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How can I calculate the eigenvectors of the following matrix?

$$\begin{bmatrix}1& 3\\3& 2\end{bmatrix}$$

I calculated the eigenvalues. I got

$$\lambda_1 = 4.541381265149109$$

$$\lambda_2 = -1.5413812651491097$$

But, now I don't know how to get the eigenvectors. When I create a new matrix after I subtracted Lambda value from all the members of the matrix on the main diagonal and tried to solve the homogeneous system of equations, I get only null vector for both $\lambda_1$ and $\lambda_2$....

When I used this website for calculating eigenvalues and eigenvectors. I got these eigenvectors

$(0.6463748961301958, 0.7630199824727257)$ for $\lambda_1$

$(-0.7630199824727257, 0.6463748961301957)$ for $\lambda_2$

.... but have no idea how to calculate them by myself... Is it even possible? ....or it's possible to calculate it numerically?

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    $\begingroup$ Why are you writing everything in decimals? $\endgroup$ – Chris Eagle Jul 13 '13 at 16:20
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    $\begingroup$ Can you find the characteristic equation, and solve it exactly, with square-roots? You might have to do some algebra, instead of using the calculator :( $\endgroup$ – Empy2 Jul 13 '13 at 16:20
  • $\begingroup$ @user1111261 my point to you is that even the eigenvalues must be determined numerically in general, so don't be surprised the same goes for e-vectors. $\endgroup$ – James S. Cook Jul 13 '13 at 16:30
  • $\begingroup$ @user1111261: In addition to the answers to not use decimals, you can also use the general expression for a 2x2 and it's eigenvalues and eigenvectors. Regards $\endgroup$ – Amzoti Jul 13 '13 at 16:37
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Indeed, as Chris Eagle and Michael pointed out to you, calculators are not always your best friend.

Instead, if you do your maths with the characteristic equation, you'll find out that the eigenvalues look nicer this way:

$$ \lambda = \frac{3 \pm \sqrt{37}}{2} $$

And it's not at all impossible to find the eigenvectors. For instance the one with the $+$ sign, you could start like this:

$$ \begin{pmatrix} 1 - \dfrac{3 + \sqrt{37}}{2} & 3 & \vert & 0 \\ 3 & 2 - \dfrac{3 + \sqrt{37}}{2} & \vert & 0 \end{pmatrix} $$

Hint: After some easy simplifications, you'll find out that it's very useful to multiply one of the rows by $1 - \sqrt{37}$ and that you can write the corresponding eigenvector as simple as this: $(1 - \sqrt{37}, -6)$.

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We're given the matrix $A = \begin{bmatrix}1& 3\\3& 2\end{bmatrix}$.

$\det(A) = 2-9 = -7,\;\;\operatorname{Trace}(A) = 3, ;$ and that gives us the characteristic polynomial $$\lambda^2 - 3\lambda - 7$$

Now, when finding eigenvalues: obtain exact values (don't use decimal approximations unless explicitly asked to do so, and even then, when your aim is to find eigenvectors with your eigenvalues, use exact values): and here you can simply use the quadratic formula to find the roots of the characteristic polynomial, i.e., the eigenvalues: $$\lambda^2 - 3x\lambda - 7 = 0 \iff \lambda_i = \dfrac {3\pm \sqrt{37}}{2}$$

Now using the exact representation of your eigenvalues, compute the desired eigenvectors, by solving $$(A-\lambda_{i}I)=0 \quad\text{for each}\;\lambda_i$$

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  • $\begingroup$ Reason 1 for using exact values when possible: Errors tend to accumulate through a calculation, so waiting until the last minute to approximate will give the most precise approximation. Reason 2: sometimes, the exact form will simplify in some nice fashion (as Augustí Roig indicates). Long decimal approximations don't do this. Sometimes this just makes life easier; sometimes it helps reveal important underlying structure you've missed. $\endgroup$ – dfeuer Jul 13 '13 at 16:43
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    $\begingroup$ Reason 3: Since decimal approximations are not the actual eigenvalues, there will be no eigenvectors for them. That's (probably) what caused the "only null vectors" issue mentioned in the question. $\endgroup$ – Andreas Blass Jul 13 '13 at 17:43
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No, you cannot in general find the $\lambda$ without a numerical method. The characteristic equation is an $n$-th order polynomial equation and we know (have proven by Galois theory etc...) that it is not possible to find closed-form algebraic solutions in all cases (there is no analogue of the quadratic equation for 5-th and higher order polynomial equations, although many particular problems admit integer or rational solutions...).

So, yes, in general some numerical method is probably a good idea.

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The characteristic equation is $$\begin{gather} \left|\matrix{ 1-\lambda & 3 \\ 3 & 2-\lambda }\right| = 0 \\ (1-\lambda)(2-\lambda)-9 = 0\\ \lambda^2-3\lambda-7 = 0. \end{gather}$$

Solving, $$\lambda = \frac{3 \pm \sqrt{37}}{2}$$

Now we solve for eigenvectors: $$\pmatrix{ 1-\lambda & 3 \\ 3 & 2 - \lambda }v = 0.$$

I really don't want to take the time to row-reduce that mess, but you get the idea.

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For simple matrices it's better to find eigenvalues explicitely. In your case that characteristic polynomial is $x^2-3x-7$, which gives you the eigenvalues $\frac{3\pm\sqrt{37}}{2}$. Now given these eigenvalues it's easy to find eigenvectors.

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The characteristic polynomial is $\chi(t)=t^{2}-3t-7$, for which you need the quadratic formula. This yields eigenvalues of $\lambda_{1}=\frac{3+\sqrt{37}}{2}$ and $\lambda_{2}=\frac{3-\sqrt{37}}{2}$. To find the eigenvectors, solve $(A-\lambda_{i}I)=0$, obtaining a vector for each eigenvalue. It might be a bit messy with the improper fraction, but it isn't that bad with a $2\times 2$ matrix.

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Now I get it.... After I try to solve the system of 2 equations for Lambda 1.... I get:

v1 = 0.84712708838 * v2
v1 = 0.84712708838 * v2

So, the system has infinitely many solutions...

So, I can e.g. write that (0.84712708838 , 1) is the eigenvector for Lambda 1.

Similarly for Lambda 2.... it also has infinitely many solutions...

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By definition an eigenvector for eigenvalue $\lambda$, is a member of nullspace of $A - \lambda I$, if you can compute nullspaces, you can compute eigenvectors.

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