*Previously* we saw that the determinant of a matrix is constant on the conjugacy class of that matrix, so the determinant of a linear map #L:V\to V# is well defined as the determinant of a matrix determining #L#. The characteristic polynomial has the same property:

If #V# is a finite-dimensional vector space and #L:V\to V# a linear map, then the characteristic polynomial #p_A(x)# of the matrix #A = L_\alpha# does not depend on the choice of the basis #\alpha# for #V#.

We can speak of the **characteristic polynomial** of #L#. We use the notation #p_L(x)#.

In particular the trace and determinant of a matrix #A# determining #L# are numbers that do not depend on the choice of the matrix #A#. Hence, we can speak of the **trace** and the **determinant** of the linear map #L#. We also write #\text{tr}(L)# instead of # \text{tr}(A)# and #\det(L)# instead of #\det(A)#.

If #\beta # is a different basis, then #L_{\alpha}# and #L_{\beta}# are *conjugate*: the invertible matrix #T = {}_\beta I_\alpha# satisfies \(L_\beta = T\, L_\alpha T^{-1}\), so \[\begin{array}{rcl}p_{L_\beta}(x)&=&\det(L_\beta-x\cdot I)\\&=&\det\left( T\, L_\alpha T^{-1}-x\cdot I\right) \\& =&\det\left( T\,\left( L_\alpha -x\cdot I\right) \,T^{-1}\right)\\ &=& \det( T)\cdot \det(L_\alpha-I)\cdot \det(T^{-1}) \\ &=& \det(L_\alpha-x\cdot I)\cdot \det( T\cdot T^{-1}) \\ &=& \det(L_\alpha-x\cdot I)\\&=&p_{L_\alpha}(x)\end{array}\]This calculation shows that #L_{\alpha}# and #L_{\beta}# have the same characteristic polynomial. According to *Characteristic polynomial of a matrix* #L_{\alpha}# and #L_{\beta}# then also have the same determinant and trace, being equal to the constant term, respectively #(-1)^{n-1}# times the coefficient of #x^{n-1}# in their characteristic polynomial of degree #n#.

Let #P_2# be the vector space of all polynomials of degree no higher than #2# and let #L:P_2\to P_2# be the linear map defined by \[L(p(x)) = x\cdot\frac{\dd}{\dd x}(p(x))\]The images of the elements of the basis #\alpha = \basis{1,x,x^2}# are\[\begin{array}{rcl}L(1)&=&0\\ L(x) &=& x\\ L(x^2) &=& 2x^2\end{array}\]The matrix of #L# in relation to #\alpha# is the diagonal matrix with #0,1,2# on the diagonal. Hence, the characteristic polynomial of #L# is \[p_L(x)=-x\cdot (1-x)\cdot (2-x)=-x^3+3x^2+2x\]

Consider the matrices\[A = \matrix{0&1\\ 0&0}\phantom{xxx}\text{ and }\phantom{xxx}B = \matrix{0&0\\ 0&0}\]Both matrices have the characteristic polynomial #x^2#. But they are not conjugate, since the conjugacy class of the zero matrix #B# only consists of #B# itself while #A\ne B#. Hence, the conjugacy class of a matrix is not uniquely defined by the characteristic polynomial. In other words, it may happen that the characteristic polynomials of two linear maps #V\to V# are equal, but no two bases can be found for #V#, such that the matrix of the linear map with respect to one basis is equal to the matrix of the other linear map with respect to the second basis for #V#.

Let #P_2# be the vector space of all polynomials of degree at most #2# in #x# and let #L:P_2\to P_2# be the linear map defined by \[L(p(x)) = (3 x+2)\cdot\frac{\dd}{\dd x}(p(x))\]

Determine the characteristic polynomial of #L# as a function of #t#.

Solution \( p_L(t)=\) \(-t^3+9 t^2-18 t\)

The images of the elements of the basis #\alpha = \basis{1,x,x^2}# are \[\begin{array}{rcl}L(1)&=&0\\ L(x) &=& 3 x+2\\ L(x^2) &=& 6 x^2+4 x\end{array}\] This shows that the matrix of #L# with respect to #\alpha# is

\[L_\alpha = \matrix{0 & 2 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & 6 \\ }\] The characteristic polynomial of #L# is \[ p_L(t)=\det(L_\alpha-t\cdot I_3) = \left(6-t\right)\cdot \left(t-3\right)\cdot t=-t^3+9 t^2-18 t\]