Home / A-Level Further Maths 9231 / Matrices: systems and eigenvalues / Eigenvalues and eigenvectors
Cambridge International AS & A Level · Topic 2.2
A-Level Further Maths 9231 / Topic 2.2 / AS & A Level

The directions a matrix only stretches.

Most vectors change direction when a matrix acts on them, but a few special ones are only scaled. Find the scale factors from the characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\), find the vectors by solving \((\mathbf{A} - \lambda\mathbf{I})\mathbf{e} = \mathbf{0}\), and use \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) to prove facts about \(\mathbf{A}^{n}\), \(\mathbf{A}^{-1}\) and \(\mathbf{A} + k\mathbf{I}\).

Topic 2.2All matrices: systems and eigenvalues lessons →
The key idea

A non-zero vector \(\mathbf{e}\) is an eigenvector of the square matrix \(\mathbf{A}\), with eigenvalue \(\lambda\), when \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\). Rearranged, \((\mathbf{A} - \lambda\mathbf{I})\mathbf{e} = \mathbf{0}\) with \(\mathbf{e} \ne \mathbf{0}\), which is possible only when \(\mathbf{A} - \lambda\mathbf{I}\) is singular. So the eigenvalues are the roots of the characteristic equation \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\): a quadratic for a \(2 \times 2\) matrix, a cubic for a \(3 \times 3\). Each eigenvalue then gives its eigenvector from \((\mathbf{A} - \lambda\mathbf{I})\mathbf{e} = \mathbf{0}\), and any non-zero multiple of an eigenvector is an eigenvector too.

x y y = x y = −2x e v Ae = 5e Av rows of A: (4, 1), (2, 3) e and v images under A invariant lines Ae = (5, 5): same line, ×5 Av = (4, 2): pushed off the x-axis, so v is not an eigenvector eigenvalues 5 and 2; (1, −2) lies on y = −2x
Fig. 1: the matrix A with rows (4, 1) and (2, 3) sends e = (1, 1) to (5, 5), five times as long and on the same line y = x, so e is an eigenvector with eigenvalue 5. The vector v = (1, 0) goes to (4, 2), off its own line, so v is not an eigenvector. The second eigenvector (1, -2), eigenvalue 2, lies along y = -2x; both dashed lines are invariant lines of the transformation
Section 01

The characteristic equation of a 2 × 2 matrix.

Start from \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) and write \(\lambda\mathbf{e}\) as \(\lambda\mathbf{I}\mathbf{e}\), so that \((\mathbf{A} - \lambda\mathbf{I})\mathbf{e} = \mathbf{0}\). If \(\mathbf{A} - \lambda\mathbf{I}\) had an inverse, multiplying by it would give \(\mathbf{e} = \mathbf{0}\), which is not allowed. So a non-zero \(\mathbf{e}\) exists only when \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\). For \(\mathbf{A} = \begin{pmatrix}a&b\\c&d\end{pmatrix}\) this is \((a - \lambda)(d - \lambda) - bc = 0\), a quadratic in \(\lambda\).

StepWhat you do
1. Characteristic equation\(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\): subtract \(\lambda\) from each diagonal entry only
2. Eigenvaluessolve the quadratic; this course only needs real, distinct roots
3. Eigenvectorsfor each \(\lambda\), solve \((\mathbf{A} - \lambda\mathbf{I})\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{0}\); the two equations are multiples of each other, so use one
4. Checkmultiply out \(\mathbf{A}\mathbf{e}\) and confirm it equals \(\lambda\mathbf{e}\)

An eigenvector is a direction, not a single vector: if \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) then \(\mathbf{A}(k\mathbf{e}) = k\mathbf{A}\mathbf{e} = \lambda(k\mathbf{e})\) for any \(k \ne 0\). So \(\begin{pmatrix}1\\-2\end{pmatrix}\), \(\begin{pmatrix}-1\\2\end{pmatrix}\) and \(\begin{pmatrix}3\\-6\end{pmatrix}\) all describe the same eigenvector; give the simplest.

