Home page for accesible maths 7 Eigenvalues and eigenvectors 7.2 How to find the eigenvalues and eigenvectors 7.4 Subspaces and dimension

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7.3. Eigenvalues of a $2\times 2$ matrix

In this section we will look a little closer at the case of $2\times 2$ matrices in general. Since they are so small, it is possible to give general statements about their eigenvalues and eigenvectors. We will also make some remarks about the case when $A$ is a symmetric matrix.

First, we express the characteristic polynomial in terms of the determinant and the trace. Recall from Exercise 1.6.15 that the trace of a matrix $A$ (written $\operatorname{tr}\nolimits A$ ) equals the sum of the entries $a_{ii}$ along its diagonal.

Theorem 7.3.1.

Let $A\in\operatorname{M}_{{2}}({{\mathbb{R}}})$ . The characteristic polynomial of $A$ is

c_{A}(\lambda)=\lambda^{2}-\operatorname{tr}\nolimits(A)\lambda+\det A.

Furthermore:

(i)

If $(\operatorname{tr}\nolimits A)^{2}-4\det A>0$ , then $A$ has two distinct real eigenvalues.
(ii)

If $(\operatorname{tr}\nolimits A)^{2}-4\det A=0$ , then $A$ has a unique real eigenvalue.
(iii)

If $(\operatorname{tr}\nolimits A)^{2}-4\det A<0$ , then $A$ has no real eigenvalues.

Proof.

Write $A$ in the form $A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ . To find the characteristic polynomial, we expand the following determinant:

c_{A}(\lambda)=\det(A-\lambda I_{2})=\begin{vmatrix}a-\lambda&b\\ c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^{2}-(a+d)\lambda+(% ad-bc).

Since $a+d=\operatorname{tr}\nolimits A$ and $ad-bc=\det A$ , the result follows.

For the statement about the number of real eigenvalues, recall that quadratic polynomials have real roots if and only if their discriminant is $\geq 0$ . In this case, the discriminant is $(\operatorname{tr}\nolimits A)^{2}-4\det A$ . ∎

Remark 7.3.2.

If we were also considering complex eigenvalues, instead of just the real eigenvalues, then the characteristic equation would always have solutions. More generally, the Fundamental Theorem of Algebra says that any non-constant polynomial has a complex root.

Theorem 7.3.3.

Let $A\in\operatorname{M}_{{2}}({{\mathbb{R}}})$ and suppose that $A$ is symmetric. Then, $A$ has at least one real eigenvalue. More specifically:

(i)

If $A$ has just one eigenvalue, then it is a scalar matrix (i.e. $A=aI_{2}$ for some $a\in{\mathbb{R}}$ ).
(ii)

If $A$ has two distinct eigenvalues, then the eigenspaces are two perpendicular lines that pass through the origin.

Proof.

Since $A$ is symmetric, we can write it in the form $A=\begin{pmatrix}a&b\\ b&d\end{pmatrix}$ for some $a,b,d\in{\mathbb{R}}$ . Then its characteristic polynomial has discriminant

(\operatorname{tr}\nolimits A)^{2}-4\det A=(a+d)^{2}-4(ad-b^{2})=a^{2}+2ad+d^{% 2}-4ad+4b^{2}=(a-d)^{2}+4b^{2}.

Since $a, b, d$ are all real numbers, the discriminant is always $\geq 0$ , so there is always at least one real eigenvalue.

If there is only one eigenvalue, then $(a-d)^{2}+4b^{2}=0$ , which is true if and only if $a-d=b=0$ ; in other words $A=\begin{pmatrix}a&0\\ 0&a\end{pmatrix}$ as required.

In the second case, we omit the proof that the eigenspaces are perpendicular lines. This is not hard to prove with the theory developed in MATH220. Thus, instead of a formal proof, we illustrate the theorem by examples. ∎

Example 7.3.4.

Let $A=\begin{pmatrix}3&2\\ 2&3\end{pmatrix}$ . From Theorem 7.3.1, we find

$\lambda=\frac{\operatorname{tr}\nolimits A\pm\sqrt{(\operatorname{tr}\nolimits A% )^{2}-4\det A}}{2}=\frac{6\pm 4}{2}=\begin{cases}1\\ 5.\end{cases}$

Now for each eigenvalue, we determine its eigenspace:

$\lambda=1$ : We seek the reduced echelon form of $A-I_{2}=\begin{pmatrix}2&2\\ 2&2\end{pmatrix}$ , which is $\begin{pmatrix}1&1\\ 0&0\end{pmatrix}$ . Thus, the eigenvectors are all the non-zero vectors of the form $\begin{pmatrix}x\\ -x\end{pmatrix}$ with $x\in{\mathbb{R}},x\neq 0$ . That is, the eigenspace is

$V_{1}=\left\{\begin{pmatrix}x\\ -x\end{pmatrix}\in{\mathbb{R}}^{2}\right\}.$

$\lambda=5$ : The reduced echelon form of $A-5I_{2}=\begin{pmatrix}-2&2\\ 2&-2\end{pmatrix}$ is $\begin{pmatrix}1&-1\\ 0&0\end{pmatrix}$ . Thus, the eigenvectors are all the non-zero vectors of the form $\begin{pmatrix}x\\ x\end{pmatrix}$ with $x\in{\mathbb{R}},x\neq 0$ . That is, the eigenspace is

$V_{5}=\left\{\begin{pmatrix}x\\ x\end{pmatrix}\in{\mathbb{R}}^{2}\right\}.$

Observe that $V_{1}$ is the line $y=-x$ , whereas $V_{5}$ is the line $y=x$ . They are perpendicular to each other as expected from Theorem 7.3.3.

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7.3. Eigenvalues of a 2×2 matrix

Theorem 7.3.1.

Proof.

Remark 7.3.2.

Theorem 7.3.3.

Proof.

7.3. Eigenvalues of a $2\times 2$ matrix