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 $



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.

  • 1
    $\begingroup$ If $\mathbf v$ is an eigenvector then so is $\alpha \mathbf v$ $\endgroup$ – Doug M May 4 '18 at 23:43
  • $\begingroup$ Your answer is collinear to the official answer, so it's fine too. $\endgroup$ – Bernard May 4 '18 at 23:44

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

  • $\begingroup$ So, the method what I did to find eigenvectors was correct. Can you please suggest any easy method to find the eigenvectors $\endgroup$ – tien lee May 4 '18 at 23:50

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

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


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.

  • $\begingroup$ Can you please suggest any altenate method to find the eigenvectors $\endgroup$ – tien lee May 4 '18 at 23:51
  • $\begingroup$ In your context of linear algebra, I would honestly refrain from doing so: There are numerical algorithms for computation in specific settings (matrices with a certain "shape") but these will likely not help you with the sort of problem you posted. If you really want to dive in deep, check out the english wikipedia on "eigenvalue algorithm" and "eigenvalue decomposition". I know it is not cool to link the wiki, but the issue really becomes complicated and you would need some interest in numerics to make it worthwhile to proceed here. The wiki gives you a good idea of the required backgrounds. $\endgroup$ – mol3574710n0fN074710n May 5 '18 at 0:01
  • $\begingroup$ As for linear algebra I can say: "Back then" we always did it the way you just demonstrated and we only started bothering about the numerical stuff way later... $\endgroup$ – mol3574710n0fN074710n May 5 '18 at 0:02
  • $\begingroup$ Also worth to note: These algorithms are not "easy" at all (especially analysing their numerical properties). They are just more feasible with matrices of say $n = 10^6$ and are often specialized for the specific type (shape) of the matrix. $\endgroup$ – mol3574710n0fN074710n May 5 '18 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.