Eigenvalues/Eigenvectors of a $3\times3$ Matrix, and the Associated Linear System: 04-BS-1, December 2014, Q7

Question 7: Eigenvalues/Eigenvectors of a $3\times3$ Matrix, and the Associated Linear System (a) 6, (b) 6, (c) 8 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 $A=\begin{pmatrix}3&2&0\\0&1&0\\-10&-4&-2\end{pmatrix}$; candidate eigenvector $(1,-1,-2)^T$; claimed eigenvalue $3$; the linear system $\mathbf x'=A\mathbf x$.

Find. (a) The eigenvalue for $(1,-1,-2)^T$. (b) An eigenvector for $\lambda=3$. (c) The general solution $\mathbf x(t)$.

Approach. (a)–(b) Directly verify $A\mathbf v=\lambda\mathbf v$ (for (a)) and solve $(A-3I)\mathbf v=0$ (for (b)). (c) $A$ is $3\times3$, so a third eigenvalue is needed; use $\operatorname{tr}(A)=$ sum of eigenvalues to find it without a full cubic, then assemble the general solution as a linear combination of $e^{\lambda_it}\mathbf v_i$.

(a) Verify the given eigenvector. $$A\begin{pmatrix}1\\-1\\-2\end{pmatrix}=\begin{pmatrix}3(1)+2(-1)+0(-2)\\0(1)+1(-1)+0(-2)\\-10(1)-4(-1)-2(-2)\end{pmatrix}=\begin{pmatrix}1\\-1\\-2\end{pmatrix}=1\cdot\begin{pmatrix}1\\-1\\-2\end{pmatrix}.$$ So $(1,-1,-2)^T$ is an eigenvector with eigenvalue $\boxed{\lambda_1=1}$.
(b) Confirm $\lambda=3$ and find its eigenvector. Solve $(A-3I)\mathbf v=0$, i.e. $\begin{pmatrix}0&2&0\\0&-2&0\\-10&-4&-5\end{pmatrix}\mathbf v=0$. Row 1 gives $y=0$; row 3 then gives $-10x-5z=0\Rightarrow z=-2x$. Taking $x=1$: $$\boxed{\lambda_2=3,\quad\mathbf v_2=(1,0,-2)^T}$$ (check: $A(1,0,-2)^T=(3,0,-6)^T=3(1,0,-2)^T$, confirming $\lambda=3$ is indeed an eigenvalue.)
(c) Find the third eigenvalue via the trace. $\operatorname{tr}(A)=3+1-2=2=\lambda_1+\lambda_2+\lambda_3=1+3+\lambda_3\Rightarrow\lambda_3=-2$. Solving $(A+2I)\mathbf v=0$: $\begin{pmatrix}5&2&0\\0&3&0\\-10&-4&0\end{pmatrix}\mathbf v=0$ gives $y=0$ (row 2), then $5x=0\Rightarrow x=0$ (row 1), with $z$ free: $$\mathbf v_3=(0,0,1)^T.$$
(c) Assemble the general solution. With three independent real eigenpairs, $$\mathbf x(t)=\boxed{C_1e^{t}\begin{pmatrix}1\\-1\\-2\end{pmatrix}+C_2e^{3t}\begin{pmatrix}1\\0\\-2\end{pmatrix}+C_3e^{-2t}\begin{pmatrix}0\\0\\1\end{pmatrix}}.$$

Quantity	Result
(a) Eigenvalue for $(1,-1,-2)^T$	$\lambda_1=1$
(b) Eigenvector for $\lambda=3$	$(1,0,-2)^T$
Third eigenvalue (via trace)	$\lambda_3=-2$, eigenvector $(0,0,1)^T$
(c) General solution	$\mathbf x(t)=C_1e^{t}(1,-1,-2)^T+C_2e^{3t}(1,0,-2)^T+C_3e^{-2t}(0,0,1)^T$

Concept & Theory

Topic. Eigenvalue verification, eigenvector computation, and the trace shortcut for a $3\times3$ system of linear ODEs.

Key relations. $A\mathbf v=\lambda\mathbf v$ verifies a claimed eigenpair directly without computing the characteristic polynomial; $\operatorname{tr}(A)=\sum\lambda_i$ lets a missing eigenvalue be found by subtraction once the others are known. For $\mathbf x'=A\mathbf x$ with $n$ independent real eigenpairs $(\lambda_i,\mathbf v_i)$, the general solution is $\mathbf x(t)=\sum_iC_ie^{\lambda_it}\mathbf v_i$.

Why this works. Direct substitution is always faster than expanding a $3\times3$ characteristic determinant when an eigenpair is already suspected; the trace identity follows from the fact that both $\operatorname{tr}(A)$ and $\sum\lambda_i$ equal the sum of the diagonal entries of $A$'s (upper-triangular) Jordan/triangularized form.

Common pitfall. Re-deriving the full characteristic polynomial when a direct substitution check (a)/(b) or the trace shortcut (c) is far faster and less error-prone.

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

Question 7 of 8: Eigenvalues/Eigenvectors of a $3\times3$ Matrix, and the Associated Linear System

Question 7: Eigenvalues/Eigenvectors of a $3\times3$ Matrix, and the Associated Linear System (a) 6, (b) 6, (c) 8 marks