Characteristic polynomial of a linear map

Matrix calculus: Matrices and coordinate transformations

Characteristic polynomial of a linear map

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 $n$ .

Example 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 $p_L(x)=-x\cdot (1-x)\cdot (2-x)=-x^3+3x^2+2x$

UniquenessConsider 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 $L(p(x)) = (-2 x-9)\cdot\frac{\dd}{\dd x}(p(x))$
Determine the characteristic polynomial of $L$ as a function of $t$ .

$p_L(t)=$ $-t^3-6 t^2-8 t$

The images of the elements of the basis $\alpha = \basis{1,x,x^2}$ are $\begin{array}{rcl}L(1)&=&0\\ L(x) &=& -2 x-9\\ L(x^2) &=& -4 x^2-18 x\end{array}$ This shows that the matrix of $L$ with respect to $\alpha$ is
$L_\alpha = \matrix{0 & -9 & 0 \\ 0 & -2 & -18 \\ 0 & 0 & -4 \\ }$ The characteristic polynomial of $L$ is $p_L(t)=\det(L_\alpha-t\cdot I_3) = \left(-t-2\right)\cdot t\cdot \left(t+4\right)=-t^3-6 t^2-8 t$

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