GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 21

Question

The value of $'x'$ for which all the eigen values of the matrix given below $\begin{bmatrix}
2 & 3+i & 6\\
x & 3 & 2\\
1 & 4 & 3\\
\end{bmatrix}$ are real is ________ $(i = \sqrt{-1})$

• A. $3 + i$
• B. $1 - 3i$
• C. $3 - i$
• D. $1 + 3i$

Comments

if I take determinant of A by substituting the vaue of x then I am not getting real no. as determinant must be real right? coz eigen values are real..

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if eigen  values are real then it can not be guranteed that determinant of the matrix will be real as because matrix itself contains complex entries.

Answer 1

Eigen values of real Hermition (a complex square matrix that is equal to its own conjugate transpose) matrix are real.   
    
$\text{For Real Symmetric matrix}: A = A^{T}$    
    
$\text{For Hermition matrix}: A = A^{\theta} = (\;\overline{A}\;)^{T}$

Let $A = \begin{bmatrix}
2 & 3+i & 6 \\
x & 3 & 2 \\
1 & 4 & 3
\end{bmatrix}$
    
$\implies \overline{A} = \begin{bmatrix}
2 & 3-i & 6 \\
\overline{x} & 3 & 2 \\
1 & 4 & 3
\end{bmatrix}$    
    
$\implies A^{\theta} = (\;\overline{A} \;)^{T} = \begin{bmatrix}
2 & \overline{x} & 1 \\
3-i & 3 & 4 \\
6 & 2 & 3
\end{bmatrix}$        
    
 $\text{For Hermition matrix}: A^{\theta} = (\;\overline{A}\;)^{T} = A$
 
$\implies \begin{bmatrix}
2 & \overline{x} & 1 \\
3-i & 3 & 4 \\
6 & 2 & 3
\end{bmatrix} =  \begin{bmatrix}
2 & 3+i & 6 \\
x & 3 & 2 \\
1 & 4 & 3
\end{bmatrix}$    
    
Here, $x = 3 - i.$
    
So, the correct answer is $(C).$

Comments

Sir, the given matrix is not Hermition to begin with irrespective of the value of x. We can see this in the above solution as well. The value $A^\theta[1][3] = 1  $ and $A[1][3] = 6$ which clearly are not equal.

One way to solve this question is to find the determinant and equate the imaginary part to 0 as this ensures the eigenvalues are real.

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How can this be an Hermitian Matrix? It is not satisfying A = Transpose Conjuagte(A).

Can someone please clarify?

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How someone know in exam that it is Hermitian Matrix? and clearly the condition for Hermitian Matrix is not satisfied in the solution provided