The geometry (link to 1.4). For a \(2 \times 2\) transformation, \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) says every point on the line through the origin in the direction of \(\mathbf{e}\) is sent to another point on the same line, scaled by \(\lambda\). So the eigenvectors give the invariant lines through the origin, as in Fig. 1. When \(\lambda = 1\) the line is a line of invariant points.

Worked example Eigenvalues and eigenvectors of a 2 × 2 matrix
Find the eigenvalues and corresponding eigenvectors of \(\mathbf{A} = \begin{pmatrix}4&1\\2&3\end{pmatrix}\).
  1. \(\det(\mathbf{A} - \lambda\mathbf{I}) = \begin{vmatrix}4 - \lambda&1\\2&3 - \lambda\end{vmatrix} = (4 - \lambda)(3 - \lambda) - 2\).
  2. Expand: \(\lambda^{2} - 7\lambda + 12 - 2 = \lambda^{2} - 7\lambda + 10 = 0\), so \((\lambda - 2)(\lambda - 5) = 0\) and \(\lambda = 2\) or \(\lambda = 5\).
  3. \(\lambda = 2\): \(\mathbf{A} - 2\mathbf{I} = \begin{pmatrix}2&1\\2&1\end{pmatrix}\), so \(2x + y = 0\). Take \(x = 1\), \(y = -2\): \(\mathbf{e} = \begin{pmatrix}1\\-2\end{pmatrix}\).
  4. \(\lambda = 5\): \(\mathbf{A} - 5\mathbf{I} = \begin{pmatrix}-1&1\\2&-2\end{pmatrix}\), so \(-x + y = 0\) and \(\mathbf{e} = \begin{pmatrix}1\\1\end{pmatrix}\).
  5. Check: \(\begin{pmatrix}4&1\\2&3\end{pmatrix}\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}2\\-4\end{pmatrix} = 2\begin{pmatrix}1\\-2\end{pmatrix}\) and \(\begin{pmatrix}4&1\\2&3\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}5\\5\end{pmatrix}\).
Answer: \(\lambda = 2\) with \(\begin{pmatrix}1\\-2\end{pmatrix}\); \(\lambda = 5\) with \(\begin{pmatrix}1\\1\end{pmatrix}\) (or any non-zero multiples)
Worked example Checking a vector, then the invariant lines
The matrix \(\mathbf{M} = \begin{pmatrix}1&4\\2&3\end{pmatrix}\) represents a transformation of the plane. (a) Show that \(\begin{pmatrix}2\\-1\end{pmatrix}\) is an eigenvector of \(\mathbf{M}\) and state its eigenvalue. (b) Find the other eigenvalue and hence the invariant lines of the transformation that pass through the origin.
  1. (a) \(\begin{pmatrix}1&4\\2&3\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}2 - 4\\4 - 3\end{pmatrix} = \begin{pmatrix}-2\\1\end{pmatrix} = -1\begin{pmatrix}2\\-1\end{pmatrix}\), so it is an eigenvector with eigenvalue \(-1\).
  2. (b) \((1 - \lambda)(3 - \lambda) - 8 = \lambda^{2} - 4\lambda - 5 = (\lambda + 1)(\lambda - 5) = 0\), so the other eigenvalue is \(5\).
  3. \(\lambda = 5\): \(\mathbf{M} - 5\mathbf{I} = \begin{pmatrix}-4&4\\2&-2\end{pmatrix}\), so \(x = y\) and \(\mathbf{e} = \begin{pmatrix}1\\1\end{pmatrix}\).
  4. Each eigenvector gives an invariant line through the origin: direction \(\begin{pmatrix}2\\-1\end{pmatrix}\) is the line \(y = -\tfrac{1}{2}x\), and \(\begin{pmatrix}1\\1\end{pmatrix}\) is \(y = x\).
  5. On \(y = x\) points move 5 times further from the origin; on \(y = -\tfrac{1}{2}x\) each point is sent to the point opposite it through the origin.
Answer: (a) eigenvalue \(-1\)   (b) eigenvalue \(5\) with \(\begin{pmatrix}1\\1\end{pmatrix}\); invariant lines \(y = x\) and \(y = -\tfrac{1}{2}x\)
Your turn
Find the eigenvalues and corresponding eigenvectors of \(\begin{pmatrix}2&3\\4&1\end{pmatrix}\).
Examiner trap

