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*3` 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\\eq 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.