Any Vector is a Linear Combination of Basis Vectors Uniquely

Problem 151

Let $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a basis for a vector space $V$ over a scalar field $K$. Then show that any vector $\mathbf{v}\in V$ can be written uniquely as
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3,\]
where $c_1, c_2, c_3$ are scalars.

Since $B$ is a basis for $V$, any vector $\mathbf{v} \in V$ is a linear combination of basis vectors in $B$.
Thus, there exist scalars $c_1, c_2, c_3 \in K$ such that
\[\mathbf{v}=c_1\mathbf{v}_+c_2\mathbf{v}_2+c_3\mathbf{v}_3.\]
Hence such an expression as a linear combination of basis vectors exists.

We now show that such representation of $\mathbf{v}$ is unique.
Suppose we have another representaion
\[\mathbf{v}=d_1\mathbf{v}_+d_2\mathbf{v}_2+d_3\mathbf{v}_3\]
for some scalars $d_1, d_2, d_3 \in K$.

Then we have
\begin{align*}
c_1\mathbf{v}_+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{v}=d_1\mathbf{v}_+d_2\mathbf{v}_2+d_3\mathbf{v}_3.
\end{align*}
Thus, we obtain
\[(c_1-d_1)\mathbf{v}_+(c_2-d_2)\mathbf{v}_2+(c_3-d_3)\mathbf{v}_3=\mathbf{0}.\]
Since $B$ is a basis, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.

Hence the coefficients of the above linear combination must be all zero, and thus we obtain
\[c_1-d_1=0, \quad c_2-d_2=0, \quad c_3-d_3=0,\]
equivalently, we have
\[c_1=d_1, \quad c_2=d_2, \quad c_3=d_3.\]
Therefore two representations of the vector $\mathbf{v}$ are the same, and thus the representation of $\mathbf{v}$ as a linear combination of basis vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ is unique.

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