📂Linear Algebra

The Relationship Between the Characteristic Polynomials of a Linear Transformation and the Map into the Quotient Space

Theorem¹

Let $V$ be an $n$-dimensional vector space. Let $T : V \to V$ be a linear transformation, $W \le V$ a $T$-invariant subspace, $T|_{W}$ the restriction map, and $\overline{T}$ the linear transformation on the quotient space.

$$ T|_{W} : W \to W \\ \overline{T} : V/W \to V/W $$

Let $f(t), g(t), h(t)$ be the characteristic polynomials of $T, T|_{W}, \overline{T}$, respectively. Then the following holds.

$$ f(t) = g(t)h(t) $$

Corollary

• The characteristic polynomial of $T|_{W}$ divides the characteristic polynomial of $T$

Proof

Let $\gamma = \left\{ v_{1}, \dots, v_{k} \right\}$ be an ordered basis of $W$. Let $\beta = \left\{ v_{1}, \dots, v_{k}, v_{k+1}, \dots, v_{n} \right\}$ be a basis of $V$ extended from $\gamma$. Then a basis of the quotient space is $\alpha = \left\{ v_{k+1} + W, \dots, v_{n}+W \right\}$. Moreover, the following holds.

$$ \begin{bmatrix} T \end{bmatrix}_{\beta} = \begin{bmatrix} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} & A \\ O & B \end{bmatrix} $$

Now we will show that $B = \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha}$. First, rewriting the matrix above,

$$ \begin{bmatrix} T \end{bmatrix}_{\beta} = \left[ \begin{array}{c|c} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} & \begin{array}{ccc} t_{1,k+1} & \cdots & t_{1n} \\ \vdots & \ddots & \vdots \\ t_{k,k+1} & \cdots & t_{kn} \end{array} \\ \hline O & \begin{array}{ccc} t_{k+1,k+1} & \cdots & t_{k+1,n} \\ \vdots & \ddots & \vdots \\ t_{n,k+1} & \cdots & t_{nn} \end{array} \end{array} \right] $$

Let us find the components of the $k+1$-th column of $\begin{bmatrix} T \end{bmatrix}_{\beta}$. To find the matrix representation, we look at which elements the basis is mapped to. Suppose $Tv_{k+1}, \dots, Tn_{n}$ are expressed as the following linear combinations.

$$ Tv_{k+1} = \sum_{i=1}^{n} a_{i,k+1}v_{i},\quad \dots,\quad Tv_{n} = \sum_{i=1}^{n} a_{in}v_{i} $$

Then

$$ \begin{bmatrix} T \end{bmatrix}_{\beta} = \left[ \begin{array}{c|c} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} & \begin{array}{ccc} a_{1,k+1} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{k,k+1} & \cdots & a_{kn} \end{array} \\ \hline O & \begin{array}{ccc} a_{k+1,k+1} & \cdots & a_{k+1,n} \\ \vdots & \ddots & \vdots \\ a_{n,k+1} & \cdots & a_{nn} \end{array} \end{array} \right] $$

Now let us find $\begin{bmatrix}\ \overline{T}(v_{k+1} + W) \end{bmatrix}_{\alpha}$. Since $v_{1}, \dots, v_{k} \in W$, we have $av_{i} \in W\ (1 \le i \le k)$ and $av_{i} + W = W\ (1 \le i \le k)$. Since $W$ is the zero vector in $V/W$,

$$ \begin{align*} \overline{T}(v_{k+1} + W) &= T(v_{k+1}) + W \\ &= \left( \sum_{i=1}^{n}a_{i,k+1}v_{i} \right) + W \\ &= \left( a_{1,k+1}v_{1} + W \right) + \cdots + \left( a_{k,k+1}v_{k} + W \right) \\ &\quad + \left( a_{k+1,k+1}v_{k+1} + W \right) + \cdots + \left( a_{n,k+1}v_{n} + W \right)\\ &= \left( a_{k+1,k+1}v_{k+1} + W \right) + \cdots + \left( a_{n,k+1}v_{n} + W \right)\\ &= a_{k+1,k+1}\left( v_{k+1} + W \right) + \cdots a_{n,k+1}\left( v_{n} + W \right)\\ &= \sum\limits_{i=k+1}^{n}a_{i,k+1}\left( v_{i} + W \right) \end{align*} $$

so $\begin{bmatrix}\ \overline{T}(v_{k+1} + W) \end{bmatrix}_{\alpha} = \begin{bmatrix} a_{k+1,k+1} \\ \vdots \\ a_{n,k+1}\end{bmatrix}$. Therefore, the following holds.

$$ \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha} = \begin{bmatrix} \begin{bmatrix}\ \overline{T}(v_{k+1} + W) \end{bmatrix}_{\alpha} & \cdots & \begin{bmatrix}\ \overline{T}(v_{n} + W) \end{bmatrix}_{\alpha}\end{bmatrix} = \begin{bmatrix} a_{k+1,k+1} & \cdots & a_{k+1,n} \\ \vdots & \ddots & \vdots \\ a_{n,k+1} & \cdots & a_{nn} \end{bmatrix} $$

Therefore, letting $A = \begin{bmatrix} a_{1,k+1} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{k,k+1} & \cdots & a_{kn} \end{bmatrix}$,

$$ \begin{bmatrix} T \end{bmatrix}_{\beta} = \begin{bmatrix} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} & A \\ O & \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha} \end{bmatrix} $$

$$ \implies \begin{bmatrix} T \end{bmatrix}_{\beta} -\lambda I = \begin{bmatrix} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} - \lambda I & A \\ O & \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha} - \lambda I \end{bmatrix} $$

Determinant of a block matrix

Let $A = \begin{bmatrix} A_{1} & A_{2} \\ O & A_{3} \end{bmatrix}$ be a block matrix. Then the following holds.

$$ \det A = \det A_{1} \det A_{3} $$

Therefore,

$$ f(t) = \det \left( \begin{bmatrix} T \end{bmatrix}_{\beta} -\lambda I \right) = \det \left( \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} - \lambda I \right) \det \left( \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha} - \lambda I \right) = g(t)h(t) $$

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