Question

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
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 1

Correct Answer - Option 2 : λ1 = 1, (1, 0, -1); λ2 = 2, (2, -2, -1); λ3 = 3, (1, -2, -1)

Concept: 

The equation of the form \(| A - λ{I}|\) = 0 is called the characteristic equation where,

A is a square matrix 

\(λ\) is an eigen value and, 

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

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 \( \begin{bmatrix} {1} & {-1} & 0 \\[0.3em] 2& 3 & 2 \ \\[0.3em] 1 & 1 & 2 \\[0.3em] \end{bmatrix} \)= A (Say)

Consider,  ( A - λI ) = 0

⇒  \( \begin{bmatrix} {1} & {-1} & 0 \\[0.3em] 2& 3 & 2 \ \\[0.3em] 1 & 1 & 2 \\[0.3em] \end{bmatrix} \) - \(λ\) \( \begin{bmatrix} {1} & {0} & 0 \\[0.3em] 0& 1 & 0 \ \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \)

⇒ \( \begin{bmatrix} {1-λ} & {-1} & 0 \\[0.3em] 2& 3-λ & 2 \ \\[0.3em] 1 & 1 & 2-λ \\[0.3em] \end{bmatrix}\) =  \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \)

⇒\((1-λ)[(3-λ)(2-λ)-2]+1[(4-2λ)-2]+0 = 0\) 

⇒ \((1-λ)[λ^2-5λ+4]+2(1-λ) = 0\)

⇒ \((1-λ)[λ^2-5λ+6] = 0\)

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

⇒ \(λ_1=1,λ_2=2,λ_3=3\) .

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

Let X = \( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) be an eigen vector.

Consider , 

 (A - λI). X= 0

corresponding to \(λ _1\) eigen value  we have ,

\( \begin{bmatrix} {1-λ_1} & {-1} & 0 \\[0.3em] 2& 3-λ_1 & 2 \ \\[0.3em] 1 & 1 & 2-λ_1 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

⇒\( \begin{bmatrix} {1-1} & {-1} & 0 \\[0.3em] 2& 3-1 & 2 \ \\[0.3em] 1 & 1 & 2-1 \\[0.3em] \end{bmatrix}\). \( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

⇒\( \begin{bmatrix} {0} & {-1} & 0 \\[0.3em] 2& 2 & 2 \ \\[0.3em] 1 & 1 & 1 \\[0.3em] \end{bmatrix}\). \( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

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

⇒ - x₂ = 0

2x₁ + 2x₂ + 2x₃ = 0  and 

 

Therefore x₂ = 0  so we have x₁+x₃ = 0  i.e.  \(x_1\) and  \(x_3\) are dependent on each other.

so let x₁ = 1  then,  x₃ = -1

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

Similarly, corresponding to  λ₂ eigen value  we have,

\( \begin{bmatrix} {1-λ_2} & {-1} & 0 \\[0.3em] 2& 3-λ_2 & 2 \ \\[0.3em] 1 & 1 & 2-λ_2 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \)

⇒\( \begin{bmatrix} {1-2} & {-1} & 0 \\[0.3em] 2& 3-2 & 2 \ \\[0.3em] 1 & 1 & 2-2 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

⇒\( \begin{bmatrix} {-1} & {-1} & 0 \\[0.3em] 2& 1 & 2 \ \\[0.3em] 1 & 1 & 0 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

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

⇒ -x1 - x2 = 0 

2x1 + x2 + 2x3 = 0  and

x1 + x2 = 0

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

so, let x₁ = 2  then, x₂ = - 2 

and  2x1 + x2 + 2x3 = 0

⇒ x₃ = -1 

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

Finally ,corresponding to  λ3  eigen value  we have,

\( \begin{bmatrix} {1-λ_3} & {-1} & 0 \\[0.3em] 2& 3-λ_3 & 2 \ \\[0.3em] 1 & 1 & 2-λ_3 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

⇒\( \begin{bmatrix} {1-3} & {-1} & 0 \\[0.3em] 2& 3-3 & 2 \ \\[0.3em] 1 & 1 & 2-3 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

⇒\( \begin{bmatrix} {-2} & {-1} & 0 \\[0.3em] 2& 0 & 2 \ \\[0.3em] 1 & 1 & -1 \\[0.3em] \end{bmatrix}\).\( \begin{bmatrix} {x_1} \\[0.3em] x_2 \\[0.3em] x_3 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} {0} \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] \end{bmatrix} \) 

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

⇒2x₁ - x₂ = 0

2x₁ + 2x3 = 0 

x₁ + x₂ -x₃ = 0

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

Let x₁ = 1 then x₃ = -1 

and x1 + x2 - x3 = 0 

⇒ x₂ = -2

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

Hence, the correct answer is option 2)