# The Vector $S^{-1}\mathbf{v}$ is the Coordinate Vector of $\mathbf{v}$

## Problem 632

Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.
Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.

Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.

Contents

## Proof.

We first express the vector $\mathbf{v}$ as a linear combination of the basis vectors
$\mathbf{v}=c_1\mathbf{v}_+c_2 \mathbf{v}_2.$ This expression is unique and the coordinate vector of $\mathbf{v}$ with respect to the basis $B$ is defined to be
$[\mathbf{v}]_B =\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}.$

Let
$S^{-1}\mathbf{v}= \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}.$ Or equivalently,
$\mathbf{v}=S\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}.$ Our goal is to show that $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$.

We have
\begin{align*}
c_1\mathbf{v}_+c_2 \mathbf{v}_2&=\mathbf{v}=S\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=x_1\mathbf{v}_1+x_2\mathbf{v}_2.
\end{align*}

Hence
$(x_1-c_1)\mathbf{v}_1+(x_2-c_2)\mathbf{v}_2=\mathbf{0}.$ As $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is linearly independent, we obtain $x_1=c_1$ and $x_2=c_2$.

Therefore,
$S^{-1}\mathbf{v}=\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}=[\mathbf{v}]_B.$

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