Find a corresponding eigenvector for each eigenvalue

Question

Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$

(a) Find the eigenvalues of A.

(b) Find a corresponding eigenvector for each eigenvalue in part (a).

My attempt

a) Eigenvalues: $$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-13\lambda +40=0 \Rightarrow \lambda_1=5, \lambda_2=8.$$

b) If $\lambda=5$, then

$\begin{pmatrix} 6-5 & 1 \\ 2 & 7-5 \end{pmatrix}$ = $\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\Rightarrow $ Assuming this as B.

Then $ B\bar { x } =\bar { 0 } $,

$\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\begin{pmatrix}X_1\\X_2\end{pmatrix}=0$

$\begin{pmatrix} 1 & 1 & 0 \\ 2 & 2 & 0 \end{pmatrix}$ By doing row reduction $=>$ $R_2->R_2-2R_1$

$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ By doing row reduction $

$X_1+X_2=0$

$X_1=-X_2$

Let $X_2=1$, then $X_1=-1$

$\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$

But, the answer is $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}1\\-1\end{pmatrix}$..

I verified many times but I ended up getting the answer as $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$

Can anyone please verify which is the correct answer.

Answer 1

Both answers are correct, that is,$$A.\begin{pmatrix}1\\-1\end{pmatrix}=5\begin{pmatrix}1\\-1\end{pmatrix}\iff A.\begin{pmatrix}-1\\1\end{pmatrix}=5\begin{pmatrix}-1\\1\end{pmatrix}.$$

Answer 2

Both answers are correct. Eigenvectors that are multiples of each other share the same eigenvalue for a particular matrix.

$$Ae=\lambda e\\ A(ke)=k(Ae)=k(\lambda e)=\lambda (ke)$$

Answer 3

If $v$ is an eigenvector to your matrix, so is $\xi \cdot v$ for any $\xi \in \mathbb{R} \setminus \left\{ 0\right\}$. This can be extended to a complex setting.

Moreover if $u$ and $v$ are eigenvectors to the same eigenvalue, all their linear combinations (except $0$) are as well eigenvectors to that eigenvalue. This is why there is the concept of "eigenspaces".

In conclusion: Never mind the sign of an eigenvector or be startled when your calculation yields a scalar multiple.