V$ be the $m\times3$ matrix whose columns are $\mathbf v_1,\mathbf v_2,\mathbf v_3$, and let $\mathbf c$ be the vector whose components are the constants $c_1,c_2,c_3$. Now consider the matrix equation

$\mathbf V\mathbf c=\mathbf 0$


Multiplying both sides by $\mathbf V^\top$, we have

$\mathbf V^\top(\mathbf V\mathbf c)=(\mathbf V^\top\mathbf V)\mathbf c=\mathbf 0$

More explicitly, we're writing

$\mathbf V=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}$

Multiply both sides by $\mathbf V^\top$ and the left hand side can be written as

$\mathbf V^\top\mathbf V=\begin{bmatrix}{\mathbf v_1}^\top\\{\mathbf v_2}^\top\\{\mathbf v_3}^\top\end{bmatrix}\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}=\begin{bmatrix}{\mathbf v_1}^\top\mathbf v_1&{\mathbf v_1}^\top\mathbf v_2&{\mathbf v_1}^\top\mathbf v_3\\{\mathbf v_2}^\top\mathbf v_1&{\mathbf v_2}^\top\mathbf v_2&{\mathbf v_2}^\top\mathbf v_3\\{\mathbf v_3}^\top\mathbf v_1&{\mathbf v_3}^\top\mathbf v_2&{\mathbf v_3}^\top\mathbf v_3\end{bmatrix}$

We're told that ${\mathbf v_i}^\top\mathbf v_j=0$ whenever $i\neq j$, so we're left with

$\mathbf V^\top\mathbf V=\begin{bmatrix}\|\mathbf v_1\|^2&0&0\\0&\|\mathbf v_2\|^2&0\\0&0&\|\mathbf v_3\|^2\end{bmatrix}$

Each of $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are nonzero, which means their norms are nonzero, which necessarily implies that $\mathbf c=0$, and so the vectors $\mathbf v_1,\mathbf v_2,\mathbf v_3$ must necessarily be linearly independent.