I am working on a past Algebra exam paper and have come across a problem which requires me to write the linear operator associated to a given matrix $M$ in the standard basis of $\mathbb{R}^4$.
What does it mean to 'write a linear operator IN a given basis'?
Thank you.
NB: I have looked at this link, but I am not sure if the concept of writing a matrix in a given basis is synonymous to the concept of writing a linear operator in a given basis.
EDIT:
I think I should be more specific.
The statement of the problem I am working on is as follows:
Let $M$ be the matrix:
$$\begin{bmatrix}
-2 & 3 & 7 & -3 \\
-6 & 1 & 16 & 1 \\
-2 & 1 & 6 & -1 \\
-2 & -1 & 6 & 3 \\
\end{bmatrix}$$
Write the linear operator associated to $M$ in the standard basis of $\mathbb{R}^4$.
The model solution is as follows:
The linear operator associated to $M$ is the linear map given by
$$M
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix} =
\begin{bmatrix}
-2x_1 + 3x_2 + 7x_3 -3x_4 \\
-6x_1 + x_2 + 16x_3 + x_4 \\
-2x_1 + x_2 + 6x_3 – x_4 \\
-2x_1 – x_2 + 6x_3 + 3x_4
\end{bmatrix} $$
My current issue is that I don't fully understand how the model solution is an example of the linear operator written in the standard basis of $\mathbb{R}^4$.
In particular, I'm still trying to grasp the concept of writing a linear operator IN a given basis. What does this really mean?
Best Answer
A linear operator can be written as a matrix in a given basis.
For example, suppose we have the linear operator, T, from $R^2$ to $R^2$ that maps (x, y) to T(x, y)= (x- y, 2y). Since that is from $R^2$ to $R^2$, in can be written as a 2 by 2 matrix: $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$. If we use the "standard basis" for $R^2$, (1, 0) and (0, 1), then (x, y)= x(1,0)+ y(0, 1) so $\begin{bmatrix} x \\ y\end{bmatrix}$ is the representation in the standard basis. The operation, in matrix form is $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} ax+ by \\ cx+ dy\end{bmatrix}$.
We want that to be $T(x, y)= \begin{bmatrix} x- y \\ 2y\end{bmatrix}$ so we must have ax+ by= x- y and cx+ dy= 2y. That is two equations for the four unknowns, a, b, c, and d, but remember this must be true for all x and y. In particular, taking x= 1, y= 0 we get a(1)+ b(0)= a= 1- 0, so a= 1, and c(1)+ d(0)= 2(0), so c= 0. Taking x= 0, y= 1, we get a(0)+ b(1)= 0- 1, so b= -1, and c(0)+ d(1)= 2(1) so d= 2. The matrix representing linear operator, T, in this particular basis, is $\begin{bmatrix} 1 & -1 \\ 0 & 2\end{bmatrix}$. This particular choice of "x= 0, y= 1" and "x= 1, y= 0" makes the calculations especially easy since (0, 1) and (1, 0) are the basis vectors. Notice that $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}$ and $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d \end{bmatrix}$. That is, applying the linear operator to each basis vector in turn, then writing the result as a linear combination of the basis vectors gives us the columns of the matrices as those coefficients.
For another example, let the vector space be the set of all polynomials of degree at most 2 and the linear operator, D, be the differentiation operator. That is, any such "vector" can be written as $P= ax^2+ bx+ c$ and $DP= 2ax+ b$. If we take $\{x^2,x, 1\}$ as basis $ax^2+ bx+ c$ will be written as $\begin{bmatrix} a \\ b \\ c\end{bmatrix}$. Applying the derivative operator to the first "basis vector", $x^2= 1x^2+ 0x+ 0$, gives $2x= 0x^2+ 2x+ 0$ so the first column of the matrix representation is $\begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}$. Applying the derivative operator to the second "basis vector", $x= 0x^2+ 1x+ 0$, gives $1= 0x^2+ 0x+ 1$ so the second column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$. Finally, applying the derivative operator to the third "basis vector", $1= 0x^2+ 0x+ 1$, gives $0= 0x^2+ 0x+ 0$ so the third column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$. The matrix representing the derivative operator in this basis is $\begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$