Subtract \(\lambda\) from the diagonal entries only: \(\mathbf{A} - \lambda\mathbf{I}\), not \(\mathbf{A} - \lambda\) taken from every entry. Watch the sign when you expand: \(\det\begin{pmatrix}4 - \lambda&1\\2&3 - \lambda\end{pmatrix}\) is \((4 - \lambda)(3 - \lambda) - 2\), and the constant term of the quadratic is \(\det\mathbf{A} = 10\), not \(12\). And the zero vector is never an eigenvector, even though \(\mathbf{A}\mathbf{0} = \lambda\mathbf{0}\) for every \(\lambda\).

Section 02

Three by three: a cubic and a vector product.

For a \(3 \times 3\) matrix, \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) is a cubic. Expand along a row (choose one with zeros if you can), collect powers of \(\lambda\), and look for a root among the factors of the constant term by the factor theorem. Divide out that factor and solve the remaining quadratic. Each eigenvector then satisfies three equations \((\mathbf{A} - \lambda\mathbf{I})\mathbf{e} = \mathbf{0}\), of which only two are independent.

The vector product shortcut. Each row \(\mathbf{r}\) of \(\mathbf{A} - \lambda\mathbf{I}\) satisfies \(\mathbf{r} \cdot \mathbf{e} = 0\), so \(\mathbf{e}\) is perpendicular to every row. Take two rows that are not parallel: their vector product is perpendicular to both, so it is an eigenvector.

\[\mathbf{e} = \mathbf{r}_{1} \times \mathbf{r}_{2} = \begin{pmatrix}a_{2}b_{3} - a_{3}b_{2}\\a_{3}b_{1} - a_{1}b_{3}\\a_{1}b_{2} - a_{2}b_{1}\end{pmatrix}\]

If those two rows are parallel the product is \(\mathbf{0}\); use a different pair. Either method is fine, but always check \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) at the end: a wrong eigenvalue makes the rows independent and no eigenvector exists, which shows up at once.

A useful check on the cubic: the sum of the eigenvalues equals the sum of the diagonal entries of \(\mathbf{A}\) (the coefficient of \(\lambda^{2}\)), and their product equals \(\det\mathbf{A}\) (the constant term, with \(\lambda = 0\)).

