GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 14

Answer 1

Option a :  Given A is matrix which rotate the x vector by 30 every time means it changes the direction . So , Ax = kx ( would not be satisfy for the real value of k ) the case would happen when you find out the Eigen value it will be there but that will be imaginary . ( Statement is true ) 

Option b : Eigen value will be diagonal entries only in case of Lower Triangular Matrix  , Upper Triangular Matrix , Diagonal Matrix . Statement state for every matrix that is false . ( Statement is false ) 

Option c : Singular Matrix have determinant zero . Here 3*3 matrix have Eigen value 1 , 2, 3, and their product is not zero . Hence the Matrix is not Singular (Statement is false ) .

Option d : Defination :  Eigenvectors corresponding to distinct Eigenvalues are orthogonal in symmetric matrix .  And in question it state that all its eigen vector are orthogonal it's false . May be the two eigen vector derive from the same eigen value it may be possible and their dot product may not result to zero . That's why all eigen vector need not to be orthogonal . ( Statement is False ) 

Option : B , C , D are right choice .

Comments

I think option a) is wrong ......

Suppose A i.e matrix of rotation is real sq matrix of order n, then it can't have all complex eigen values if order of A is odd since complex eigen values occur in conjugate pairs.

You may refer to following references :

https://www.math.fsu.edu/~quine/MB_10/7_rotations.pdf

https://www.quora.com/How-do-I-prove-that-1-is-one-of-the-eigen-values-in-a-rotation-matrix

Answer 2

@Sachin Mittal 1 for option D won;t the stmt is true or there is difference in symmetric or real symmetric matrix because in real symmetric matrix u have said that n orthogonal eigen vectors there 

 

Comments

in a symmetric matrix, only eigenvectors of different eigenvalues are orthogonal not every eigenvector.

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will you help me to explain option a ????? I can’t understand why it’s true . 

@Sachin Mittal 1 sir help me out !!!!!!

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@~LOSER 

Ax = $\lambda$x means Ax gives a vector that is parallel to x by factor $\lambda$. If A is such that it rotates every vector x by 30 degrees then no Ax will be in parallel to x.

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@Sachin Mittal 1 the interpretation of option A would be same if it is written as “for every vector x, the angle between ….”?

Answer 3

In general, as it was pointed out in the lecture, the eigen vectors of real symmetric matrices are linearly independent even if their eigen values shall be repeated. If the vectors are linearly independent then  they would be orthogonal too.

The question talks about symmetric matrices in general so it is false.

For real symmetric matrices the eigen vectors corresponding to distinct eigen values are orthogonal.

Proof: Let l1 and l2 be 2 distinct eigen values with their vectors v1 and v2 for a matrix A.

A v1 = l1 v1--------1

A v2 = l2 v2---------2

premultiply equation 1 by v2T(v2 Transpose) in both lhs and rhs

v2T A v1 = v2T l1 v1 = l1 v2T v1---------3

Taking transpose of Av2  and l2 v2 and post multiplying by v1 in equation 2 for both lhs and rhs

v2T AT v1 = v2T l2 v1 = l2 v2T v1

v2T A v1 = l2 v2T v1 (as AT = A for symmetric matrix)--------4

from 3 and 4

l1 v2T v1 = l2 v2T v1

l1 v2T v1 - l2 v2T v1 = 0

(l1 - l2)v2T v1 = 0

l1 is not equal to l2 so l1 - l2 is not 0. So v2T v1 is equal to 0. Hence the eigen vectors are orthogonal.