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UPSC 2014 Maths Optional Paper 1 Q3c-i — Step-by-Step Solution

8 marks · Section A

Eigenvalues and eigenvectors · Linear Algebra · asked 10× in 14 yrs · Read the full method →

Question

Let $A=\begin{bmatrix}-2 & 2 & -3\\ 2 & 1 & -6\\ -1 & -2 & 0\end{bmatrix}$. Find the eigen values of $A$ and the corresponding eigen vectors.

Technique

Characteristic polynomial → factor → eigenvalues; null space of $A-\lambda I$ for each eigenvalue.

Solution

Step 1 — Characteristic polynomial

$\det(A-\lambda I)=\det\begin{pmatrix}-2-\lambda & 2 & -3\\ 2 & 1-\lambda & -6\\ -1 & -2 & -\lambda\end{pmatrix}$.

Expand along row 1:

Factor $\lambda^{2}-\lambda-12=(\lambda-4)(\lambda+3)$:

Setting $=0$: eigenvalues are $\lambda=5$ (simple) and $\lambda=-3$ (double).

Step 2 — Eigenvector for $\lambda=5$

$A-5I=\begin{pmatrix}-7 & 2 & -3\\ 2 & -4 & -6\\ -1 & -2 & -5\end{pmatrix}$.

Row reduce ($R_1\leftrightarrow -R_3$, then $R_2\to R_2-2R_1$, $R_3\to R_3+7R_1$):

From row 2: $y=-2z$. From row 1: $x=-2y-5z=4z-5z=-z$.

Eigenvector: $(-z,-2z,z)=-z(1,2,-1)$. Standard form: $\boxed{\;v_1=(1,2,-1)\;\text{for }\lambda=5.}$

Step 3 — Eigenvectors for $\lambda=-3$

$A+3I=\begin{pmatrix}1 & 2 & -3\\ 2 & 4 & -6\\ -1 & -2 & 3\end{pmatrix}$.

$R_2\to R_2-2R_1=(0,0,0)$, $R_3\to R_3+R_1=(0,0,0)$. So rank 1, leaving one equation:

Two free variables ⇒ two independent eigenvectors:

• $y=1,z=0$: $(-2,1,0)$.
• $y=0,z=1$: $(3,0,1)$.

$\boxed{\;v_2=(-2,1,0),\;v_3=(3,0,1)\;\text{for }\lambda=-3.}$

The eigenspace for $\lambda=-3$ is 2-dimensional (algebraic = geometric multiplicity 2 ⇒ $A$ is diagonalisable).

Summary