← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q2c — Step-by-Step Solution

12 marks · Section A

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

Question

Find the eigen values and eigen vectors of the matrix:

Technique

Direct expansion of $\det(A-\lambda I)$; integer-root search; back-solve $(A-\lambda I)x=0$ for each eigenvalue.

Solution

Note. $A$ is symmetric (real), so eigenvalues are real and there is an orthogonal basis of eigenvectors. The trace is $7$, the determinant we compute below.

Step 1 — Characteristic polynomial

Expand along row 1:

Compute each bracket:

• $(5-\lambda)(1-\lambda)-1=5-5\lambda-\lambda+\lambda^2-1=\lambda^2-6\lambda+4$.
• $(1-\lambda)-3=-\lambda-2$.
• $1-3(5-\lambda)=1-15+3\lambda=3\lambda-14$.

So

Expand $(1-\lambda)(\lambda^2-6\lambda+4)=\lambda^2-6\lambda+4-\lambda^3+6\lambda^2-4\lambda=-\lambda^3+7\lambda^2-10\lambda+4$.

Add $10\lambda-40$: $p(\lambda)=-\lambda^3+7\lambda^2-10\lambda+4+10\lambda-40=-\lambda^3+7\lambda^2-36$.

So $p(\lambda)=-\lambda^3+7\lambda^2-36$, or equivalently $\lambda^3-7\lambda^2+36=0$.

Step 2 — Solve $\lambda^3-7\lambda^2+36=0$

Try integer roots dividing $36$: $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\dots$.

$\lambda=-2$: $-8-28+36=0$ ✓.

Factor: $\lambda^3-7\lambda^2+36=(\lambda+2)(\lambda^2-9\lambda+18)=(\lambda+2)(\lambda-3)(\lambda-6)$.

Eigenvalues: $\lambda=-2,\;3,\;6$. (Trace $=-2+3+6=7$ ✓; product $=-2\cdot 3\cdot 6=-36$, and $\det A$ should equal $-(-36)=36$.)

Check determinant directly: $\det A=1(5-1)-1(1-3)+3(1-15)=4+2-42=-36$. But eigenvalue product is $-36$, and the standard relation is $\det A=\prod\lambda_i$. So $\det A=-36$ ✓.

Step 3 — Eigenvector for $\lambda=-2$

$(A+2I)x=0$: $\begin{pmatrix}3 & 1 & 3\\ 1 & 7 & 1\\ 3 & 1 & 3\end{pmatrix}x=0$.

Row 3 = Row 1, drop. From $3x+y+3z=0$ and $x+7y+z=0$. Subtract $3\times(2)$ from $(1)$: $3x+y+3z-3x-21y-3z=0\Rightarrow -20y=0\Rightarrow y=0$. Then $3x+3z=0\Rightarrow z=-x$.

Eigenvector: $v_1=(1,0,-1)$.

Step 4 — Eigenvector for $\lambda=3$

$(A-3I)x=0$: $\begin{pmatrix}-2 & 1 & 3\\ 1 & 2 & 1\\ 3 & 1 & -2\end{pmatrix}x=0$.

Try $R_1+R_3$: $(1,2,1)x=0$ — same as $R_2$! So rows 1 and 3 sum to $R_2$; two independent equations: $-2x+y+3z=0$ and $x+2y+z=0$.

From (2): $x=-2y-z$. Sub into (1): $-2(-2y-z)+y+3z=0\Rightarrow 4y+2z+y+3z=0\Rightarrow 5y+5z=0\Rightarrow y=-z$.

Then $x=-2(-z)-z=2z-z=z$. So $(x,y,z)=(z,-z,z)=z(1,-1,1)$.

Eigenvector: $v_2=(1,-1,1)$.

Step 5 — Eigenvector for $\lambda=6$

$(A-6I)x=0$: $\begin{pmatrix}-5 & 1 & 3\\ 1 & -1 & 1\\ 3 & 1 & -5\end{pmatrix}x=0$.

From $R_2$: $x-y+z=0$, so $y=x+z$. Sub into $R_1$: $-5x+x+z+3z=0\Rightarrow -4x+4z=0\Rightarrow z=x$. Then $y=2x$.

Eigenvector: $v_3=(1,2,1)$.

Step 6 — Verify orthogonality (since $A$ is symmetric)

• $v_1\cdot v_2=(1)(1)+(0)(-1)+(-1)(1)=0$ ✓.
• $v_1\cdot v_3=(1)(1)+(0)(2)+(-1)(1)=0$ ✓.
• $v_2\cdot v_3=(1)(1)+(-1)(2)+(1)(1)=0$ ✓.

Summary

Answer