How to calculate the determinant of a linear map?
I am having a hard time trying to figure out how to calculate the determinant of this linear map:
Let $P_1$ be a vector space over a field $R$. Define a linear map $l :P_1 \rightarrow P_1$ where $l(am_1+bm_0):=(a+b)m_1+( −3 a−5 b )m_0$ for $a,b \in R$.
How should I proceed? It is linear map that maps something from the polynomial vector space $P_1$ to $P_1$. But what would be the approach to calculate the determinant? I don't want the answer, I just want to know how to proceed.
(Going off your comment) To calculate the matrix representation of a linear map $f : V \rightarrow W$ of vector spaces $V$ and $W$ with bases $v_1, ..., v_n$ and $w_1, ..., w_m$, the matrix will be the unique $A_{i,j}$ such that $f(v_j) = \sum_{i = 1}^m A_{i, j} w_i$.
In your example we have $P_1$ a two dimensional vector space with basis $m_0, m_1$. $f(m_1) = m_1 - 3 m_0$ and $f(m_0) = m_1 - 5 m_0$. So $A_{00} = -5$, $A_{10} = 1$, $A_{01} = -3$, and $A_{11} = 1$. So your matrix looks like this:
$$\begin{bmatrix} -5 & -3 \\ 1 & 1 \end{bmatrix} $$
You may know, there is a correspondence between linear maps and matrices. Linear maps are determined by what they do to basis elements, and matrices are a way of encoding the values of a linear map on a basis.
The determinant for 2-by-2 matrices has a simple formula: $A_{00} A_{11} - A_{10} A_{01}$. So here, that's $-2$.
Notice how the way the linear map was given did not tell you right away what the $A_{ij}$ were. To do this, you must plug in a basis element like $m_0$ and express it as a linear sum of the basis elements in the range (here that's $m_0$ and $m_1$). That is what gives you the coefficients in each collumn. Another way of saying that is that each collumn tells you where the $i$th basis vector goes in the range space.
Here the formula for determinant was simple, but in general a calculation will require an inductive approach, where you calculate the determinant of $n$ $n-1 \times n -1 $ matrices to find the determinant of a matrix.