How to find 2x2 matrix with non zero elements and repeated eigenvalues? I need to find a 2x2 matrix with non zero elements that has eigenvalue = 1 repeated (double).
How can i do that?
Thanks!
 A: You can try this:
Your $2\times2$ matrix should be of the form:
$$M = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right),$$
being its characteristic polynomial:
$$p(\lambda) = |M-\lambda I_2| = \lambda^2 - (a+d) \lambda +ad- bc $$
Since you want your eigenvalues, $\lambda$, to be both $1$, your characteristic polynomial must become:
$$p(\lambda) = (\lambda-1)^2 = \lambda^2 - 2\lambda + 1,$$
so this relations must be held for $a,b,c,d$ by comparing term by term:
$$a+d = 2, \quad ad-bc = 1,$$
and every matrix $M$ satisfying this will be your answer. 
Note that $ad-bc = |M| = \lambda_1 \lambda_2 = 1$ and $a+d = \text{Tr}(M) = \lambda_1+ \lambda_2 = 2$. 
I hope this is useful to you!
Cheers!
A: The matrix you are searching for must have 
$$\chi = (\lambda-1)^2$$
as its Characteristic polynomial.
Thus your matrix given by $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ must satisfy
$$\det (\mathcal{I_2}\lambda-A) = (\lambda-a)(\lambda-d)-bc \stackrel{!}{=}(\lambda-1)^2$$
$$\Longleftrightarrow \lambda^2 -\lambda(a+d) + (ad-bc) = \lambda^2-2\lambda +1 $$
This is true iff
$$ a+d = 2 \,\, \wedge ad-bc = 1 $$
Now solving this system by Try&Error:


*

*If $a=1$ then $b$ or $c$ has to be $0$ $\longrightarrow$ not a valid solution

*If $a=2$ then $d$ has to be $0$ $\longrightarrow$ not a valid solution

*If $a=3$ then $bc$ hast to be $-4$. Chosing $b=2=-c$ gives as a solution


$$A = \begin{pmatrix} 3 & 2 \\ -2 & -1 \end{pmatrix}$$
A: Hint: Think of what the characteristic polynomial would be. How can you build a matrix with a given characteristic polynomial? (There is one exceedingly simple such $2\times 2 $ matrix.)
A: You can check that this matrix does the trick:
$A_{2,2}= \begin{pmatrix}
1/3 & 4/3\\
-1/3 & 5/3
\end{pmatrix}$
The way I got it was to use matrix similarity. This matrix is similar to a matrix with $1$ as repeated eigenvalues.
