Blogs, Linear Algebra

Standard Matrix of a Transformation

Here we will be continuing to look at linear transformations. In this case, we saw that any linear transformation between finite dimensional vector spaces could be represented using matrix multiplication. As such, we will want to be able to find matrices that would represent a given transformation.

Our Transformation

Let \(T_{1}:\mathbb{R}^{2} \to \mathbb{R}^{2}\) be given by \(T_{1}(x,y)=(2x+y,5x+3y)\) and \(T_{2}: \mathbb{R}^{2} \to \mathbb{R}^{2}\) be given by \(T_{2}(x,y)=(x+2y,3x+4y)\). Then find the standard matrix for \(T_{1}\), \(T_{2}\), \(T_{1} \circ T_{2}\) and \(T_{1}^{-1}\).

Matrix of \(T_{1}\)

In order to find the matrix for a linear transformation, we want to find the image of a basis of \(\mathbb{R}^{2}\). Since we are finding the standard matrix, we want to find the image of the standard basis \(\left\{(1,0),(0,1)\right\}\). We first find the image of \((1,0)\).
\begin{align*}
T(1,0)&=(2*1+0,5*(1)+3(0)) \\
&=(2,5).
\end{align*}
We will then find the image of \((0,1)\). Here we get
\begin{align*}
T(0,1)&=(2(0)+1,5(0)+3(1)) \\
&=(1,3).
\end{align*}

Now that we have the image of the basis elements, we can find the standard matrix by turning these vectors into columns of a matrix. This then gives us the matrix
\begin{align*}
A_{1}=\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\end{align*}
where
\begin{align*}
T_{1}(\mathbf{v})=A_{1}\mathbf{v}.
\end{align*}

Matrix of \(T_{2}\)

Now that we have found the standard matrix for \(T_{1}\), we use a similar process to find the standard matrix for \(T_{2}\). We, therefore, start by finding the image of the standard basis. Recall that \(T_{2}(x,y)=(x+2y,3x+4y)\). We then have
\begin{align*}
T_{2}(1,0)&=(1+2(0),3(1)+4(0)) \\
&=(1,3).
\end{align*}
We also find that
\begin{align*}
T_{2}(0,1)&=(1(0)+2(1),3(0)+4(1)) \\
&=(2,4).
\end{align*}

Hence, we find that the standard matrix is
\begin{align*}
A_{2}=\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}.
\end{align*}
This tells us that
\begin{align*}
T_{2}(\mathbf{v})=A_{2}\mathbf{v}.
\end{align*}

Matrix of \(T_{1} \circ T_{2}\)

In order to find the standard matrix for \(T_{1} \circ T_{2}\), we could find the composition of the functions, then follow the same process we had for the last two. However, we also know that, in order to find the standard matrix of \(T_{1} \circ T_{2}\), we can just find \(A_{1} \cdot A_{2}\). Therefore, the standard matrix will be
\begin{align*}
\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}=\begin{bmatrix}
5 & 8 \\
14 & 22
\end{bmatrix}
\end{align*}

Hence, we have that
\begin{align*}
T_{1} \circ T_{2} (\mathbf{v})=\begin{bmatrix} 5 & 8 \\ 14 & 22 \end{bmatrix} \mathbf{v}.
\end{align*}

Matrix of \(T_{1}^{-1}\)

Again, we could find the matrix of the inverse of \(T_{1}\) by first finding the inverse of the transformation. However, we also know that we can do so by finding the inverse of the matrix \(A_{1}\). In order to do this, we can use a calculator. However, we will augment the matrix with the identity and row reduce to find the inverse. We then get
\begin{align*}
\begin{bmatrix}
2 & 1 & 1 & 0\\
5 & 3 & 0 & 1
\end{bmatrix} \to \begin{bmatrix}
1 & .5 & .5 & 0\\
5 & 3 & 0 & 1
\end{bmatrix} \\
\to \begin{bmatrix}
1 & .5 & .5 & 0\\
0 & .5 & -2.5 & 1
\end{bmatrix} \\
\to \begin{bmatrix}
1 & .5 & .5 & 0\\
0 & 1 & -5 & 2
\end{bmatrix} \\
\to \begin{bmatrix}
1 & 0 & 3 & -1\\
0 & 1 & -5 & 2
\end{bmatrix}
\end{align*}

Hence, we find that
\begin{align*}
T_{1}^{-1}(\mathbf{v})=\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}.
\end{align*}

Conclusion

Here, we have seen that for linear transformations we can use matrix multiplication to define the image of vectors under the transformation. We have also seen that if we find the image of some transformations, then we can find the compositions and inverse of these by finding the multiplication or inverse of their matrices.

As always, I hope this helps you with your studies and that you have enjoyed the process. If you did, make sure to share this post with others that may find this useful as well.

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