Let $\{\mathbf{v}_1, \mathbf{v}_2\}$ be a basis of the vector space $\R^2$, where
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
1
\end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] The action of a linear transformation $T:\R^2\to \R^3$ on the basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ is given by
\begin{align*}
T(\mathbf{v}_1)=\begin{bmatrix}
2 \\
4 \\
6
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
0 \\
8 \\
10
\end{bmatrix}.
\end{align*}

Find the formula of $T(\mathbf{x})$, where
\[\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2.\]

Solution.

We give two solutions.

Solution 1 (linear combination)

Since we know the values of $T$ on the basis vectors $\mathbf{v}_1, \mathbf{v}_2$, if we express the vector $\mathbf{x}$ as a linear combination of $\mathbf{v}_1, \mathbf{v}_2$, we can find $F(\mathbf{x})$ by the linearity of the linear transformation $T$.

So let us find the scalars $c_1, c_2$ such that
\[\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2.\] We write this as
\begin{align*}
\begin{bmatrix}
x \\
y
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}.
\end{align*}

The matrix $\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}$ is invertible (as its determinant is $-2$) and its inverse matrix is
\[\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}=\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}.\]

Thus, we have
\begin{align*}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}&=\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\frac{1}{2}\begin{bmatrix}
x+y \\
x-y
\end{bmatrix}
\end{align*}
Therefore, we obtain the linear combination
\[\mathbf{x}=\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2.\]

Now we compute $T(\mathbf{x})$ as follows.
We have
\begin{align*}
T(\mathbf{x})&=T\left(\frac{1}{2}(x+y)\mathbf{v}_1+\frac{1}{2}(x-y)\mathbf{v}_2 \right)\\[6pt] &=\frac{1}{2}(x+y)T(\mathbf{v}_1)+\frac{1}{2}(x-y)T(\mathbf{v}_2) && \text{by linearity of $T$}\\[6pt] &=\frac{1}{2}(x+y)\begin{bmatrix}
2 \\
4 \\
6
\end{bmatrix}+\frac{1}{2}(x-y)\begin{bmatrix}
0 \\
8 \\
10
\end{bmatrix} \\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix}.
\end{align*}
Hence the formula is
\[T(\mathbf{x})=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix}.
\]

Solution 2 (Matrix representation)

In the second solution, we use the matrix representation for the linear transformation $T$.
Let $A$ be the matrix of $T$ with respect to the standard basis $\{\begin{bmatrix}
1 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1
\end{bmatrix}\}$ of $\R^2$.
Thus, we have $T(\mathbf{x})=A\mathbf{x}$ by definition.

To find the matrix $A$, we compute
\begin{align*}
A\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}&=A\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2 \\
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
A\mathbf{v}_1 & A\mathbf{v}_2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}
\end{align*}
It follows that we have
\begin{align*}
A&=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}^{-1}\\[6pt] &=\begin{bmatrix}
2 & 0 \\
4 & 8 \\
6 &10
\end{bmatrix}\frac{1}{2}\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & 1 \\
6 & -2 \\
8 &-2
\end{bmatrix}.
\end{align*}

We now be able to find $T(\mathbf{x})$ as follows.
We have
\begin{align*}
T(\mathbf{x})&=A\mathbf{x}\\[6pt] &=\begin{bmatrix}
1 & 1 \\
6 & -2 \\
8 &-2
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
x+y \\
6x-2y \\
8x-2y
\end{bmatrix},
\end{align*}
which is, of course, the same formula that we obtained in solution 1.

Related Question.

A similar problem for a linear transformation from $\R^3$ to $\R^3$ is given in the post “Determine linear transformation using matrix representation“.

Instead of finding the inverse matrix in solution 1, we could have used the Gauss-Jordan elimination to find the coefficients.
See the post “Give a formula for a linear transformation if the values on basis vectors are known” for a similar problem and its solution using this alternative method.

The post Give a Formula For a Linear Transformation From $\R^2$ to $\R^3$ first appeared on Problems in Mathematics.

Let $T$ be the linear transformation from the $3$-dimensional vector space $\R^3$ to $\R^3$ itself satisfying the following relations.
\begin{align*}
T\left(\, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \qquad T\left(\, \begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix} \, \right) =
\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix}, \qquad
T \left( \, \begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Then for any vector
\[\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\in \R^3,\] find the formula for $T(\mathbf{x})$.

 

 
We give two solutions using two different methods.

Solution 1 using the matrix representation.

