I have the matrix:

$$ A= \begin{bmatrix} 7 & -2 \\ 15 & -4 \\ \end{bmatrix} $$

and I am asked to find the eigenvalues and eigenvectors. I found the eigenvalues to be $\lambda = 1,2$. Now I need to find the eigenvectors:

$$(A-\lambda I)\mathbf u=\mathbf 0$$ $$ \begin{bmatrix} 5 & -2 \\ 15 & -6 \\ \end{bmatrix} \mathbf u=\mathbf 0$$ I created the augmented matrix and row reduced to get:

$$ \begin{bmatrix} 5 & -2 & 0 \\ 0 & 0 & 0\\ \end{bmatrix} $$

I set $u_2=s$ and $u_1=\frac{2}{5}s$, thus $$\mathbf u=s\begin{bmatrix} \frac{2}{5} \\ 1 \end{bmatrix}$$

However the answers say that the vector is $(2,5)$. I don't know where I went wrong.

  • 2
    $\begingroup$ You didn't go wrong. They just multiplied by 5. Since you have a constant s that won't matter. $\endgroup$ – B.A May 17 '17 at 4:09
  • $\begingroup$ Every non-zero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue. That’s why it’s a mistake to speak of the eigenvector of $2$. $\endgroup$ – amd May 17 '17 at 6:51

You mention at the end of your work that you set $u_2=s$ and $u_1=\frac{2}{5}s$. You could have just as easily set $u_2=5s$ and $u_1=2s$ and it would still have solved the equation, as setting $s$ constant with anything for which the equation $u_2=\frac{2}{5}u_1$ holds will give you a solution. So you didn't do anything wrong in your analysis, you just chose a different pair of numbers to satisfy the equation than did the answer, and you solved it in a more general sense (as a function of a constant $s$) than did the answer. Your result is correct and is essentially the same as the answer.

  • $\begingroup$ Thanks that makes sense thank you very much $\endgroup$ – Patrick Robertson May 17 '17 at 23:36
  • $\begingroup$ @Patrick if this answer satisfies you, please accept it as the answer to your question. Thank you. $\endgroup$ – user7938238 May 18 '17 at 20:07

The eigenvalue equation is $$ \mathbf{A}u = \lambda u. \tag{1} $$ Matrices are the embodiment of linear systems. So if the eigenvalue is scaled by $\alpha \lambda$, the scaled eigenvector becomes $\alpha u$.

That is, $(1)$ implies $$ \alpha \mathbf{A}u = \alpha \lambda u, \qquad \alpha \in \mathbb{C}. $$


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