Final answer:
The dimension of the range of the given linear operator L is 2. This is determined by finding the rank of the matrix associated with L, which corresponds to the dimension of its column space.
Step-by-step explanation:
Dimension of the Range of a Linear Operator
To find the dimension of the range of the linear operator L defined by its action on a vector from ℝ³ to ℝ³, we can look at the matrix that represents this operator. The matrix form of L is:
\[
\begin{bmatrix}
9 & 38 & -9 \\
0 & 19 & 0 \\
27 & 0 & -27
\end{bmatrix}
\]
To find the range of L, we need to find the span of the columns of the matrix, which essentially means we're looking for the column space. The dimension of the column space is equal to the rank of the matrix. To determine the rank, we can perform elementary row operations to bring the matrix to its row echelon form (or reduced row echelon form). Observing the matrix above, we can realize that the second column is a scalar multiple of the second row, and the third row can be generated by linearly combining the first and second row. This gives us a hint that the rank is likely to be 2 because after performing row reduction, we expect to have two non-zero rows.
The actual row reduction would confirm this, but since our problem doesn't explicitly require showing the steps, we can state that the dimension of the range of L, which equals the rank of the matrix, is 2.
This can also be reasoned by observing that no combination of the first and third columns can produce the second column, which shows that we have at least two linearly independent columns and hence, at least a dimension of 2 for the column space. Since there are only three columns in total and we see dependencies between them, the dimension cannot be 3, reinforcing that it is indeed 2.
The concept of dimensional analysis provided in the information is useful in physical contexts to check the consistency of equations involving physical quantities, but it is not directly applicable to the linear algebra problem at hand.