Hard Further Maths Further matrix algebra Questions

Challenging, exam-style Further Maths Further matrix algebra questions with worked solutions. Stretch yourself on the hardest eigenvalues, characteristic-equation, complex-eigenvalues, eigenvectors problems.

eigenvaluescharacteristic-equationcomplex-eigenvalueseigenvectorssimultaneous-equationsrepeated-eigenvalues
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The matrix M=(5222)\mathbf{M} = \begin{pmatrix}5 & 2 \\ 2 & 2\end{pmatrix} has λ=1\lambda = 1 as an eigenvalue. Which of the following describes the set of points r\mathbf{r} for which Mr=r\mathbf{M}\mathbf{r} = \mathbf{r}?
Show worked solution

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(5222)−λ(1001)=(5−λ222−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}5 & 2 \\ 2 & 2\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}5 - \lambda & 2 \\ 2 & 2 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣5−λ222−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}5 - \lambda & 2 \\ 2 & 2 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Expand the 2×22\times2 determinant

    (5−λ)(2−λ)−(2)(2)=0\left(5-\lambda\right)\left(2-\lambda\right) - \left(2\right)\left(2\right) = 0

    Use ad−bcad-bc on the matrix A−λI\mathbf{A}-\lambda\mathbf{I}.

  4. Multiply out and collect the terms

    det⁡(A−λI)=λ2−7λ+6\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \lambda^{2} - 7 \lambda + 6

    This is the characteristic polynomial in its expanded form.

  5. Interpret Mr=r\mathbf{M}\mathbf{r} = \mathbf{r}

    Mr=r=1⋅r\mathbf{M}\mathbf{r} = \mathbf{r} = 1\cdot\mathbf{r}

    A fixed point is an eigenvector with eigenvalue 11 (or the origin).

  6. Find the eigenvector for λ=1\lambda = 1

    (M−I)v=0⇒v=(1−2)\left(\mathbf{M}-\mathbf{I}\right)\mathbf{v} = \mathbf{0} \Rightarrow \mathbf{v} = \begin{pmatrix}1 \\ -2\end{pmatrix}

    Every multiple of this vector is fixed by M\mathbf{M}.

  7. Convert the direction into a line

    y=−2xy = -2x

    The whole line through the origin in this direction consists of invariant points.

  8. Test a point of the line

    M(1−2)=(1−2)\mathbf{M}\begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}1 \\ -2\end{pmatrix}

    The point is mapped to itself, confirming that it is invariant.

  9. Contrast with the other eigenvalue

    λ=6⇒the line y=12x is invariant but its points are not fixed\lambda = 6 \Rightarrow \text{the line } y = \frac{1}{2}x \text{ is invariant but its points are not fixed}

    Points on that line are scaled by 66, so only the origin stays put.

  10. Note that the origin alone is not the answer

    λ=1 is an eigenvalue⇒a whole line is fixed\lambda = 1 \text{ is an eigenvalue} \Rightarrow \text{a whole line is fixed}

    A non-trivial eigenvector for λ=1\lambda = 1 exists, so more than the origin is fixed.

  11. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  12. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  13. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  14. Recall the shortcut for a 2×22\times2 characteristic equation

    λ2−(tr⁡A)λ+det⁡A=0\lambda^{2} - \left(\operatorname{tr}\mathbf{A}\right)\lambda + \det\mathbf{A} = 0

    The coefficients are the trace and the determinant, which is a quick check on the algebra.

  15. Select the correct description

    y=−2xy = -2x

    This is the line of invariant points of the transformation.

Answer
The line of invariant points y=−2xy = -2x
Question 2
9 markschallenging
The matrix A=(400012021)\mathbf{A} = \begin{pmatrix}4 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1\end{pmatrix}. Using the Cayley-Hamilton theorem, which of the following is equal to A−1\mathbf{A}^{-1}?
Show worked solution

Worked solution

  1. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(400012021)−λ(100010001)=(4−λ0001−λ2021−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}4 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}4 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 2 \\ 0 & 2 & 1 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  2. Set the determinant equal to zero

    det⁡(A−λI)=∣4−λ0001−λ2021−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}4 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 2 \\ 0 & 2 & 1 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  3. Expand the determinant along the first row

    (4−λ)∣1−λ221−λ∣−(0)∣0201−λ∣+(0)∣01−λ02∣=0\left(4 - \lambda\right)\begin{vmatrix}1 - \lambda & 2 \\ 2 & 1 - \lambda\end{vmatrix} - \left(0\right)\begin{vmatrix}0 & 2 \\ 0 & 1 - \lambda\end{vmatrix} + \left(0\right)\begin{vmatrix}0 & 1 - \lambda \\ 0 & 2\end{vmatrix} = 0

    The place signs along the first row are +, −, ++,\,-,\,+.

