Verifying an Eigenvector and Finding a Second Eigenpair: 04-BS-1, December 2016, Q4

Question 4: Verifying an Eigenvector and Finding a Second Eigenpair (a) 6, (b) 14 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. A $4\times4$ matrix $A$ and a specific vector $\mathbf x$ to test as an eigenvector.

Find. (a) The eigenvalue associated with $\mathbf x$. (b) An eigenvector for the eigenvalue $\lambda=3$.

Approach. (a) Compute $A\mathbf x$ directly and check it is a scalar multiple of $\mathbf x$. (b) Row-reduce $A-3I$ to find its null space.

(a) Compute $A\mathbf x$. With $\mathbf x=(2,0,-1,0)^T$, row by row: $$A\mathbf x=\begin{pmatrix}1(2)+1(0)+6(-1)-1(0)\\-1(2)+2(0)-2(-1)+1(0)\\1(2)-1(0)+0(-1)+1(0)\\1(2)+1(0)+2(-1)+2(0)\end{pmatrix}=\begin{pmatrix}-4\\0\\2\\0\end{pmatrix}=-2\begin{pmatrix}2\\0\\-1\\0\end{pmatrix}=-2\mathbf x.$$ So $\mathbf x$ is an eigenvector with eigenvalue $$\boxed{\lambda=-2}$$
(b) Form $A-3I$. $$A-3I=\begin{pmatrix}-2&1&6&-1\\-1&-1&-2&1\\1&-1&-3&1\\1&1&2&-1\end{pmatrix}$$ (Note row 2 and $-$row 4 are identical, so the system has a redundant equation and a nontrivial null space, confirming $3$ is indeed an eigenvalue.)
(b) Solve $(A-3I)\mathbf v=0$ using rows 1, 3, 4. Row 3: $v_1=v_2+3v_3-v_4$. Substituting into row 4 ($v_1+v_2+2v_3-v_4=0$): $2v_2+5v_3-2v_4=0$. Substituting the row-3 expression into row 1 ($-2v_1+v_2+6v_3-v_4=0$) collapses to $-v_2+v_4=0\Rightarrow v_2=v_4$.
(b) Finish solving. Combining $v_2=v_4$ with $2v_2+5v_3-2v_4=0$ gives $5v_3=0\Rightarrow v_3=0$. Then $v_1=v_2+0-v_4=v_2-v_4=0$ (since $v_2=v_4$). With $v_3=0,v_1=0$ and $v_2=v_4$ free, taking $v_2=v_4=1$: $$\boxed{\mathbf v=(0,1,0,1)^T}$$ (Check: $A\mathbf v=(1(0)+1(1)+6(0)-1(1),\,-1(0)+2(1)-2(0)+1(1),\,1(0)-1(1)+0(0)+1(1),\,1(0)+1(1)+2(0)+2(1))=(0,3,0,3)=3\mathbf v$ ✓.)

Quantity	Result
(a) Eigenvalue for $\mathbf x=(2,0,-1,0)^T$	$\lambda=-2$
(b) Eigenvector for $\lambda=3$	$(0,1,0,1)^T$

Concept & Theory

Topic. Verifying a given eigenvector and finding an eigenvector for a given eigenvalue, without computing the full characteristic polynomial.

Key relations. $\mathbf x$ is an eigenvector of $A$ with eigenvalue $\lambda$ iff $A\mathbf x=\lambda\mathbf x$. A scalar $\lambda$ is an eigenvalue iff $A-\lambda I$ is singular, i.e. $(A-\lambda I)\mathbf v=0$ has a nontrivial solution.

Why this works. Both parts avoid the $4\times4$ characteristic polynomial entirely: part (a) is a direct multiplication-and-compare, and part (b) only needs to exhibit one nontrivial null-space vector of $A-3I$ — the redundant row (row 2 $=-$row 4) is itself the evidence of singularity, since a $4\times4$ matrix with a repeated row (up to sign) cannot have full rank.

Common pitfall. Arithmetic slips in the row-by-row matrix–vector product for $A\mathbf x$; in part (b), not noticing the redundant row and instead grinding through a full $4\times4$ Gaussian elimination (still correct, just slower and more error-prone); reporting an eigenvector without checking it against the original $A$ (not $A-3I$) as a final verification.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 8 (§8.1 Eigenvalues, Eigenvectors).

Question 4 of 8: Verifying an Eigenvector and Finding a Second Eigenpair

Question 4: Verifying an Eigenvector and Finding a Second Eigenpair (a) 6, (b) 14 marks