Worked example The characteristic equation of a 3 × 3 matrix
Find the eigenvalues of \(\mathbf{A} = \begin{pmatrix}1&1&1\\2&1&-1\\2&0&2\end{pmatrix}\).
  1. Expand \(\det(\mathbf{A} - \lambda\mathbf{I})\) along the first row: \[\begin{aligned}&\begin{vmatrix}1 - \lambda&1&1\\2&1 - \lambda&-1\\2&0&2 - \lambda\end{vmatrix}\\ &= (1 - \lambda)\big[(1 - \lambda)(2 - \lambda) - 0\big]\\ &\quad - 1\big[2(2 - \lambda) + 2\big]\\ &\quad + 1\big[0 - 2(1 - \lambda)\big]\end{aligned}\]
  2. Simplify: \[\begin{aligned}&(1 - \lambda)^{2}(2 - \lambda) - 6 + 2\lambda - 2 + 2\lambda\\ &= (2 - 5\lambda + 4\lambda^{2} - \lambda^{3}) - 8 + 4\lambda\\ &= -\lambda^{3} + 4\lambda^{2} - \lambda - 6\end{aligned}\]
  3. So the characteristic equation is \(\lambda^{3} - 4\lambda^{2} + \lambda + 6 = 0\).
  4. Try factors of 6: \(\lambda = -1\) gives \(-1 - 4 - 1 + 6 = 0\), so \((\lambda + 1)\) is a factor.
  5. Divide: \(\lambda^{3} - 4\lambda^{2} + \lambda + 6 = (\lambda + 1)(\lambda^{2} - 5\lambda + 6) = (\lambda + 1)(\lambda - 2)(\lambda - 3)\).
  6. Check: \(-1 + 2 + 3 = 4\), the sum of the diagonal \(1 + 1 + 2\); and \((-1)(2)(3) = -6 = \det\mathbf{A}\).
Answer: \(\lambda = -1\), \(2\), \(3\)
Worked example Three eigenvectors, two methods
Find an eigenvector of \(\mathbf{A} = \begin{pmatrix}1&1&1\\2&1&-1\\2&0&2\end{pmatrix}\) for each of its eigenvalues \(-1\), \(2\) and \(3\).
  1. \(\lambda = 2\), by equations: \(\mathbf{A} - 2\mathbf{I} = \begin{pmatrix}-1&1&1\\2&-1&-1\\2&0&0\end{pmatrix}\). The third row gives \(2x = 0\), so \(x = 0\); the first gives \(y + z = 0\). Take \(\mathbf{e} = \begin{pmatrix}0\\1\\-1\end{pmatrix}\).
  2. \(\lambda = -1\), by vector product: rows of \(\mathbf{A} + \mathbf{I}\) are \((2, 1, 1)\) and \((2, 2, -1)\). \[\begin{pmatrix}2\\1\\1\end{pmatrix} \times \begin{pmatrix}2\\2\\-1\end{pmatrix} = \begin{pmatrix}-1 - 2\\2 + 2\\4 - 2\end{pmatrix} = \begin{pmatrix}-3\\4\\2\end{pmatrix}\]
  3. \(\lambda = 3\): rows of \(\mathbf{A} - 3\mathbf{I}\) are \((-2, 1, 1)\) and \((2, -2, -1)\). \[\begin{pmatrix}-2\\1\\1\end{pmatrix} \times \begin{pmatrix}2\\-2\\-1\end{pmatrix} = \begin{pmatrix}-1 + 2\\2 - 2\\4 - 2\end{pmatrix} = \begin{pmatrix}1\\0\\2\end{pmatrix}\]
  4. Check one: \(\mathbf{A}\begin{pmatrix}-3\\4\\2\end{pmatrix} = \begin{pmatrix}-3 + 4 + 2\\-6 + 4 - 2\\-6 + 0 + 4\end{pmatrix} = \begin{pmatrix}3\\-4\\-2\end{pmatrix} = -1\begin{pmatrix}-3\\4\\2\end{pmatrix}\).
Answer: \(\lambda = -1\): \(\begin{pmatrix}-3\\4\\2\end{pmatrix}\); \(\lambda = 2\): \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\); \(\lambda = 3\): \(\begin{pmatrix}1\\0\\2\end{pmatrix}\)
Your turn
Show that \(1\) is an eigenvalue of \(\mathbf{B} = \begin{pmatrix}3&-1&1\\-1&2&0\\-1&1&1\end{pmatrix}\) and find a corresponding eigenvector.
Examiner trap

The middle component of the vector product is the one that goes wrong: it is \(a_{3}b_{1} - a_{1}b_{3}\), the reverse of the pattern of the other two. A sign slip there gives a vector that fails \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\), so always check. And do not use rows of \(\mathbf{A}\) itself: the rows must come from \(\mathbf{A} - \lambda\mathbf{I}\) for the eigenvalue you are working on.

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What is the characteristic equation of \(\begin{pmatrix}3&1\\2&2\end{pmatrix}\)?

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Which of these is an eigenvector of \(\begin{pmatrix}4&1\\2&3\end{pmatrix}\)?

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Given that \(\mathbf{A}\mathbf{e} = 4\mathbf{e}\) with \(\mathbf{e} \ne \mathbf{0}\), which statement is true?

