This note covers eigenvalues and eigenvectors of square matrices: what they are, how to find them using the characteristic equation, and how they allow a matrix to be diagonalised so that high powers are easy to compute. They describe the directions a matrix leaves unchanged and the long-term behaviour of repeated processes.
Key ideas
Most vectors change direction when you multiply them by a matrix. For a square matrix \(A\), some special non-zero vectors \(\mathbf{v}\) are only stretched or shrunk, so that
\[ A\mathbf{v}=\lambda\mathbf{v} \]The number \(\lambda\) is an eigenvalue of \(A\) and \(\mathbf{v}\) is a corresponding eigenvector. An eigenvector must be non-zero. If \(\mathbf{v}\) is an eigenvector, so is any non-zero multiple of it, so eigenvectors are only defined up to a scale factor.
Finding eigenvalues. Rearrange \(A\mathbf{v}=\lambda\mathbf{v}\) as \((A-\lambda I)\mathbf{v}=\mathbf{0}\). A non-zero solution \(\mathbf{v}\) exists only when \(A-\lambda I\) is singular, that is when
\[ \det(A-\lambda I)=0 \]This is the characteristic equation. For a 2×2 matrix it is the quadratic \(\lambda^{2}-(\operatorname{tr}A)\lambda+\det A=0\), where the trace \(\operatorname{tr}A\) is the sum of the diagonal entries. The eigenvalues are its roots.
Useful checks. The sum of the eigenvalues equals the trace, and the product of the eigenvalues equals the determinant. For a triangular matrix (all zeros above or below the diagonal) the eigenvalues are simply the diagonal entries.
Finding eigenvectors. For each eigenvalue \(\lambda\), solve \((A-\lambda I)\mathbf{v}=\mathbf{0}\). The rows of \(A-\lambda I\) will be multiples of each other (in 2×2), so you get one independent equation and can choose one component freely. Any non-zero choice is acceptable. Your GDC can also return eigenvalues and eigenvectors directly, which is a good way to check hand work, though the calculator may scale the vector differently.
Diagonalisation. If a 2×2 matrix has two distinct eigenvalues \(\lambda_1\) and \(\lambda_2\) with eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\), then form \(P\) with these eigenvectors as its columns and \(D=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}\). Then
\[ A=PDP^{-1}\qquad\text{and}\qquad A^{n}=PD^{n}P^{-1} \]The order of the columns in \(P\) must match the order of the eigenvalues in \(D\). Powers of a diagonal matrix are easy: \(D^{n}=\begin{pmatrix}\lambda_1^{n}&0\\0&\lambda_2^{n}\end{pmatrix}\).
Long-term behaviour. For a process \(\mathbf{x}_{n+1}=A\mathbf{x}_n\), the eigenvalue with the largest absolute value, the dominant eigenvalue, controls the long-term growth factor per step. After many steps the vector points almost along the corresponding eigenvector. If the dominant eigenvalue is larger than 1 the quantity grows, and if it lies between \(-1\) and \(1\) the quantity shrinks towards zero.
Why it works. Applying \(A\) twice to an eigenvector gives \(A^{2}\mathbf{v}=A(\lambda\mathbf{v})=\lambda^{2}\mathbf{v}\), and in general \(A^{n}\mathbf{v}=\lambda^{n}\mathbf{v}\). Diagonalisation uses this: change to the eigenvector directions, where \(A\) simply scales, apply the powers there, and then change back.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(A\mathbf{v}=\lambda\mathbf{v}\), \(\mathbf{v}\ne\mathbf{0}\) | Defining equation of an eigenvalue \(\lambda\) and eigenvector \(\mathbf{v}\). |
| \(\det(A-\lambda I)=0\) | Characteristic equation. Solve it to find the eigenvalues. |
| \(\lambda^{2}-(\operatorname{tr}A)\lambda+\det A=0\) | Characteristic equation of a 2×2 matrix. |
| \((A-\lambda I)\mathbf{v}=\mathbf{0}\) | Solve for each \(\lambda\) to get its eigenvectors. |
| \(A=PDP^{-1}\) | Diagonalisation. Columns of \(P\) are eigenvectors, \(D\) holds the eigenvalues. |
| \(A^{n}=PD^{n}P^{-1}\) | Efficient way to calculate powers of \(A\). |
| \(\sum\lambda=\operatorname{tr}A\), \(\prod\lambda=\det A\) | Quick checks on eigenvalues. |
Worked examples
Example 1 Find the eigenvalues and eigenvectors of \(A=\begin{pmatrix}4&1\\2&3\end{pmatrix}\).
The trace is \(7\) and the determinant is \(4\times3-1\times2=10\). The characteristic equation is \(\lambda^{2}-7\lambda+10=0\).
Factorise: \((\lambda-5)(\lambda-2)=0\), so \(\lambda=5\) or \(\lambda=2\). (Check: \(5+2=7\) and \(5\times2=10\).)
For \(\lambda=5\): \(A-5I=\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\). The equation \(-x+y=0\) gives \(y=x\), so \(\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}\).
