Linear Operator Determinant

Topics: Algebra - Linear Transformation

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(definition)

Let $T:V\to V$. We define the determinant of the linear operator $T$:

$$\det (T):=\det ([T{]}_{\beta })$$

…where the basis choice is irrelevant.

Example

Let $T:P_2(\mathbb{R})\to P_2(\mathbb{R})$ be defined by $T(f)=f^{\prime }$.

How can we calculate $\det (T)$?

We’ll find $[T{]}_{\beta }$ with $\beta =\{1,x,x^2\}$.

We calculate $T(\beta )$ then build the matrix with the coordinate vectors:

• $T(1)=0=0(1)+0x+0x^2$
• $T(x)=1=1(1)+0x+0x^2$
• $T(x^2)=2x=0(1)+2x+0x^2$

Then:

$$[T{]}_{\beta }=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right)$$

And from there, we can calculate its determinant $\det ([T{]}_{\beta })=0$, so we finally have that $\det (T)=0$.

Properties of the Linear Operator Determinant §

1. $T$ is invertible $⟺$ $\det (T)\ne 0$
2. If $T$ is invertible, then $\det (T^{-1})=[det(T){]}^{-1}$
3. $\det (TU)=\det (T)\det (U)$, where $U:V\to W$
4. If $\lambda$ is a scalar and $\beta$ a basis for $V$, then:
   • $\det (T-\lambda I_V)=\det (A-\lambda I)$
   • …where $A=[T{]}_{\beta }$