Try this problem from IIT JAM 2017 exam (Problem 58) and know how to evaluate Eigen value of a Matrix.
Let $\alpha, \beta, \gamma, \delta$ be the eigenvalues of the matrix
$$
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
Then find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$
Linear Algebra
Matrix
Eigen Values
But try the problem first...
Answer: $6$
IIT JAM 2017 , Problem 58
First hint
Characteristic Equation of a matrix $A$ of order $n$ is defined by $|A-xI|=0$, where $x$ is scalar and $I$ is the identity matrix of order $n$.
The roots of the characteristic equation are called the Eigen Values of that matrix.
Now it is easy. Give it a try.
Second Hint
Let
$$
A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
Then its characteristic equation is $|A-xI|=0$ for some scalar $x$ and $I$ the identity matrix of order $4$
i.e., $
\begin{vmatrix}
-x & 0 & 0 & 0 \\
1 & -x & 0 & -2 \\
0 & 1 & -x & 1 \\
0 & 0 & 1 & 2-x
\end{vmatrix}
=0$
$
\Rightarrow -x\begin{vmatrix}
-x & 0 & -2 \\
1 & -x & 1 \\
0 & 1 & 2-x
\end{vmatrix}
=0 $
$\Rightarrow -x[-x\{(-x)(2-x)\}-1]-2(1-0)]=0$
$\Rightarrow x^{4}-2 x^{3}-x^{2}+2 x=0$
$\Rightarrow x(x^{3}-2 x^{2}-x+2)=0$
$\Rightarrow x(x-1)(x-2)(x+1)=0$
Final Step
Then the eigen values are : $0,1,-1,2$
Then, $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} =0^2+1^2+(-1)^2+2^2=6$ [ANS]
Try this problem from IIT JAM 2017 exam (Problem 58) and know how to evaluate Eigen value of a Matrix.
Let $\alpha, \beta, \gamma, \delta$ be the eigenvalues of the matrix
$$
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
Then find the value of $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$
Linear Algebra
Matrix
Eigen Values
But try the problem first...
Answer: $6$
IIT JAM 2017 , Problem 58
First hint
Characteristic Equation of a matrix $A$ of order $n$ is defined by $|A-xI|=0$, where $x$ is scalar and $I$ is the identity matrix of order $n$.
The roots of the characteristic equation are called the Eigen Values of that matrix.
Now it is easy. Give it a try.
Second Hint
Let
$$
A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
Then its characteristic equation is $|A-xI|=0$ for some scalar $x$ and $I$ the identity matrix of order $4$
i.e., $
\begin{vmatrix}
-x & 0 & 0 & 0 \\
1 & -x & 0 & -2 \\
0 & 1 & -x & 1 \\
0 & 0 & 1 & 2-x
\end{vmatrix}
=0$
$
\Rightarrow -x\begin{vmatrix}
-x & 0 & -2 \\
1 & -x & 1 \\
0 & 1 & 2-x
\end{vmatrix}
=0 $
$\Rightarrow -x[-x\{(-x)(2-x)\}-1]-2(1-0)]=0$
$\Rightarrow x^{4}-2 x^{3}-x^{2}+2 x=0$
$\Rightarrow x(x^{3}-2 x^{2}-x+2)=0$
$\Rightarrow x(x-1)(x-2)(x+1)=0$
Final Step
Then the eigen values are : $0,1,-1,2$
Then, $\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2} =0^2+1^2+(-1)^2+2^2=6$ [ANS]
