$f \in \mathrm{End} (\mathbb{C^2})$ $f(e_1)=e_1+e_2$ $f(e_2)=e_2-e_1$. Eigenvalues of f and the bases of the associated eigenspaces

Question

Let $f \in \mathrm{End} (\mathbb{C^2})$ be defined by its image on the standard basis $(e_1,e_2)$:

$f(e_1)=e_1+e_2$

$f(e_2)=e_2-e_1$

I want to determine all eigenvalues of f and the bases of the associated eigenspaces.

First of all how does the transformation matrix of $f$ look like? Is it

$\begin{pmatrix}1 &-1 \\1 &1 \end{pmatrix}$?

Answer 1

If one represents the standard basis $e_1$, $e_2$ in the usual form

$e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag 1$

$e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 2$

and writes the matrix of $f$ as

$[f] = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}, \tag 3$

then we have, since

$f(e_1) = e_1 + e_2, \; f(e_2) = e_2 - e_1, \tag 4$

$\begin{pmatrix} 1 \\ 1 \end{pmatrix} = e_1 + e_2 = [f]e_1 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \alpha \\ \gamma \end{pmatrix}, \tag 5$

and

$\begin{pmatrix} -1 \\ 1 \end{pmatrix} = e_2 - e_1 = [f]e_2 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \beta \\ \delta \end{pmatrix}, \tag 6$

from which it immediately follows that

$\alpha = \gamma = \delta = 1, \tag 7$

$\beta = -1; \tag 8$

thus

$[f] = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}, \tag 9$

as anticipated by our OP user567319. Once we have (9), it is an easy matter to find the eigenvalues of $f$, sincd they must satisfy

$0 = \det([f] - \lambda I) = \det \left ( \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} - \lambda I \right ) = \det \left (\begin{bmatrix} 1 - \lambda & -1 \\ 1 & 1 - \lambda \end{bmatrix} \right )$ $= (1 - \lambda)^2 + 1 = \lambda^2 - 2\lambda + 2; \tag{10}$

it follows from (10), using the quadratic formula, that

$\lambda = \dfrac{1}{2}(2 \pm \sqrt{-4}) = \dfrac{1}{2}(2 \pm 2 i) = 1 \pm i; \tag{11}$

it is now an easy matter to find the eigenvectors, satisfying as they do

$\begin{pmatrix} \lambda \mu \\ \lambda \nu \end{pmatrix} = \lambda \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{pmatrix} \mu - \nu \\ \mu + \nu \end{pmatrix}, \tag{12}$

whence

$\lambda \mu = \mu - \nu, \tag{13}$

$\lambda \nu = \mu + \nu; \tag{14}$

from (13),

$(1 - \lambda) \mu = \nu; \tag{15}$

it follows that, taking $\mu = 1$, the eigenvectors are

$\begin{pmatrix} 1 \\ \mp i \end{pmatrix} = \begin{pmatrix} 1 \\ 1 - \lambda \end{pmatrix}, \; \lambda = 1 \pm i; \tag{16}$

we check the case $\lambda = 1 + i$:

$\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} 1 + i \\ 1 - i \end{pmatrix} = (1 + i)\begin{pmatrix} 1 \\ -i \end{pmatrix}; \tag{17}$

a check of the case $\lambda = 1 - i$ follows from this by complex conjugation, since $[f]$ is a real matrix.

Answer 2

Yes, the matrix of $f$ is $$ A=\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} $$ because $f(e_1)=1e_1+1e_2$ and $f(e_2)=(-1)e_1+1e_2$ and you fill the columns with the coefficients of the linear combinations.

The characteristic polynomial is $p(X)=(1-X)^2+1=X^2-2X+2$, so the eigenvalues are $\lambda_1=1+i$ and $\lambda_2=1-i$.

You need to find a vector in the null space of $A-(1+i)I$: Gaussian elimination yields $$ \begin{pmatrix} -i & -1 \\ 1 & -i \end{pmatrix} \xrightarrow{R_1\gets i R_1} \begin{pmatrix} 1 & -i \\ 1 & -i \end{pmatrix} \xrightarrow{R_2\gets R_2-R_1} \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix} $$ Since you want to solve the system $(A-(1+i)I)x=0$, a nonzero solution is $$ \begin{pmatrix} i \\ 1 \end{pmatrix} $$ Do similarly for the other eigenvalue.