  4. Evaluate the three 2×22\times2 minors

    ∣1−λ221−λ∣=λ2−2λ−3,∣0201−λ∣=0,∣01−λ02∣=0\begin{vmatrix}1 - \lambda & 2 \\ 2 & 1 - \lambda\end{vmatrix} = \lambda^{2} - 2 \lambda - 3,\qquad \begin{vmatrix}0 & 2 \\ 0 & 1 - \lambda\end{vmatrix} = 0,\qquad \begin{vmatrix}0 & 1 - \lambda \\ 0 & 2\end{vmatrix} = 0

    Each minor is an ordinary 2×22\times2 determinant, here a polynomial in λ\lambda.

  5. Multiply out and collect the terms

    det⁡(A−λI)=−λ3+6λ2−5λ−12\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = - \lambda^{3} + 6 \lambda^{2} - 5 \lambda - 12

    This is the characteristic polynomial in its expanded form.

  6. Multiply through by −1-1 to make the equation monic

    λ3−6λ2+5λ+12=0\lambda^{3} - 6 \lambda^{2} + 5 \lambda + 12 = 0

    An odd-sized determinant produces −λ3-\lambda^{3}; changing every sign puts the equation in the required form.

  7. Apply the Cayley-Hamilton theorem

    A3−6A2+5A+12I=0\mathbf{A}^{3}- 6\mathbf{A}^{2}+ 5\mathbf{A}+ 12\mathbf{I} = \mathbf{0}

    The matrix satisfies its own characteristic equation.

  8. Multiply through by A−1\mathbf{A}^{-1}

    every power drops by 1, and IA−1=A−1\text{every power drops by }1\text{, and }\mathbf{I}\mathbf{A}^{-1} = \mathbf{A}^{-1}

    This is legitimate because det⁡A=−12≠0\det\mathbf{A} = -12 \neq 0.

  9. Make A−1\mathbf{A}^{-1} the subject

    A−1=−112A2+12A−512I\mathbf{A}^{-1} = -\frac{1}{12}\mathbf{A}^{2}+ \frac{1}{2}\mathbf{A}- \frac{5}{12}\mathbf{I}

    Divide by the constant term of the characteristic equation and change the signs.

  10. Check against the inverse computed directly

    A−1=(14000−1323023−13)\mathbf{A}^{-1} = \begin{pmatrix}\frac{1}{4} & 0 & 0 \\ 0 & - \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{2}{3} & - \frac{1}{3}\end{pmatrix}

    Evaluating the chosen expression gives exactly this matrix.

  11. Check the product

    AA−1=(100010001)\mathbf{A}\mathbf{A}^{-1} = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}

    The candidate expression really is a two-sided inverse.

  12. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  13. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  14. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  15. Recall the shortcut for a 2×22\times2 characteristic equation

    λ2−(tr⁡A)λ+det⁡A=0\lambda^{2} - \left(\operatorname{tr}\mathbf{A}\right)\lambda + \det\mathbf{A} = 0

    The coefficients are the trace and the determinant, which is a quick check on the algebra.

  16. Select the correct expression

    A−1=−112A2+12A−512I\mathbf{A}^{-1} = -\frac{1}{12}\mathbf{A}^{2}+ \frac{1}{2}\mathbf{A}- \frac{5}{12}\mathbf{I}

    Only this combination evaluates to the inverse of A\mathbf{A}.

Answer
−112A2+12A−512I-\frac{1}{12}\mathbf{A}^{2}+ \frac{1}{2}\mathbf{A}- \frac{5}{12}\mathbf{I}
Question 3
9 markschallenging
Given A=(4213)\mathbf{A} = \begin{pmatrix}4 & 2 \\ 1 & 3\end{pmatrix}, find the eigenvalues of 2A2−3I2\mathbf{A}^{2}-3\mathbf{I}, giving them in increasing order.
Show worked solution

Worked solution

  1. Form the characteristic equation

    det⁡(A−λI)=0⇒λ2−7λ+10=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0 \Rightarrow \lambda^{2} - 7 \lambda + 10 = 0

    Expanding det⁡(A−λI)\det\left(\mathbf{A}-\lambda\mathbf{I}\right) and writing the result with leading coefficient 11 gives this equation.

  2. Solve the characteristic equation

    (λ−5)(λ−2)=0⇒λ=2, 5\left(\lambda - 5\right) \left(\lambda - 2\right) = 0 \Rightarrow \lambda = 2,\ 5

    Factorising picks out the eigenvalues; a repeated factor gives a repeated eigenvalue.

