GO Classes 2024 | Weekly Quiz 7 | Linear Algebra | Question: 8

Answer 1

We are given that $A$ is a $3$x$3$ real symmetric matrix with distinct Eigen values thus, corresponding to them we will obtain three non-zero Linearly Independent Eigen vectors. We are given two of those vectors.

Since option (D) is a zero vector and option (B) is a vector which is same as $x_{1}$ , both options can be eliminated.

Option (A) and (C) look promising since, they are Linearly Independent to both (A) and (C) but we are given that  $A$ is a real symmetric matrix. 

Eigenvectors corresponding to distinct Eigenvalues are orthogonal in symmetric matrix 

Therefore, we should check whether the vectors are orthogonal to each other. 

The value of the dot product of Eigen vectors corresponding to any pair of different eigen values of a symmetric positive definite matrix is zero.

For (A) i.e. $x_{3}$ $= \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}$

$x_{3}^{T}$ $x_{1}$ $ = \begin{bmatrix} 1& 0&-1 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} \ =0$

$x_{3}^{T}$ $x_{2}$ $=  \begin{bmatrix} 1& 0&-1 \end{bmatrix}  \cdot \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}  =0$

Thus, Option (A) is correct.

 

For (C) i.e. $x_{3}$ $= \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$

$x_{3}^{T}$ $x_{1}$ $= \begin{bmatrix} -1& 0&1 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} =0$

$x_{3}^{T}$ $x_{2}$ $= \begin{bmatrix} -1& 0&1 \end{bmatrix} \cdot \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}  =0$

Thus, Option (C) is correct.

 

A simple observation of the options would lead us to the same conclusion as well. Only (A) and (C) are rotating the transformation matrix $A$. 

Comments

@amichopra Eigen vector correspond to different eigen value will be always orthogonal to each other, whether matrix is symmetric or not, but eigen vectors correspond to any eigen value  in symmatric matrix wil be always orthogonal to each other.

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@Anuj Negi That is incorrect.

Refer theorem 2 here https://www.quandt.com/papers/basicmatrixtheorems.pdf

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@amichopra Thanku Bhai, got it.

Answer 2

Here each eigenvalues are different so they will give us independent eigenvectors.

Option A : x3 is independent from x1 and x2, so true.

Option B : x3 is clearly dependent on x1, false.

Option C : x3 is independent from x1 and x2, true.

Option D : Eigenvectors are by definition nonzero, hence false. (but eigen values can be zero or nonzero)