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When determining the eigenvectors from matrix $A$: $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} $$

I found the eigenvalue to be $ \lambda = 1 $

calculating $ (A - 1*\lambda) $ gives me the matrix : $$ \left[ \begin{array}{cc|c} 0&0&0\\ 0&0&0 \end{array} \right] $$

Which eigenvector would this produce?

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    $\begingroup$ Compute the eigenvalues again. $\endgroup$ – Git Gud Mar 19 '16 at 22:19
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    $\begingroup$ How did you find the eigenvalue $-1$? $\endgroup$ – J.-E. Pin Mar 19 '16 at 22:19
  • $\begingroup$ What is the solution space for the system related to $\;\det(A-1I)\;$ . The whole space, right? $\endgroup$ – DonAntonio Mar 19 '16 at 22:23
  • $\begingroup$ The matrix is diagonal.....so the eigenvectors and eigenvalues should be obvious $\endgroup$ – ClassicStyle Mar 19 '16 at 22:27
  • $\begingroup$ @TylerHG Not to someone who is new to the subject. $\endgroup$ – Arthur Mar 19 '16 at 23:20
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For $\lambda=1$, the Eigenvector equation is

$$0v=0.$$

This clearly means that any vector is a solution.

By the way, this can be found directly by identifying

$$Av=\lambda v$$ and $$Iv=v.$$

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You made a small error when calculating your eigenvalues. If you solve for the the determinant equal to zero you end up with the following equation: $(1-\lambda)(1-\lambda) = 0$.

Solving this you get both eigenvalues of $\lambda_1 = \lambda_2 = 1$. You can see from this how a diagonal matrix greatly simplifies your calculations. In a diagonal matrix the diagonal terms are your eigenvalues. Try recalculcating your eigenvectors with these eigenvalues. You will get $[0,\ 1]^T$ and $[1,\ 0]^T$.

Hope this helps.

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This is fairly obvious, and can be solved with a bit of intuition without even touching an equation

The basis vector $\vec{i}$ equals $[1,0]^T$ and the basis vector $\vec{j}$ equals $[0,1]^T$. This means this matrix $(\vec i,\vec j)$ is the identity matrix. Nothing is changed because no vector is changed; every (nonzero) vector is an eigenvector with an eigenvalue of 1.

(sorry, I don't know how to use mathJax- I'm just a 14yr-old learning linear algebra through the youtube channel 3blue1brown. they have an excellent tutorial that will give you a visual intuition for what's going on) Here is the playlist: https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab00

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  • $\begingroup$ Welcome to Math.SE. I've taken the liberty to use some MathJax and $\LaTeX$ formatting, and hope that it will encourage you to use similar formatting in future posts to make your points. See this introduction. $\endgroup$ – hardmath May 9 '18 at 15:01

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