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$\left. \begin{array} { l } { \text { If A is a skew-symmetric matrix of order n, then number of linearly } } \\ { \text { independent eigen vectors of } \left( \mathrm { A } + \mathrm { A } ^ { \mathrm { T } } \right) = } \end{array} \right.$

A) 0

B) 1

C) n -1

D) n

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I know it can be solved using

#indpendent eigen vectors = n   - rank(matrix)

I want to now how the above formula derived !!

1 Answer

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Given that $A$ is skew-symmetric matrix of order $n.$,then $A^{T}$ is also be order $n.$

We know that for skew-symmetric $A^{T}=-A$

                                                    $\Rightarrow A^{T}+A=[ 0 ]$

Let's take order $n=3$

$A=\begin{bmatrix} 0 &1 &-2 \\ -1 &0 &3 \\ 2 &-3 &0 \end{bmatrix}$

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

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

$A+A^{T}=\begin{bmatrix} 0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{bmatrix}$

Charactricstic equation for  matrix $A+A^{T}:$

$|(A+A^{T})-\lambda I|=0$       where $I=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$  Identity matrix.

$\Rightarrow \begin{vmatrix} -\lambda & 0 &0 \\ 0 & -\lambda &0 \\ 0&0 & -\lambda \end{vmatrix}=0$   where $\lambda$ is the Eigen Value.

$\Rightarrow  -\lambda(\lambda^{2}-0)=0$

$\Rightarrow  -\lambda^{3}=0$

$\Rightarrow  \lambda^{3}=0$

So,Eigen values are $\lambda_{1}=0,\lambda_{2}=0,\lambda_{3}=0$

We know that $AX=\lambda X$

$\Rightarrow(AX-\lambda X)=[0]$

$\Rightarrow((A+A^{T})X-\lambda X)=[0]$     In this question $A=A+A^{T}$

$\Rightarrow((A+A^{T})-\lambda I)X=[0]$ ------->(1) 

Now,$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}-0.\begin{bmatrix} 1 &0 &0 \\ 0&1 &0 \\ 0 & 0 &1 \end{bmatrix}$  where $\lambda=0$ and $I$ is the identity matrix.

$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}-\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}$  where $\lambda=0$ and $I$ is the identity matrix.

$(A+A^{T})-\lambda I=\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}$

Now from the equation $(1),$

$((A+A^{T})-\lambda I)X=[ 0 ]$

$\Rightarrow\begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0 & 0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\ z \end{bmatrix}=\begin{bmatrix} 0\\0 \\ 0 \end{bmatrix}$  ,where $X=\begin{bmatrix} 0\\0 \\ 0 \end{bmatrix}$ Eigen Vector

   Rank$=$Order( Dimension of the matrix)$-$Nullity

Rank of a matrix $A+A^{T} =0$ and number of unknowns(variables) are $3$.

So$,r=0,n=3$

clearly see $r<n$ This is the condition for Infinite many numbers of solutions.

we can assign $n-r=3-0=3$ linealrly independent values to the $3$ unknowns(variables).

let say $x=k_{1},y=k_{2},z=k_{3}$

So,The number of Linealrly independent Eigen Vector is $3$.

edited by

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