For the operator T on R3 defined by T(x, y, z) = (x - y, 2x + 3y + 2z, x + y + 2z), all eigenvalues and a basis for each eigenspace as

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  1. λ1 = 1, (1, 1, -1); λ2 = 2, (2, 2, -1); λ3 = 3, (1, -2, -1)
  2. λ1 = 1, (1, 0, -1); λ2 = 2, (2, -2, -1); λ3 = 3, (1, -2, -1)
  3. λ1 = 1, (1, 0, 1); λ2 = 2, (2, -2, 1); λ3 = 3, (1, -2, -1)
  4. λ1 = 1, (1, 0, -1); λ2 = 2, (2, 2, -1); λ3 = 3, (1, -2, 1)

Answer (Detailed Solution Below)

Option 2 : λ1 = 1, (1, 0, -1); λ2 = 2, (2, -2, -1); λ3 = 3, (1, -2, -1)
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Detailed Solution

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Concept

The equation of the form |AλI| = 0 is called the characteristic equation where,

A is a square matrix 

λ is an eigenvalue and, 

I is an identity matrix, and the non-trivial solution of this equation is the eigenvalues.

The eigenspace is the space generated by the eigenvectors corresponding to the same eigenvalue - that is, the space of all vectors that can be written as a linear combination of those eigenvectors i.e, If λ  is any eigen value ,A is any square matrix, I is an identity matrix and X is any eigen vector then we have  | A - λI |.X = 0.

Calculation :

First ,we shall find all eigen values using the equation  | A - λI | = 0 .

we have T(x, y, z) = (x - y, 2x + 3y + 2z, x + y + 2z)

then the matrix form of this  can be written as [110232 112]= A (Say)

Consider,  |A - λI| = 0

⇒ |1λ1023λ2 112λ| =  0

(1λ)[(3λ)(2λ)2]+1[(42λ)2]+0=0 

⇒ (1λ)[λ25λ+4]+2(1λ)=0

⇒ (1λ)[λ25λ+6]=0

So on solving the above quadratic equation we get the following eigen values 

⇒ λ1=1,λ2=2,λ3=3 .

So the corresponding eigen space of the three eigen values can be found by the equation |A - λI | .X = 0.

Let X = [x1x2x3] be an eigen vector.

Consider, 

 (A - λI)X= 0

corresponding to λ1 eigenvalue we have,

[1λ11023λ12 112λ1].[x1x2x3] = [000] 

[11102312 1121][x1x2x3] = [000] 

[010222 111][x1x2x3] = [000] 

On doing matrix multiplication we get the following set of equations,

⇒ - x2 = 0

2x1 + 2x2 + 2x= 0 

Therefore x2 = 0  so we have x1+x3 = 0  i.e.  x1 and  x3 are dependent on each other.

so let x1 = 1  then,  x3 = -1

 eigenspace corresponding to the eigen value λ1 = 1 is ( 1, 0, -1 )

Similarly, corresponding to  λ2 eigen value  we have,

[1λ21023λ22 112λ2].[x1x2x3] = [000]

[12102322 1122].[x1x2x3] = [000] 

[110212 110].[x1x2x3] = [000] 

On doing matrix multiplication we get the following set of equations,

⇒ -x- x2 = 0 

2x1 + x2 + 2x= 0  and

x1 + x2 = 0

Therefore x2 = - x1 i.e.   xand  x2 are dependent on each other.

so, let x1 = 2  then, x2 = - 2 

and  2x1 + x2 + 2x= 0

⇒ x3 = -1 

 eigenspace corresponding to the eigen value  λ2 = 2 is ( 2, -2, -1).

Finally, corresponding to  λ3  eigenvalue  we have,

[1λ31023λ32 112λ3].[x1x2x3] = [000] 

[13102332 1123].[x1x2x3] = [000] 

[210202 111].[x1x2x3] = [000] 

On doing matrix multiplication we get the following set of equations,

⇒2x- x2 = 0

2x1 + 2x3 = 0 

x+ x-x3 = 0

Therefore   x3 = - x i.e.  x1 and  x3 are dependent on each other.

Let x1 = 1 then x3 = -1 

and x1 + x2 - x= 0 

⇒ x= -2

 eigenspace corresponding to the eigen value λ3 = 3   is (1,-2,-1) .

Hence, the correct answer is option 2)

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