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$$\left[\begin{array}{ccc} 4 & 0 \\ 2 & 2 \end{array}\right]$$

I am trying to find the two eigenvalues associated with this matrix. So I find the eigenvalues $\lambda$ that make $\det(A-\lambda I)=0$. $$(4-\lambda)(2-\lambda)=0.$$

With $\lambda=4$, I got $\left(\begin{array}{ccc} 0 & 0 \\ 2 & -2 \end{array}\right)\times \left(\begin{array}{ccc} u_1\\ u_2\end{array}\right)=\left(\begin{array}{ccc} 0\\ 0\end{array}\right)$. So eigenvector $u$ is $[1,1]$.

But the problem is with $\lambda=2$, I got $\left(\begin{array}{ccc} 2 & 0 \\ 2 & 0 \end{array}\right)\times \left(\begin{array}{ccc} v_1\\ v_2\end{array}\right)=\left(\begin{array}{ccc} 0\\ 0\end{array}\right)$. So $v_1$ has to be $0$, while $v_2$ can be anything.

What is eigenvector $v$?

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    $\begingroup$ You already said the answer: set the first entry to 0 and the section be one to anything. $\endgroup$ – Zach Boyd Oct 6 '17 at 2:57
  • $\begingroup$ I have edited the matrix next to the eigenvector associated with $\lambda=4$, hope that is what you meant to put. $\endgroup$ – Ahmed S. Attaalla Oct 6 '17 at 2:57
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You've found the eigenvectors, for $\lambda =2$ it is $\begin{pmatrix}0 \\ v_2 \end{pmatrix}$ for any $v_2 \neq 0$. Eigenvectors are never unique, as any scalar multiple of an eigenvector is still an eigenvector. If all you need is one eigenvector, you can take $v_2$ to be anything so $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ works.

Why is any scalar multiple of an eigenvector also an eigenvector? Suppose $v$ is an eigenvector of an operator $A$ corresponding to an eigenvalue $\lambda$, then by definition, $v\neq 0$ and $v$ satisfies the eigenvalue equation$$Av=\lambda v.$$ Now let $a\neq 0$ be a constant, then $$A(av)=a(Av)=a(\lambda v)=\lambda(av)$$ so $av$ also satisfies the eigenvalue equation and $av\neq 0$, hence $av$ is also an eigenvector of $A$ for any $a \neq 0$.

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All eigenvectors have the property that any scalar multiple is also an eigenvector with the same eigenvalue. So your first answer for $\lambda = 4$ could be rephrased "$(a,a),$ where $a$ is anything." Similarly your second answer "$(0,v_2)$ where $v_2$ is anything" makes perfect sense.

(Actually, a caveat: "anything" cannot be zero since eigenvectors must be nonzero by definition.)

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