  3. Apply the eigenvalue to a polynomial in A\mathbf{A}

    Av=λv⇒(2A2−3I)v=(2λ2−3)v\mathbf{A}\mathbf{v} = \lambda\mathbf{v} \Rightarrow \left(2\mathbf{A}^{2}-3\mathbf{I}\right)\mathbf{v} = \left(2\lambda^{2}-3\right)\mathbf{v}

    Each power of A\mathbf{A} acting on an eigenvector contributes the matching power of λ\lambda, so v\mathbf{v} is still an eigenvector.

  4. Substitute each eigenvalue in turn

    λ=2⇒μ=5,λ=5⇒μ=47\lambda = 2 \Rightarrow \mu = 5,\qquad \lambda = 5 \Rightarrow \mu = 47

    The eigenvectors are unchanged, so these are all the eigenvalues of the new matrix.

  5. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(4213)−λ(1001)=(4−λ213−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}4 & 2 \\ 1 & 3\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}4 - \lambda & 2 \\ 1 & 3 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  6. Set the determinant equal to zero

    det⁡(A−λI)=∣4−λ213−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}4 - \lambda & 2 \\ 1 & 3 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  7. Expand the 2×22\times2 determinant

    (4−λ)(3−λ)−(2)(1)=0\left(4-\lambda\right)\left(3-\lambda\right) - \left(2\right)\left(1\right) = 0

    Use ad−bcad-bc on the matrix A−λI\mathbf{A}-\lambda\mathbf{I}.

  8. Check the sum of the eigenvalues against the trace

    2+5=7=tr⁡A2 + 5 = 7 = \operatorname{tr}\mathbf{A}

    The trace of A\mathbf{A} is 77, which agrees.

  9. Check the product of the eigenvalues against the determinant

    (2)×(5)=10=det⁡A\left(2\right) \times \left(5\right) = 10 = \det\mathbf{A}

    The determinant of A\mathbf{A} is 1010, which agrees.

  10. Form the matrix explicitly as a check

    2A2−3I=(33281419)2\mathbf{A}^{2}-3\mathbf{I} = \begin{pmatrix}33 & 28 \\ 14 & 19\end{pmatrix}

    Its trace is 5252 and its determinant is 235235, matching the sum and the product of the new eigenvalues.

  11. Check the new eigenvalues against the new trace

    5+47=525 + 47 = 52

    The sum of the new eigenvalues equals the trace of the new matrix.

  12. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  13. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  14. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  15. State the eigenvalues in increasing order

    5, 475,\ 47

    These are the eigenvalues of 2A2−3I2\mathbf{A}^{2}-3\mathbf{I}.

Answer
5, 475,\ 47
Question 4
9 markschallenging
The matrix M=(120k31011)\mathbf{M} = \begin{pmatrix}1 & 2 & 0 \\ k & 3 & 1 \\ 0 & 1 & 1\end{pmatrix} has λ=4\lambda = 4 as an eigenvalue. Find the value of the constant kk.
Show worked solution

Worked solution

  1. Use the condition for λ\lambda to be an eigenvalue

    det⁡(M−λI)=0\det\left(\mathbf{M}-\lambda\mathbf{I}\right) = 0

    An eigenvalue makes M−λI\mathbf{M}-\lambda\mathbf{I} singular.

  2. Substitute λ=4\lambda = 4

    M−(4)I=(−320k−1101−3)\mathbf{M}-\left(4\right)\mathbf{I} = \begin{pmatrix}-3 & 2 & 0 \\ k & -1 & 1 \\ 0 & 1 & -3\end{pmatrix}

    The unknown kk is carried through the subtraction unchanged.

  3. Set the determinant equal to zero

    ∣−320k−1101−3∣=6k−6=0\begin{vmatrix}-3 & 2 & 0 \\ k & -1 & 1 \\ 0 & 1 & -3\end{vmatrix} = 6 k - 6 = 0

    Expanding the determinant gives an equation in kk alone.

  4. Solve for kk

    k=1k = 1

    This is the only value of kk for which λ=4\lambda = 4 is an eigenvalue.

  5. Check the answer by substituting back

    M=(120131011)⇒det⁡(M−(4)I)=0\mathbf{M} = \begin{pmatrix}1 & 2 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 1\end{pmatrix} \Rightarrow \det\left(\mathbf{M}-\left(4\right)\mathbf{I}\right) = 0

    The determinant is indeed zero, so λ=4\lambda = 4 is an eigenvalue.

