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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.

User Buggy B
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1 Answer

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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.
User Fojeeck
by
7.5k points
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