IIT Madras MS DSAI Written Test 2024

Question

Given, that the eigen values of a 2 x 2 matrix are -1,1 and its singular values are 1,0. What is the rank of the matrix?
a) rank is 0
b) rank is 1
c) Such a matrix can't exist
d) rank is 2

Answer 1

$\Rightarrow $  Rank of  a matrix  is Non Zero Singular Values .

Proof- http://faculty.washington.edu/trogdon/105A/html/Lecture25.html

$\Rightarrow $ But here non zero eigen values are also given which means Full rank must be there .

From both of these we can Conclude that  No such Matrix exist. So option $C$ should be Ans.

Answer 2

Assume matrix to be

$\begin{pmatrix} a &b \\ c&-a \end{pmatrix}$

"-a" because sum of trace is sum of eigenvaues which is zero

then determinant will be ${a^2}$+bc=1 (Product of eigenvalues)

then for symmetric values they are eigenvalues of ${A^T}$A doing that gives

$\begin{pmatrix} {a^2 + b^2} & ac-ab\\ ac-ab & {c^2+a^2} \end{pmatrix}$

Determinant of this will be 0

${(a^2+b^2)(c^2+a^2)-(ac-ab)^2=0}$

simplifing this will give ${(a^2 + bc)^2=0}$

which is a contradiction because that should be 1. Hence the answer is C

 

Answer 3

Singular values = square root of eigen values , singular values are positive . If you take square root of -1 you get i which is not real positive. So such a matrix cannot exist