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I am trying to find the eigenvalues for this $4\times4$ matrix $A$, where $A$ is

\begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}

I was wondering if there is an easy way to do it or do I have to find the determinant the long way?

Bérénice
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Itsnhantransitive
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  • I think that you definitely have to compute the determinant. – Crostul Apr 23 '17 at 23:07
  • My professor gave me a hint. He said he rigged this matrix so that there is an easy way to find the characteristic polynomial. I cannot think of a way... – Itsnhantransitive Apr 23 '17 at 23:08
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    Well, this is a block matrix, so the characteristic polynomial is $$((2-x)^2-1)^2$$ Compute the $2 \times 2$ determinants of the two $2 \times 2$ submatrices (first two rows and columns, last two rows and columns), and multiply them. – Crostul Apr 23 '17 at 23:10
  • Ah I see. So we don't even have to worry about the

    \begin{bmatrix} 1 & 1 \ 1 & 1 \ \end{bmatrix} ?

     – Itsnhantransitive Apr 23 '17 at 23:15
  • Yup. Look up "determinant of block matrix" for references. – Crostul Apr 23 '17 at 23:18
  • So would the determinant be the same if the matrix was this? \begin{bmatrix} 2 & 1 & -1 & -1 \ 1 & 2 & 1 & 1 \ 0 & 0 & 2 & 1 \ 0 & 0 & 1 & 2 \ \end{bmatrix} – Itsnhantransitive Apr 23 '17 at 23:22
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    I have posted a solution to the same problem a few days ago. https://math.stackexchange.com/questions/2239762/diagonalization-help-me-to-to-find-out-which-one/2239798#2239798 Also, you should look at the comments following the post. – Jacky Chong Apr 23 '17 at 23:23
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    Yes. Those four entries are irrelevant and don't affect the determinant. – Crostul Apr 23 '17 at 23:23
  • At the same time, when you see that much zeros in a matrix, the time to find a clever method, you could have calculated it... – zwim Apr 24 '17 at 00:22

1 Answers1

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Result 1: Eigenvalues of the matrix in a triangular form is the entries of the principal diagonal. For example, if we have the following matrix (Upper triangular)

\begin{bmatrix} 2 & 1 & -1 & -1 \\ 0 & 2 & 1 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}

Then eigenvalues of this matrix are given by the entries of principal diagonal which are $2, 2, ,2 ,2$. That is $2$ with algebraic multiplicity $4$.

Generalized form of above result can be summarized from the below example which may help you to solve your problem.

Consider the following matrix

\begin{bmatrix} A_1 & B \\ \mathcal{O} & A_2 \\ \end{bmatrix}

where $A_1$ and $A_2$ are square matrices and $\mathcal{O}$ is the matrix with zero entries.

Eigenvalue of above matrix can be obtained by computing the eigenvalues of matrices $A_1$ and $A_2$. Since Characteristic polynomial of this matrix is the product of Characteristic polynomial of $A_1$ and $A_2$.

In your problem you can take $A_1 $ to be

\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}

and

$A_2 $ to be

\begin{bmatrix} 2 & 1 \\ 1 & 2 \\ \end{bmatrix}

Srijan
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