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.

Respuesta :

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

[tex]\mathbf V\mathbf c=\mathbf 0[/tex]


Multiplying both sides by [tex]\mathbf V^\top[/tex], we have

[tex]\mathbf V^\top(\mathbf V\mathbf c)=(\mathbf V^\top\mathbf V)\mathbf c=\mathbf 0[/tex]

More explicitly, we're writing

[tex]\mathbf V=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\mathbf v_3\end{bmatrix}[/tex]

Multiply both sides by [tex]\mathbf V^\top[/tex] and the left hand side can be written as

[tex]\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}[/tex]

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

[tex]\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}[/tex]

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