# Linear Algebra: $2\times 2$ matrix yields only 1 eigenvalue

When trying to find the eigenvalues and eigen vectors of $$A$$ ($$2\times 2$$), I get only 1 eigenvalue, perhaps with a multiplicity of 2?

Let $$A=\begin{pmatrix} 5 & 1 \\ -4 & 1\end{pmatrix}$$

I begin by taking the $$\det(A- \lambda I)$$

Thus $$\det(A-\lambda I)=\begin{vmatrix} 5-\lambda & 1 \\ -4 & 1-\lambda\end{vmatrix}$$, factoring out I get $$(\lambda-3)^2$$ with $$\lambda=3$$ as my only eigenvalue.

Where did I go wrong and where do I go from here? How can the multiplicity of my eigen value help me here?

• Is $(\lambda -3)^2=0$ really yields $\lambda=3$ or it should be like $\lambda=3,3$ – Sujit Bhattacharyya Nov 5 '18 at 16:27

You can find one eigenvector associated to $$\lambda =3$$ for example

$$v=\binom 1{-2}$$

and that is it.

There is only one eigenvector and it is OK.

They say the algebraic multiplicity of $$\lambda =3$$ is $$2$$ but the the geometric multiplicity is $$1$$

• I guess my question is how did you get [1; -2] if the A-lambda*identity matrix is [2 1; 0 0][x;y] – APorter1031 Nov 5 '18 at 16:37
• @APorter1031 The first equation is the only nontrivial one and it reads $2x+y=0$, picking $x=1$ (a completely arbitrary choice) gives $y=-2$. – Ian Nov 5 '18 at 16:39
• Thank you so much! – APorter1031 Nov 5 '18 at 16:39

This makes sense! A $$2\times 2$$ matrix need not necessarily have 2 eigenvalues. What matters is that the sum of the algebraic multiplicities is 2, which it is, so it seems as if your answer is absolutely correct.

Your work is correct, $$\lambda = 3$$ is the only eigenvalue with multiplicity $$2$$. If this seems weird, consider the following example:

$$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}.$$

It has only one eigenvalue, $$\lambda$$, with multiplicity $$2$$; you can just read it from diagonal entries.

In general, $$n\times n$$ complex matrix has precisely $$n$$ eigenvalues, but counted with multiplicities. It's not required that they are distinct. It's just like any polynomial of degree $$n$$ has precisely $$n$$ complex roots, but they don't have to be distinct.

Note, however, that it's possible that real matrix actually has no real eigenvalues. Just like there are polynomials with no real roots.