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in Linear Algebra by Loyal (6.6k points) | 230 views

Ans 3??

$\begin{bmatrix} 1\\0 \end{bmatrix}$

$\begin{bmatrix} 0\\1 \end{bmatrix}$

$\begin{bmatrix} 1\\1 \end{bmatrix}$

@srestha , $3^{rd}$ one is linear combination of $1^{st}$ and $2^{nd}$

u mean adding two matrix, we get 3rd one


but we can use it as a eigen vector too
@srestha , yes.. linearly independent eigen vectors of an Identity matrix are same as its column/row vectors.

If we have an identity matrix $I_{n}$ then no. of linearly independent vectors = $n$. These linearly independent eigen vectors are same as its column/row vectors.

srestha mam linearly independent eigen vectors of identity matrix = 2 

Eigen value = $\lambda =1,1$

$\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$ $\begin{bmatrix} x1\\ x2 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0 \end{bmatrix}$


Rank of matrix = 0

Number of unknowns = 2

no of unknowns - rank = 2

'2' linear independent solution from which we get infinite number of solution



You are right

$r(I)=0$ and Number of unknowns$=2$

Here$,(I)<UK(0<2)$, this is the condition for Infinite many numbers of solutions, and we can assign$UK-r(I)=2-0=2$ linearly independent values to different variables.

Let say$x_{1}=k_{1}$ and $x_{2}=k_{2}$ 

So$,X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} k_{1}\\k_{2} \end{bmatrix}$

You can clearly see,there are $two$ linearly independent eigen vectors.


Can u specify

which are those two linearly independent eigen vectors are?
Can a 2x2 matrix have 3 linearly independent eigen vectors?

@Vikas Verma 

No,maximum $2$ are possible,in $2\times 2$ Matrix


@srestha Ma'am

see my comment, and Magma comment, I hope you got it

2 linear independent solution means

$\begin{bmatrix} 0&0 \\ 0 & 0 \end{bmatrix}$ $\begin{bmatrix} X1\\ X2 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0 \end{bmatrix}$


from this we get equation

0.X1 +  0.X2  = 0

Now you  put any  real number in X1 , X2 it satisfy the equation always

therefore X1 = K1

                  X2 = K2

where K1 and K2 are two arbitrary number K1,K2 $\epsilon$ R

therefore from K1 and K2 we get infinite number of solution

formula is $A-\lambda I=0$


So, equation will be

$\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$ $\begin{bmatrix} X_{1}\\ X_{2} \end{bmatrix}$=1.$\begin{bmatrix} X_{1}\\ X_{2} \end{bmatrix}$

Now we need to find $X_{1}$, $X_{2}$


@srestha Ma'am

I don't understand what you write,and it look like wrong, but i can explain

see this it might be useful


we don't need to find out 2 Eigen vectors,

$\lambda_1 = 1$

$\lambda_2 = 1$

Algebraic multiplicity of '1' = 2

Geometric multiplicity of '1' = 2

1 Answer

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Best answer

We know that for Eigenvector  $AX=\lambda X$------------------>$(1)$ Where A=Given Matrix and $X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}$ Eigen Vectors

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

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

For Eigen Values we write the characteristic  equation$:|A-\lambda I|=0$----------->$(3)$

Now,Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

First we can find the Eigen values:

Here in this question $A=I$

So,we can write the equation$:|I-\lambda I|=0$------->$(4)$

now we can find $(I-\lambda I)=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}-\lambda \begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$

$(I-\lambda I)=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}- \begin{bmatrix} \lambda & 0\\ 0 &\lambda \end{bmatrix}$

$(I-\lambda I)=\begin{bmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{bmatrix}$

Now from the equation $(4),$

$|I-\lambda I|=\begin{vmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{vmatrix}=0$

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


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

We have Another Method,to finding the Eigen Values:

Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

Let suppose Eigen Values are $\lambda_{1},\lambda_{2}$

Important Property of finding the Eigen Values:
$(1)$ Sum of All Eigen Values$=$Sum of All Leading Diagonal Elements(Trace of the Matrix)
$(2)$ Product of All Eigen Values $=Det(A)=|A|$


and $\lambda_{1}.\lambda_{2}=|I|=(1-0)=1$------------->$(6)$

Now we can find the values of $\lambda_{1},\lambda_{2}$

We know that if $a,b$ are the root the quadratic equation,so we can construct the quadratic equation


If $\lambda_{1},\lambda_{2}$ are the root of the quadratic eqation we can write the quadratic equation

In our question we have the value of $\lambda_{1}+\lambda_{2}=2,\lambda_{1}.\lambda_{2}=1$







We have Another Method,to finding the Eigen Values:

       Given that Matrix $I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$

In the case of Triangular matrix(Lower triangular matrix or Upper triangular matrix),Leading diagonal elements itself are Eigen Values.

We all know that Identity Matrix is the triangular matrix $($ Symmetric matrix also $A^{T}=A$$)$

So,Eigen Values are $\lambda_{1}=1,\lambda_{2}=1$

Now, from the equation $(2)$

                        $(A-\lambda I) X=[0]$

Here $A=I,$      $(I-\lambda I) X=[0]$

                        $(I-\lambda I)X=\begin{bmatrix} 1-\lambda & 0\\ 0 &1-\lambda \end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$

Here $\lambda=1$

$\Rightarrow \begin{bmatrix} 1-1 & 0\\ 0 &1-1 \end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$

$\Rightarrow \begin{bmatrix}0 & 0\\ 0 &0\end{bmatrix}.\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$

Rank of the matrix $,r(I)=0$ ,Number of unknowns $=2$

Here$,r(I)<UK (0<2),$ this is the condition for Infinite many numbers of solutions, and we can assign $UK−r(I)=2−0=2$ linearly independent values to different variable  for finding the $x_{1},x_{2}$      $[$UK---->Number of Unknown$]$

Here $x_{1}=k_{1},x_{2}=k_{2}$

Now  $,0.x_{1}+0.x_{2}=0$------------>$(7)$

Now,$X=\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}=\begin{bmatrix} k_{1}\\k_{2} \end{bmatrix} = k_1 \begin{bmatrix} 1\\ 0 \end{bmatrix} + k_2 \begin{bmatrix} 0\\ 1 \end{bmatrix}$

You can clearly see,there are two linearly independent eigen vector                   

by Boss (43.7k points)
edited by
@Lakshman  x1=k1 and x2=k2 , then how only 2 independent eigen vectors will be there ? it will be infinite , because k1 and k2 can be anything, please explain ?

In given question, they simply ask, how many linearly independent eigenvector

So,it is linearly independent eigenvector= Number of Unknowns$-$Rank(I)=2-0=2

If  Rank(I)<UK (0<2)

This is the case of infinitely many numbers of solution

if we want to find out the eigenvector we can solve the further

if you put $ k_1=5,k_2=10$

at a time you able to find only two eigenvectors, and they are linearly independent

clearly, see both are independent of each other




So the linearly independent eigenvectors are  k1$\begin{pmatrix} 1\\ 0 \end{pmatrix}$  and   k2$\begin{pmatrix} 0\\ 1 \end{pmatrix}$    for any values of k1, k2.  ?

Yes any value of $k$


hope you got the answer now.

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