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The number of linearly independent eigen vector for eigen value 1

$\begin{bmatrix} 1 & 3 & 2 \\ 0 & 4 & 2\\ 0 &-3 & -1 \end{bmatrix}$Matrix 

After Substituing eigen value 1

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

Rank of Matrix = 1 

The number of linearly independent Vector should be equal to Matrix Rank = 1 ? The ans given is No Of Unknown - Rank ( 3-1 = 2)

asked in Linear Algebra by Active (3.1k points)
retagged by | 339 views
+1
it should be n(order of matrix) - rank of matrix A-$\lambda$.I

= 3 - 1      (since 1 is the rank of matrix A-$\lambda$.I)

= 2
0
Rank of A- lmdba I is 2 here. So 3-2=1.

1 Answer

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Given that $A=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}$

We know that $AX=\lambda X$----->$(1),$Where $X=\begin{bmatrix} x\\y \\z \end{bmatrix}$

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

$\Rightarrow (A-\lambda I) X=[0]$------->$(2),$Where $I=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$

Now,we can find $A-\lambda I=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}-\lambda \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$

$\Rightarrow A-\lambda I=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}-\begin{bmatrix} \lambda &0 &0 \\ 0 &\lambda &0 \\ 0 &0 &\lambda \end{bmatrix}$

$\Rightarrow A-\lambda I=\begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}$

Now,put the value in equation $(2)$

$(A-\lambda I)X=\begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

$For$ Eigen value $\lambda=1,$we can find the eigen vector

$\begin{bmatrix} 1-1&3 &2 \\ 0 &4-1 &2 \\ 0 &-3 &-1-1 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

$\begin{bmatrix} 0&3 &2 \\ 0 &3&2 \\ 0 &-3 &-2 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Now we can perform some operation into the matrix,which is not effect to the rank,

$R_{3}\rightarrow R_{3}+R_{2}$

$R_{2}\rightarrow R_{2}-R_{1}$

$\begin{bmatrix} 0&3 &2 \\ 0 &0&0 \\ 0 &0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Rank of the matrix$=1$ and Number of variables(Unknowns$=3$)

$r(A)=1,UK=3$

$r(A)<UK$,So,we can assign $3-1=2$ values to the two variables and solve the equation and get the solutions.

So, Linearly independent Eigen vector$=3-1=2$

Now,i can extend a bit,so it will be usefull.

When $R(A)<UK$,So,this is the case of infinite many number of solutions.

and We can assign the $UK-r(A)=3-1=2$ different values to the different variable and solve the equations.

$\Rightarrow\begin{bmatrix} 0&3 &2 \\ 0 &0&0 \\ 0 &0 &0 \end{bmatrix}.\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} 0\\0 \\0 \end{bmatrix}$

Let we can assign $x=k_{1}$ and $y=k_{2}$

Now $,0.x+3y+2z=0$

Here$,0.k_{1}+3.K_{2}+2z=0$

$\Rightarrow 0+3k_{2}+2z=0$

$\Rightarrow 3k_{2}+2z=0$

$\Rightarrow 2z=-3k_{2}$

$\Rightarrow z=-\frac{3}{2}k_{2}$

So,Eigen Vector $X=\begin{bmatrix} x\\y \\z \end{bmatrix}=\begin{bmatrix} k_{1}\\k_{2} \\-\frac{3}{2}k_{2} \end{bmatrix} =$ $k_1 \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} + k_2 \begin{bmatrix} 0\\ 1\\ \frac{-3}{2} \end{bmatrix}$

You can clearly see$,x,y$ are independent and $z$ are dependent on $y$ 

For Finding the all Eigen Values,we can do something better,

                                        $|A-\lambda I|=0$

                   $\Rightarrow\begin{vmatrix} 1-\lambda &3 &2 \\ 0 &4-\lambda &2 \\ 0 &-3 &-1-\lambda \end{vmatrix}=0$

                  $\Rightarrow (1-\lambda)\begin{vmatrix} 4-\lambda& 2\\-3 &-1-\lambda \end{vmatrix}=0$

                 $\Rightarrow (1-\lambda)[(4-\lambda)(-1-\lambda)-(-3)(2)]=0$

                 $\Rightarrow (1-\lambda)[(4-\lambda)(-1-\lambda)+6]=0$

                 $\Rightarrow (1-\lambda)[-4-4\lambda+\lambda+\lambda^{2}+6]=0$

                $\Rightarrow (1-\lambda)[\lambda^{2}-3\lambda+2]=0$

                $\Rightarrow (1-\lambda)[\lambda^{2}-2\lambda-\lambda+2]=0$

               $\Rightarrow (1-\lambda)[\lambda(\lambda-2)-1(\lambda-2)]=0$

              $\Rightarrow (1-\lambda)[(\lambda-2)(\lambda-1)]=0$

              $\Rightarrow (1-\lambda)(\lambda-1)(\lambda-2)=0$

              $\Rightarrow (-1) (\lambda-1)(\lambda-1)(\lambda-2)=0$

             $\Rightarrow (\lambda-1)(\lambda-1)(\lambda-2)=0$

              So$,\lambda=1,1,2$

We can also write like this $\lambda_{1}=1,\lambda_{2}=1,\lambda_{3}=2$

Another Method to find the Eigen Values,

$A=\begin{bmatrix} 1 &3 &2 \\ 0 &4 &2 \\ 0 &-3 &-1 \end{bmatrix}$

Let Eigen Vlaues are $\lambda_{1},\lambda_{2},\lambda_{3}$ Because we want to find the Eigen values of $3\times3$ matrix.

Important Properties for finding the Eigen Values:

$(1)$ Sum of All Eigen Values$=$Trace of the Matrix(Sum of all Leading Diagonal Elements)

$(2)$ Product of All Eigen Values$=Det(A)=|A|$

Here,$\lambda_{1}+\lambda_{2}+\lambda_{3}=1+4-1$

    $\Rightarrow \lambda_{1}+\lambda_{2}+\lambda_{3}=4$-------------->$(3)$

and $\lambda_{1}.\lambda_{2}.\lambda_{3}=|A|$

     $\lambda_{1}.\lambda_{2}.\lambda_{3}=1(-4+6)-0+0$

$\Rightarrow \lambda_{1}.\lambda_{2}.\lambda_{3}=2$----------------->$(4)$

answered by Boss (39.6k points)
edited by
0

@Lakshman Patel RJIT

For finding rank, do we need eigen values always?

0

@ For finding only rank eigen values  not require.

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