For \(\lambda=2\): \(A-2I=\begin{pmatrix}2&1\\2&1\end{pmatrix}\). The equation \(2x+y=0\) gives \(y=-2x\), so \(\mathbf{v}_2=\begin{pmatrix}1\\-2\end{pmatrix}\).
Check: \(A\mathbf{v}_1=\begin{pmatrix}4+1\\2+3\end{pmatrix}=\begin{pmatrix}5\\5\end{pmatrix}=5\mathbf{v}_1\).
Answer: \(\lambda=5\) with eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\); \(\lambda=2\) with eigenvector \(\begin{pmatrix}1\\-2\end{pmatrix}\) (or any non-zero multiples).
Example 2 For the matrix \(A\) in Example 1, diagonalise \(A\) and use the result to find \(A^{4}\).
Take \(P=\begin{pmatrix}1&1\\1&-2\end{pmatrix}\) and \(D=\begin{pmatrix}5&0\\0&2\end{pmatrix}\), keeping the columns in the same order as the eigenvalues.
\(\det P=-2-1=-3\), so \(P^{-1}=\dfrac{1}{-3}\begin{pmatrix}-2&-1\\-1&1\end{pmatrix}=\dfrac13\begin{pmatrix}2&1\\1&-1\end{pmatrix}\).
\(D^{4}=\begin{pmatrix}625&0\\0&16\end{pmatrix}\).
\(PD^{4}=\begin{pmatrix}625&16\\625&-32\end{pmatrix}\).
\(A^{4}=PD^{4}P^{-1}=\dfrac13\begin{pmatrix}625&16\\625&-32\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}=\dfrac13\begin{pmatrix}1266&609\\1218&657\end{pmatrix}=\begin{pmatrix}422&203\\406&219\end{pmatrix}\).
Answer: \(A^{4}=\begin{pmatrix}422&203\\406&219\end{pmatrix}\). (A direct power on the GDC gives the same.)
Example 3 The populations (in thousands) of a species in two connected ponds change each year according to \(\mathbf{x}_{n+1}=B\mathbf{x}_n\) with \(B=\begin{pmatrix}1.1&0.2\\0.3&0.9\end{pmatrix}\). Find the long-term yearly growth factor and the long-term ratio of the second pond to the first.
The trace is \(2\) and the determinant is \(1.1\times0.9-0.2\times0.3=0.93\). So \(\lambda^{2}-2\lambda+0.93=0\).
\(\lambda=\dfrac{2\pm\sqrt{4-3.72}}{2}=1\pm\sqrt{0.07}\), which is \(1.2646\ldots\) or \(0.7354\ldots\)
The dominant eigenvalue is \(1.2646\ldots\), so in the long term the population grows by a factor of about \(1.26\) each year.
Eigenvector for \(\lambda=1.2646\): from the first row of \((B-\lambda I)\mathbf{v}=\mathbf{0}\), \((1.1-1.2646)x+0.2y=0\), so \(y=\dfrac{0.1646}{0.2}x=0.823x\).
Answer: long-term growth factor \(1.26\) per year (3 s.f.); the second pond's population tends to \(0.823\) times the first pond's.
Common errors
- Common error: including the zero vector as an eigenvector. It always satisfies \(A\mathbf{v}=\lambda\mathbf{v}\) but is excluded by definition.
- Common error: writing \(\det(A-\lambda)\) without the identity matrix. The \(\lambda\) must be subtracted only from the diagonal entries, as \(\lambda I\).
- Common error: placing eigenvectors in \(P\) in a different order from the eigenvalues in \(D\).
- Common error: computing \(A^{n}\) as \(PD^{n}P\) or \(P^{-1}D^{n}P\). The correct order is \(PD^{n}P^{-1}\).
- Common error: thinking that the answer is wrong because it differs from a calculator's eigenvector. Any non-zero multiple is correct.
Exam tip: after finding an eigenvalue and eigenvector, multiply \(A\mathbf{v}\) out and compare it with \(\lambda\mathbf{v}\). It takes a few seconds and catches arithmetic slips.
Check yourself
- Find the eigenvalues of \(\begin{pmatrix}2&1\\1&2\end{pmatrix}\).
Answer
Trace \(4\), determinant \(3\). \(\lambda^{2}-4\lambda+3=0\), so \((\lambda-1)(\lambda-3)=0\) and \(\lambda=1\) or \(3\). - Show that \(\begin{pmatrix}1\\1\end{pmatrix}\) is an eigenvector of \(\begin{pmatrix}2&1\\1&2\end{pmatrix}\) and state its eigenvalue.
Answer
\(\begin{pmatrix}2+1\\1+2\end{pmatrix}=\begin{pmatrix}3\\3\end{pmatrix}=3\begin{pmatrix}1\\1\end{pmatrix}\), so the eigenvalue is \(3\). - A 2×2 matrix has eigenvalues 3 and 5. State its trace and determinant.
Answer
Trace \(=3+5=8\). Determinant \(=3\times5=15\). - Write down the eigenvalues of \(\begin{pmatrix}2&7\\0&5\end{pmatrix}\).
Answer
The matrix is triangular, so the eigenvalues are the diagonal entries: \(2\) and \(5\).