  6. Find the other eigenvalues as a check

    λ=0, 1, 4\lambda = 0,\ 1,\ 4

    The full set of eigenvalues contains 44, as required.

  7. Note the alternative route

    Mv=(4)v\mathbf{M}\mathbf{v} = \left(4\right)\mathbf{v}

    One could instead impose that the system has a non-trivial solution, which leads to the same equation.

  8. Recall the definition of an eigenvalue and an eigenvector

    Av=λv,v≠0\mathbf{A}\mathbf{v} = \lambda\mathbf{v},\qquad \mathbf{v} \neq \mathbf{0}

    An eigenvector is a non-zero vector whose direction is unchanged by A\mathbf{A}.

  9. Recall the characteristic equation

    det⁡(A−λI)=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0

    This is the condition for (A−λI)v=0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0} to have a non-zero solution.

  10. Explain why the determinant must vanish

    (A−λI)v=0, v≠0\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v} = \mathbf{0},\ \mathbf{v}\neq\mathbf{0}

    If A−λI\mathbf{A}-\lambda\mathbf{I} were invertible the only solution would be v=0\mathbf{v}=\mathbf{0}, so its determinant must be zero.

  11. Recall the shortcut for a 2×22\times2 characteristic equation

    λ2−(tr⁡A)λ+det⁡A=0\lambda^{2} - \left(\operatorname{tr}\mathbf{A}\right)\lambda + \det\mathbf{A} = 0

    The coefficients are the trace and the determinant, which is a quick check on the algebra.

  12. Recall the sum of the eigenvalues

    λ1+λ2+⋯+λn=tr⁡A\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n} = \operatorname{tr}\mathbf{A}

    The trace is the sum of the leading diagonal, and it equals the sum of the eigenvalues.

  13. Recall the product of the eigenvalues

    λ1λ2⋯λn=det⁡A\lambda_{1}\lambda_{2}\cdots\lambda_{n} = \det\mathbf{A}

    The determinant is the product of the eigenvalues, counted with multiplicity.

  14. Recall that eigenvectors are fixed only up to a scalar multiple

    A(cv)=c(Av)=λ(cv)\mathbf{A}\left(c\mathbf{v}\right) = c\left(\mathbf{A}\mathbf{v}\right) = \lambda\left(c\mathbf{v}\right)

    Any non-zero multiple of an eigenvector is again an eigenvector for the same eigenvalue.

  15. Recall the diagonalisation result

    A=PDP−1\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}

    The columns of P\mathbf{P} are eigenvectors and the diagonal of D\mathbf{D} holds the matching eigenvalues.

  16. State the value of kk

    k=1k = 1

    This is the required constant.

Answer
k=1k = 1
Question 5
9 markschallenging
The matrix M=(6−22−250207)\mathbf{M} = \begin{pmatrix}6 & -2 & 2 \\ -2 & 5 & 0 \\ 2 & 0 & 7\end{pmatrix} represents a linear transformation of three-dimensional space. The plane Π\Pi through the origin is spanned by the eigenvectors of M\mathbf{M} corresponding to λ=6\lambda = 6 and λ=9\lambda = 9, and is therefore invariant. Find a Cartesian equation of Π\Pi, giving your answer in the form px+qy+rz=0px+qy+rz = 0 where pp, qq and rr are integers with no common factor.
Show worked solution

Worked solution

  1. Form the characteristic equation

    det⁡(A−λI)=0⇒λ3−18λ2+99λ−162=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = 0 \Rightarrow \lambda^{3} - 18 \lambda^{2} + 99 \lambda - 162 = 0

    Expanding det⁡(A−λI)\det\left(\mathbf{A}-\lambda\mathbf{I}\right) and writing the result with leading coefficient 11 gives this equation.

  2. Solve the characteristic equation

    (λ−9)(λ−6)(λ−3)=0⇒λ=3, 6, 9\left(\lambda - 9\right) \left(\lambda - 6\right) \left(\lambda - 3\right) = 0 \Rightarrow \lambda = 3,\ 6,\ 9

    Factorising picks out the eigenvalues; a repeated factor gives a repeated eigenvalue.

  3. Find the eigenvector for λ=6\lambda = 6

    v=(1−2−2)\mathbf{v} = \begin{pmatrix}1 \\ -2 \\ -2\end{pmatrix}

    Solve (A−(6)I)v=0\left(\mathbf{A}-\left(6\right)\mathbf{I}\right)\mathbf{v} = \mathbf{0}.

  4. Find the eigenvector for λ=9\lambda = 9

    v=(2−12)\mathbf{v} = \begin{pmatrix}2 \\ -1 \\ 2\end{pmatrix}

    Solve (A−(9)I)v=0\left(\mathbf{A}-\left(9\right)\mathbf{I}\right)\mathbf{v} = \mathbf{0}.

