Verifying an Eigenvector and an Eigenvalue, and Building Two Solutions of a Linear System: 04-BS-1, May 2015, Q3

Question 3: Verifying an Eigenvector and an Eigenvalue, and Building Two Solutions of a Linear System (a) 6, (b) 8, (c) 6 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Matrix $A=\begin{pmatrix}8&-14&-6\\4&-6&-4\\-2&-2&4\end{pmatrix}$; candidate eigenvector $(1,1,-2)^T$; claimed eigenvalue $2$; the linear system $\mathbf x'=A\mathbf x$.

Find. (a) The eigenvalue for $(1,1,-2)^T$. (b) An eigenvector for $\lambda=2$. (c) Two solutions to $\mathbf x'=A\mathbf x$.

Approach. (a) Compute $A\mathbf v$ directly and read off the scalar multiple. (b) Solve the homogeneous system $(A-2I)\mathbf v=0$. (c) Each eigenpair $(\lambda,\mathbf v)$ gives one solution $e^{\lambda t}\mathbf v$ to $\mathbf x'=A\mathbf x$; no third eigenpair is needed for just two solutions.

(a) Multiply $A$ by the candidate eigenvector. $$A\begin{pmatrix}1\\1\\-2\end{pmatrix}=\begin{pmatrix}8(1)-14(1)-6(-2)\\4(1)-6(1)-4(-2)\\-2(1)-2(1)+4(-2)\end{pmatrix}=\begin{pmatrix}6\\6\\-12\end{pmatrix}=6\begin{pmatrix}1\\1\\-2\end{pmatrix}.$$ So $(1,1,-2)^T$ is an eigenvector with eigenvalue $\boxed{\lambda_1=6}$.
(b) Set up $(A-2I)\mathbf v=0$. $$A-2I=\begin{pmatrix}6&-14&-6\\4&-8&-4\\-2&-2&2\end{pmatrix}.$$ Row 2 gives $4x-8y-4z=0\Rightarrow x=2y+z$. Row 3 gives $-2x-2y+2z=0\Rightarrow z=x+y$. Substituting $z=x+y$ into $x=2y+z$: $x=2y+x+y\Rightarrow0=3y\Rightarrow y=0$, so $z=x$. (Row 1: $6x-14(0)-6x=0$, automatically satisfied.)
(b) Report the eigenvector. With $y=0,\ z=x$, take $x=1$: $$\boxed{\mathbf v_2=(1,0,1)^T}.$$ Check: $A(1,0,1)^T=(8-6,\,4-4,\,-2+4)^T=(2,0,2)^T=2(1,0,1)^T$ — confirms $\lambda=2$ is indeed an eigenvalue.
(c) Assemble two solutions. Each eigenpair $(\lambda,\mathbf v)$ of $A$ gives a solution $\mathbf x(t)=e^{\lambda t}\mathbf v$ of $\mathbf x'=A\mathbf x$. Using (a) and (b): $$\boxed{\mathbf x_1(t)=e^{6t}\begin{pmatrix}1\\1\\-2\end{pmatrix},\qquad \mathbf x_2(t)=e^{2t}\begin{pmatrix}1\\0\\1\end{pmatrix}}.$$

Part	Result
(a) Eigenvalue for $(1,1,-2)^T$	$\lambda_1=6$
(b) Eigenvector for $\lambda=2$	$(1,0,1)^T$
(c) Two solutions	$e^{6t}(1,1,-2)^T$ and $e^{2t}(1,0,1)^T$

Concept & Theory

Topic. Direct verification of an eigenpair and construction of exponential solutions to a linear system of ODEs.

Key relations. $A\mathbf v=\lambda\mathbf v$ verifies a claimed eigenpair without computing the characteristic polynomial. For $\mathbf x'=A\mathbf x$, each eigenpair $(\lambda_i,\mathbf v_i)$ supplies one solution $\mathbf x_i(t)=e^{\lambda_it}\mathbf v_i$, since $\frac{d}{dt}\big(e^{\lambda t}\mathbf v\big)=\lambda e^{\lambda t}\mathbf v=e^{\lambda t}(A\mathbf v)=A\big(e^{\lambda t}\mathbf v\big)$.

Why this works. Substituting a candidate eigenvector is always faster than expanding a $3\times3$ characteristic determinant when an eigenpair is already suspected or partially given; the exponential-solution identity follows immediately from differentiating $e^{\lambda t}\mathbf v$ and using $A\mathbf v=\lambda\mathbf v$.

Common pitfall. Re-deriving the full characteristic polynomial to "double check" instead of trusting the direct substitution check in (a)/(b) — it is exact and strictly faster.

Source. Strang, Introduction to Linear Algebra, Ch. 5–6 (§5.1 Eigenvalues, §6.1 Systems of Differential Equations).

Question 3 of 8: Verifying an Eigenvector and an Eigenvalue, and Building Two Solutions of a Linear System

Question 3: Verifying an Eigenvector and an Eigenvalue, and Building Two Solutions of a Linear System (a) 6, (b) 8, (c) 6 marks