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UPSC 2020 Maths Optional Paper 1 Q2b — Step-by-Step Solution

20 marks · Section A

Orthogonal and unitary matrices · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

Define an $n\times n$ matrix as $A=I-2u\cdot u^T$, where $u$ is a unit column vector. (i) Examine if $A$ is symmetric. (ii) Examine if $A$ is orthogonal. (iii) Show that $\operatorname{trace}(A)=n-2$. (iv) Find $A_{3\times3}$, when $u=\begin{bmatrix}1/3\\ 2/3\\ 2/3\end{bmatrix}$.

Technique

Properties of the Householder reflector $H=I-2uu^T$ using $u^Tu=1$ and associativity/cyclicity of trace.

Solution

Throughout, $u$ is a real unit column vector, so $u^Tu=\lVert u\rVert^2=1$. The matrix $A=I-2uu^T$ is the Householder reflector. Note $uu^T$ is $n\times n$ (outer product) while $u^Tu=1$ is a scalar (inner product).

(i) Symmetry

(ii) Orthogonality

Since $A$ is symmetric, $A^TA=A^2$. Compute, using associativity and $u^Tu=1$:

The last term: $uu^Tuu^T=u(u^Tu)u^T=u(1)u^T=uu^T$, so

Hence $A^TA=I$ (and $A=A^{-1}$).

(iii) Trace

Trace is linear, and $\operatorname{tr}(uu^T)=\sum_i u_i u_i=u^Tu=1$ (or: $\operatorname{tr}(uu^T)=\operatorname{tr}(u^Tu)$ by the cyclic property). Thus

(Consistency: eigenvalues of a Householder reflector are $-1$ once and $+1$ with multiplicity $n-1$; their sum is $(n-1)-1=n-2$.)

(iv) The $3\times3$ matrix for $u=(1/3,\,2/3,\,2/3)^T$

First check $u$ is a unit vector: $\tfrac19+\tfrac49+\tfrac49=\tfrac99=1$ ✓.

Outer product:

Then

Answer