# Give a Formula for a Linear Transformation if the Values on Basis Vectors are Known ## Problem 159

Let $T: \R^2 \to \R^2$ be a linear transformation.
Let
$\mathbf{u}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ be 2-dimensional vectors.
Suppose that
\begin{align*}
T(\mathbf{u})&=T\left( \begin{bmatrix}
1 \\
2
\end{bmatrix} \right)=\begin{bmatrix}
-3 \\
5
\end{bmatrix},\\
T(\mathbf{v})&=T\left(\begin{bmatrix}
3 \\
5
\end{bmatrix}\right)=\begin{bmatrix}
7 \\
1
\end{bmatrix}.
\end{align*}
Let $\mathbf{w}=\begin{bmatrix} x \\ y \end{bmatrix}\in \R^2$.
Find the formula for $T(\mathbf{w})$ in terms of $x$ and $y$. Add to solve later

## Solution.

Note that the vectors $\mathbf{u}, \mathbf{v}$ are basis vectors for $\R^2$.
Thus we can write the vector $\mathbf{w}$ as a linear combination of $\mathbf{u}$ and $\mathbf{v}$.
Let
$a_1\mathbf{u}+a_2\mathbf{v}=\mathbf{w}.$ We want to determine $a_1$ and $a_2$.
So we consider the augmented matrix
$\left[\begin{array}{rr|r} 1 & 3 & x \\ 2 & 5 &y \end{array} \right].$ Applying the elementary row operations, we obtain a reduced row echelon form matrix for this matrix as follows. (This is the Gauss-Jordan elimination method.)
\begin{align*}
\left[\begin{array}{rr|r}
1 & 3 & x \\
2 & 5 &y
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 3 & x \\
0 & -1 &y -2x
\end{array} \right]\\
\xrightarrow{-R_2}
\left[\begin{array}{rr|r}
1 & 3 & x \\
0 & 1 &2x-y
\end{array} \right] \xrightarrow{R_1-3R_2}
\left[\begin{array}{rr|r}
1 & 0 & x-3(2x-y) \\
0 & 1 &2x-y
\end{array} \right].
\end{align*}
Therefore we have
\begin{align*}
a_1&=x-3(2x-y)=-5x+3y\\
a_2&=2x-y
\end{align*}
and the linear combination is
$\mathbf{w}=(-5x+3y)\mathbf{u}+(2x-y)\mathbf{v}.$ Now we use the linearity of the linear transformation $T$, we calculate $T(\mathbf{w})$ as follows.
\begin{align*}
T(\mathbf{w})&=T((-5x+3y)\mathbf{u}+(2x-y)\mathbf{v})\6pt] &=(-5x+3y)T(\mathbf{u})+(2x-y)T(\mathbf{v})\\[6pt] &=(-5x+3y)\begin{bmatrix} -3 \\ 5 \end{bmatrix}+(2x-y)\begin{bmatrix} 7 \\ 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 15x-9y+14x-7y \\ -25x+15y+2x-y \end{bmatrix}\\[6pt] &=\begin{bmatrix} 29x-16y\\ -23x+14y \end{bmatrix}. \end{align*} Thus the formula is \[T(\mathbf{w})=\begin{bmatrix} 29x-16y\\ -23x+14y \end{bmatrix}.

## Comment.

After you obtained the formula, check the formula with $\mathbf{w}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and see if you get $\begin{bmatrix} -3 \\ 5 \end{bmatrix}$.
If the formula gives a different vector, then you made some mistake.

To be more confident, do the same with $\mathbf{w}=\begin{bmatrix} 3 \\ 5 \end{bmatrix}$.

## Related Question/ Different methods

Instead of using Gauss-Jordan elimination to find the coefficients of the linear combination, we could use the inverse matrix.

Another method to solve this kind of problem is to use the matrix representation of the linear transformation.

See the post “Give a formula for a linear transformation from $\R^2$ to $\R^3$” for a similar problem. Two solutions are given. One uses the inverse matrix and the other uses the matrix representation of a linear transformation. Add to solve later

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