  5. Explain why the plane is invariant

    A(sv1+tv2)=(6)sv1+(9)tv2\mathbf{A}\left(s\mathbf{v}_{1}+t\mathbf{v}_{2}\right) = \left(6\right)s\mathbf{v}_{1}+\left(9\right)t\mathbf{v}_{2}

    Any combination of the two eigenvectors is mapped to another combination of the same two eigenvectors, so the plane maps into itself.

  6. Find a normal to the plane with a vector product

    n=(1−2−2)×(2−12)=(−6−63)\mathbf{n} = \begin{pmatrix}1 \\ -2 \\ -2\end{pmatrix} \times \begin{pmatrix}2 \\ -1 \\ 2\end{pmatrix} = \begin{pmatrix}-6 \\ -6 \\ 3\end{pmatrix}

    The normal is perpendicular to both spanning vectors.

  7. Simplify the normal

    n=(22−1)\mathbf{n} = \begin{pmatrix}2 \\ 2 \\ -1\end{pmatrix}

    Dividing out the common factor keeps the same direction with smaller integers.

  8. Use n⋅r=0\mathbf{n}\cdot\mathbf{r} = 0 for a plane through the origin

    2x+2y−z=02 x + 2 y - z = 0

    The plane passes through the origin, so the constant term is zero.

  9. Form the matrix A−λI\mathbf{A}-\lambda\mathbf{I}

    A−λI=(6−22−250207)−λ(100010001)=(6−λ−22−25−λ0207−λ)\mathbf{A}-\lambda\mathbf{I} = \begin{pmatrix}6 & -2 & 2 \\ -2 & 5 & 0 \\ 2 & 0 & 7\end{pmatrix} - \lambda\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}6 - \lambda & -2 & 2 \\ -2 & 5 - \lambda & 0 \\ 2 & 0 & 7 - \lambda\end{pmatrix}

    Subtracting λ\lambda from each entry of the leading diagonal is what λI\lambda\mathbf{I} does.

  10. Set the determinant equal to zero

    det⁡(A−λI)=∣6−λ−22−25−λ0207−λ∣=0\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = \begin{vmatrix}6 - \lambda & -2 & 2 \\ -2 & 5 - \lambda & 0 \\ 2 & 0 & 7 - \lambda\end{vmatrix} = 0

    A non-zero eigenvector exists only when this matrix is singular.

  11. Expand the determinant along the first row

    (6−λ)∣5−λ007−λ∣−(−2)∣−2027−λ∣+(2)∣−25−λ20∣=0\left(6 - \lambda\right)\begin{vmatrix}5 - \lambda & 0 \\ 0 & 7 - \lambda\end{vmatrix} - \left(-2\right)\begin{vmatrix}-2 & 0 \\ 2 & 7 - \lambda\end{vmatrix} + \left(2\right)\begin{vmatrix}-2 & 5 - \lambda \\ 2 & 0\end{vmatrix} = 0

    The place signs along the first row are +, −, ++,\,-,\,+.

  12. Evaluate the three 2×22\times2 minors

    ∣5−λ007−λ∣=λ2−12λ+35,∣−2027−λ∣=2λ−14,∣−25−λ20∣=2λ−10\begin{vmatrix}5 - \lambda & 0 \\ 0 & 7 - \lambda\end{vmatrix} = \lambda^{2} - 12 \lambda + 35,\qquad \begin{vmatrix}-2 & 0 \\ 2 & 7 - \lambda\end{vmatrix} = 2 \lambda - 14,\qquad \begin{vmatrix}-2 & 5 - \lambda \\ 2 & 0\end{vmatrix} = 2 \lambda - 10

    Each minor is an ordinary 2×22\times2 determinant, here a polynomial in λ\lambda.

  13. Multiply out and collect the terms

    det⁡(A−λI)=−λ3+18λ2−99λ+162\det\left(\mathbf{A}-\lambda\mathbf{I}\right) = - \lambda^{3} + 18 \lambda^{2} - 99 \lambda + 162

    This is the characteristic polynomial in its expanded form.

  14. Check the sum of the eigenvalues against the trace

    3+6+9=18=tr⁡A3 + 6 + 9 = 18 = \operatorname{tr}\mathbf{A}

    The trace of A\mathbf{A} is 1818, which agrees.

  15. State a Cartesian equation of the plane

    2x+2y−z=02 x + 2 y - z = 0

    This is the invariant plane spanned by the two eigenvectors.

Answer
2x+2y−z=02 x + 2 y - z = 0

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