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I want to find the eigenvalues of the matrix

$$ \left[ \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & a & a & 0 \\ 0 & a & a & 0 \\ 0 & 0 & 0 & b \end{array} \right] $$

Can somebody explain me the theory behind getting the eigenvalues of this $4\times4$ matrix? The way I see it is to take the $4$ $a$'s as a matrix itself and see the big matrix as a diagonal one. The problem is, I can't really justify this strategy.

Here are the eigenvalues from Wolfram Alpha.

Thanks

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Do you know how to find the eigenvalues of a $2\times2$ or $3\times3$ matrix? Even more directly, do you know what an eigenvalue is? –  JavaMan Jan 27 '13 at 20:35
    
en.wikipedia.org/wiki/… –  Belgi Jan 27 '13 at 20:36
    
For an ordinary 2x2 and 3x3 matrix, I use the characteristic equation. But I can't see how to apply that to bigger matrices –  Jeroen Jan 27 '13 at 20:42

2 Answers 2

You should think of your matrix as a block matrix consisting of the $1x1$, $0-\text{matrix}$, a $2x2$ matrix of all $a$'s and a $1x1$ matrix $b$. Your first column is $0$, which means that the vector $(1,0,0,0)$ has 0 eigenvalue. Now, the 2x2 matrix of all $a's$ by itself has eigenvectors $(1,1)$ and $(1,-1)$ which have eigenvalues $2a$ and 0 respectively. Since this $2x2$ matrix is within a $4x4$, convince yourself that the eigenvectors of the full matrix will be $(0,1,1,0)$ and $(0,1,-1,0)$. Finally, the last column has only 1 $b$ in the bottom so $(0,0,0,1)$ will be an eigenvector with eigenvalue $b$.

Note that the above works primarily because your matrix has diagonal block structure, with zeros everywhere else.

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The eigenvalues of $A$ are the roots of the characteristic polynomial $p(\lambda)=\det (\lambda I -A)$.

In this case, the matrix $\lambda I-A$ is made of three blocks along the diagonal.

Namely $(\lambda)$, $\left(\begin{matrix} \lambda-a & -a \\ -a & \lambda -a \end{matrix}\right)$, and $(\lambda -b)$.

The determinant is therefore equal to the product of the determinants of these three matrices.

So you find: $$ p(\lambda)=\lambda\cdot \lambda(\lambda-2a)\cdot (\lambda -b). $$

Now you see that your eigenvalues are $0$ (with multiplicity $2$), $2a$, and $b$.

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