The first solution uses the matrix representation of $T$.
Let $A$ be the matrix representation of the linear transformation $T$ with respect to the standard basis of $\R^3$.
Then we have $T(\mathbf{x})=A\mathbf{x}$ by definition.
We determine the matrix $A$ as follows.
We have
\begin{align*}
A\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}&= \begin{bmatrix}
A\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & A\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & A\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
T\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & T\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & T\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}.
\end{align*}
It follows that we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}.
\end{align*}

Let us find the inverse matrix by using augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
1 & 3 & 1 & 0 & 1 & 0 \\
1 & 5 & 2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 3 & 2 & -1 & 0 & 1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1-2R_2 \\R_3-3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & -1 & 2 & -3 & 1 \\
\end{array} \right] \xrightarrow{-R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1+2R_3\\ R_2-R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 &4 & -2 \\
0 & 1 & 0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right].
\end{align*}
Thus we obtain the inverse matrix
\[\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}=\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix},\] and hence we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix}
=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}.
\end{align*}
Using the relation $T(\mathbf{x})=A\mathbf{x}$, the formula for $T(\mathbf{x})$ is obtained as follows.
\begin{align*}
T(\mathbf{x})&=A\mathbf{x}\\
&=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Solution 2 using a linear combination and linearity.

The second method is to find the linear combination
\[\mathbf{x}=c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix}\] and use the linearity of the linear transformation.
To find the coefficients $c_1, c_2, c_3$, we consider the augmented matrix
\[ \left[\begin{array}{rrr|r}
1 & 2 & 0 & x \\
1 &3 & 1 & y \\
1 & 5 & 2 & z
\end{array} \right]\] and we reduce this matrix by elementary row operations.
The reduction operations are exactly the same as in solution 1 and we obtain
\begin{align*}
c_1&=-x+4y-2z\\
c_2&=x-2y+z\\
c_3&=-2x+3y-z.
\end{align*}

Therefore, we have by the linearity of $T$
\begin{align*}
T(\mathbf{x})&=T\left(\,c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=c_1T\left(\,\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}\, \right)+c_2T\left(\,\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}\, \right)+c_3T\left(\,\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=(-x+4y-2z)\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}+(x-2y+z)\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix} +(-2x+3y-z)\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Related Question.

A similar problem with a linear transformation from $\R^2$ to $\R^3$ is given in the post
“Give a formula for a linear transformation from $\R^2$ to $\R^3$“.

The post Determine linear transformation using matrix representation first appeared on Problems in Mathematics.

Let $B=\{\mathbf{v}_1, \mathbf{v}_2 \}$ be a basis for the vector space $\R^2$, and let $T:\R^2 \to \R^2$ be a linear transformation such that
\[T(\mathbf{v}_1)=\begin{bmatrix}
1 \\
-2
\end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix}
3 \\
1
\end{bmatrix}.\]

If $\mathbf{e}_1=\mathbf{v}_1+2\mathbf{v}_2 \text{ and } \mathbf{e}_2=2\mathbf{v}_1-\mathbf{u}_2$, where $\mathbf{e}_1, \mathbf{e}_2$ are the standard unit vectors in $\R^2$, then find the matrix of $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$.
 

Solution.

The matrix representation $A$ of the linear transformation $T$ with respect to the basis $\{\mathbf{e}_1, \mathbf{e}_2 \}$ is given by
\[A=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2)
\end{bmatrix}. \tag{*}\]

To find the vectors $T(\mathbf{e}_1), T(\mathbf{e}_2)$ we use the linearity of $T$.
We have
\begin{align*}
T(\mathbf{e}_1)&=T(\mathbf{v}_1+2\mathbf{v}_2)\\
&=T(\mathbf{v}_1)+2T(\mathbf{v}_2) \qquad \text{ (by the linearity of $T$)}\\
&=\begin{bmatrix}
1 \\
-2
\end{bmatrix} +2\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
7 \\
0
\end{bmatrix}.
\end{align*}

Also we have
\begin{align*}
T(\mathbf{e}_2)&=T(2\mathbf{v}_1-\mathbf{u}_2)\\
&=2T(\mathbf{v}_1)-T(\mathbf{u}_2)\\
&=2\begin{bmatrix}
1 \\
-2
\end{bmatrix}-\begin{bmatrix}
3 \\
1
\end{bmatrix}\\
&=\begin{bmatrix}
-1 \\
-5
\end{bmatrix}.
\end{align*}

Therefore the matrix $A=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2)
\end{bmatrix}$ of the linear transformation $T$ is given by
\[A=\begin{bmatrix}
7 & -1\\
0& -5
\end{bmatrix}.\] Click here if solved 17

The post Matrix Representation of a Linear Transformation of the Vector Space $R^2$ to $R^2$ first appeared on Problems in Mathematics.