Kernel, Characteristic polynomials, eigenvalues and eigenvectors

Section 1.9 Kernel, Characteristic polynomials, eigenvalues and eigenvectors

Definition 1.9.1.

Let \(T\colon M_{n\times 1}(\K)\to M_{m\times 1}(\K)\) be a linear map. The kernel of \(T\) is the following set (more precisely a linear subspace).

\begin{equation*} \ker(T)=\left\{v\in M_{n\times 1}(\K):T(v)=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\in M_{m\times 1}(\K)\right\} \end{equation*}

Note 1.9.2.

The kernel of a linear map always contains \(0\in M_{n\times 1}(\K)\text{.}\)

Definition 1.9.3.

Let \(A=(a_{ij})\in M_n(\K)\) be a square matrix. The characteristic polynomial of \(A\) is

\begin{equation*} \chi_A(x)=\det\left(xI_n-A\right)=\left|\begin{matrix}x-a_{11}\amp -a_{12}\amp\cdots\amp -a_{1n}\\-a_{21}\amp x-a_{22}\amp\cdots\amp -a_{2n}\\\vdots\amp\vdots\amp\ddots\amp\vdots\\-a_{n1}\amp -a_{n2}\amp\cdots\amp x-a_{nn}\end{matrix}\right|\in\K[x], \end{equation*}

where \(x\) is a variable and \(I_n\) is the identity matrix.

Definition 1.9.4.

Let \(A\in M_n(\K)\text{.}\) A root of the characteristic polynomial of \(A\) is called an eigenvalue of \(A\).

Definition 1.9.5.

Let \(A\in M_{n\times 1}(\K)\) matrix and \(\lambda\in\K\) be an eigenvalue of \(A\text{.}\) A nonzero \(v\in M_{n\times 1}(\K)\) is said to be an eigenvector corresponding to \(\lambda\) if \(Av=\lambda v\text{.}\)

Remark 1.9.6.

If \(v\) is an eigenvector corresponding to an eigenvalue \(\lambda\) then, for any \(\alpha\in\K\) we have the following.

\begin{equation*} A\left(\alpha v\right)=\alpha Av=\alpha(\lambda v)=\lambda\left(\alpha v\right) \end{equation*}

Thus, if \(v\) is an eigenvector corresponding to an eigenvalue \(\lambda\) then so is any nonzero scalar multiple of \(v\text{.}\)

Geometrically, if we draw a straight line through the origin in the direction of an eigenvector, then any vector on this straight line will remain on the line after the linear map corresponding to \(A\) (see Remark 1.7.7) is applied.

Note 1.9.7.

Let \(A\in M_{n\times 1}(\K)\) and \(\lambda\in\K\) be an eigenvalue of \(A\text{.}\) We descibe a method to find an eigenvector corresponding to \(\lambda\text{.}\)

Consider the following matrix \(A_\lambda\) and the linear map corresponding to \(A_\lambda\text{.}\)

\begin{equation*} A_\lambda=\lambda\cdot I_n-A\quad\text{and}\quad T_{\lambda}\colon M_{n\times 1}(\K)\to M_{n\times 1}(\K)\quad\text{given by}\quad v\mapsto A_\lambda v \end{equation*}

Suppose that \(0\neq v_0\in\ker(T_\lambda)\text{.}\) Thus we must have the following.

\begin{align*} T_\lambda(v_0) \amp=(\lambda\cdot I_n-A)v_0 \\ \amp=\lambda\cdot I_nv_0-Av_0 \\ \amp=\lambda v_0-Av_0 \end{align*}

Hence, we get that \(Av_0=\lambda v_0\text{.}\)

The above calculations show that any nonzero column vector in the kernel of \(T_\lambda\) will be an eigenvector corresponding to \(\lambda\text{.}\)

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