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What are the eigenvalues of \(\begin{pmatrix}2&5&1\\0&-1&3\\0&0&4\end{pmatrix}\)?

Section 03

Proofs from Ae = λe.

The definition \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) is enough to prove a family of results. Each proof starts from it, applies another matrix to both sides, and uses the fact that \(\lambda\) is a number, so it can be moved past any matrix: \(\mathbf{A}(\lambda\mathbf{e}) = \lambda\mathbf{A}\mathbf{e}\).

If \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\), thenEigenvalueEigenvector
\(\mathbf{A}^{n}\), \(n\) a positive integer\(\lambda^{n}\)\(\mathbf{e}\)
\(\mathbf{A}^{-1}\), when \(\mathbf{A}\) is non-singular\(\dfrac{1}{\lambda}\)\(\mathbf{e}\)
\(\mathbf{A} + k\mathbf{I}\), \(k\) a number\(\lambda + k\)\(\mathbf{e}\)
\(c\mathbf{A}\), \(c\) a number\(c\lambda\)\(\mathbf{e}\)

The eigenvector never changes, so these combine: \(\mathbf{A}^{3} - 4\mathbf{I}\) has eigenvalues \(\lambda^{3} - 4\) with the same eigenvectors. Putting \(\lambda = 0\) into \(\det(\mathbf{A} - \lambda\mathbf{I}) = 0\) shows that \(0\) is an eigenvalue exactly when \(\det\mathbf{A} = 0\), so a non-singular matrix never has eigenvalue \(0\) and \(\tfrac{1}{\lambda}\) always makes sense. Next lesson puts the eigenvectors side by side as the columns of a matrix \(\mathbf{Q}\) and writes \(\mathbf{A} = \mathbf{Q}\mathbf{D}\mathbf{Q}^{-1}\), which turns \(\mathbf{A}^{n}\) into simple arithmetic.

Worked example Powers of A, by induction
Given that \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), prove that \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^{n}\) with eigenvalue \(\lambda^{n}\), for every positive integer \(n\).
  1. Let \(P_{n}\) be the statement \(\mathbf{A}^{n}\mathbf{e} = \lambda^{n}\mathbf{e}\).
  2. \(n = 1\): \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) is given, so \(P_{1}\) is true.
  3. Assume \(P_{k}\): \(\mathbf{A}^{k}\mathbf{e} = \lambda^{k}\mathbf{e}\). Then \[\begin{aligned}\mathbf{A}^{k+1}\mathbf{e} &= \mathbf{A}(\mathbf{A}^{k}\mathbf{e}) = \mathbf{A}(\lambda^{k}\mathbf{e})\\ &= \lambda^{k}(\mathbf{A}\mathbf{e}) = \lambda^{k}\lambda\mathbf{e} = \lambda^{k+1}\mathbf{e}\end{aligned}\]
  4. So \(P_{k}\) true implies \(P_{k+1}\) true. Since \(P_{1}\) is true, by induction \(P_{n}\) is true for every positive integer \(n\).
  5. \(\mathbf{e} \ne \mathbf{0}\) (it is an eigenvector of \(\mathbf{A}\)), so \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^{n}\) with eigenvalue \(\lambda^{n}\).
Answer: \(\mathbf{A}^{n}\mathbf{e} = \lambda^{n}\mathbf{e}\) for all positive integers \(n\), as required
Worked example The inverse, then put it to work
(a) Given that \(\mathbf{A}\) is non-singular and \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) with \(\mathbf{e} \ne \mathbf{0}\), prove that \(\lambda \ne 0\) and that \(\mathbf{A}^{-1}\mathbf{e} = \tfrac{1}{\lambda}\mathbf{e}\). (b) The matrix \(\mathbf{A} = \begin{pmatrix}1&1&1\\2&1&-1\\2&0&2\end{pmatrix}\) has eigenvalues \(-1\), \(2\), \(3\). Write down the eigenvalues of \(\mathbf{A}^{-1}\) and of \(\mathbf{A}^{3} - 4\mathbf{I}\).
  1. (a) If \(\lambda = 0\) then \(\mathbf{A}\mathbf{e} = \mathbf{0}\); multiplying by \(\mathbf{A}^{-1}\) gives \(\mathbf{e} = \mathbf{0}\), a contradiction. So \(\lambda \ne 0\).
  2. Multiply \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) on the left by \(\mathbf{A}^{-1}\): \(\mathbf{e} = \lambda\mathbf{A}^{-1}\mathbf{e}\).
  3. Divide by \(\lambda \ne 0\): \(\mathbf{A}^{-1}\mathbf{e} = \tfrac{1}{\lambda}\mathbf{e}\), so \(\tfrac{1}{\lambda}\) is an eigenvalue of \(\mathbf{A}^{-1}\) with the same eigenvector.
  4. (b) \(\mathbf{A}^{-1}\): \(\tfrac{1}{-1} = -1\), \(\tfrac{1}{2}\), \(\tfrac{1}{3}\).
  5. \((\mathbf{A}^{3} - 4\mathbf{I})\mathbf{e} = \mathbf{A}^{3}\mathbf{e} - 4\mathbf{e} = (\lambda^{3} - 4)\mathbf{e}\), so the eigenvalues are \(-1 - 4 = -5\), \(8 - 4 = 4\) and \(27 - 4 = 23\).
Answer: (b) \(\mathbf{A}^{-1}\): \(-1, \tfrac{1}{2}, \tfrac{1}{3}\); \(\mathbf{A}^{3} - 4\mathbf{I}\): \(-5, 4, 23\), with the same eigenvectors as \(\mathbf{A}\)
Your turn
Prove that if \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) with \(\mathbf{e} \ne \mathbf{0}\), then \(\mathbf{e}\) is an eigenvector of \(\mathbf{A} + k\mathbf{I}\) with eigenvalue \(\lambda + k\). Hence state the eigenvalues of \(\mathbf{A} - 5\mathbf{I}\) for the matrix \(\mathbf{A}\) above.
Examiner trap

Multiply both sides on the LEFT by the same matrix: \(\mathbf{A}^{-1}(\mathbf{A}\mathbf{e}) = \mathbf{A}^{-1}(\lambda\mathbf{e})\). Never 'divide by \(\mathbf{e}\)' or cancel it: you cannot divide by a vector, and \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) does NOT mean \(\mathbf{A} = \lambda\mathbf{I}\). Also, these rules work because the SAME vector \(\mathbf{e}\) appears throughout: the eigenvalues of \(\mathbf{A} + \mathbf{B}\) are not, in general, the sums of eigenvalues of \(\mathbf{A}\) and \(\mathbf{B}\), unless they share that eigenvector.

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The matrix \(\begin{pmatrix}k&2\\3&1\end{pmatrix}\) has an eigenvalue \(4\). Find \(k\).

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The matrix \(\mathbf{A} = \begin{pmatrix}0&1&2\\0&3&2\\2&0&2\end{pmatrix}\) has an eigenvalue \(4\). Which is a corresponding eigenvector?

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A \(3 \times 3\) matrix \(\mathbf{A}\) has eigenvalues \(1\), \(-2\) and \(3\). What are the eigenvalues of \(\mathbf{A}^{2} + 2\mathbf{I}\)?

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The characteristic equation of a \(3 \times 3\) matrix is \(\lambda^{3} - 2\lambda^{2} - 5\lambda + 6 = 0\). What are its eigenvalues?

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Six original Cambridge-style questions
Eigenvalues and eigenvectors of \(2 \times 2\) and \(3 \times 3\) matrices with full working, an unknown entry found from a given eigenvalue, a proof for \(\mathbf{A}\mathbf{B}\) and \(\mathbf{A} + \mathbf{B}\), eigenvalues of \(\mathbf{A}^{-1}\) and \(\mathbf{A}^{3} - 5\mathbf{A}\), and \(\mathbf{A}^{5}\) applied to a vector through its eigenvectors.