Quick method for finding eigenvalues and eigenvectors in a symmetric $5 \times 5$ matrix?

Question

The matrix $B$:

$B = \pmatrix{ 0 & 0 & 0 & 0 & 0 \cr 0 & 8 & 0 & -8 & 0 \cr 0 & 0 & 8 & 0 & -8 \cr 0 & -8 & 0 & 8 & 0 \cr 0 & 0 & -8 & 0 & 8 \cr }$

Which has nonzero eigenvalues $\lambda_1=16$ and $\lambda_2=16$ and corresponding eigenvectors:

v$_1 = \pmatrix{ 0\cr \frac{1}{2} \sqrt2 \cr 0 \cr -\frac{1}{2} \sqrt2 \cr 0\cr }$ and v$_2 = \pmatrix{ 0\cr 0\cr \frac{1}{2} \sqrt2 \cr 0 \cr -\frac{1}{2} \sqrt2 \cr }$

What is the method for obtaining these eigenvalues and corresponding eigenvectors?

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It's a large matrix and I'm hoping there's some kind of easy trick to it. From what I can remember of eigen decomposition, normally I'd do:

$Ax = \lambda x \implies|A-\lambda I|x = 0$

$\implies \det \pmatrix{ 0-\lambda & 0 & 0 & 0 & 0 \cr 0 & 8-\lambda & 0 & -8 & 0 \cr 0 & 0 & 8-\lambda & 0 & -8 \cr 0 & -8 & 0 & 8-\lambda & 0 \cr 0 & 0 & -8 & 0 & 8-\lambda \cr }$ $\pmatrix{ x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5 \cr }$ = $\pmatrix{ 0 \cr 0 \cr 0 \cr 0 \cr 0 \cr }$

So the determinant is

$\implies -\lambda \det \pmatrix{ 8-\lambda & 0 & -8 & 0 \cr 0 & 8-\lambda & 0 & -8 \cr -8 & 0 & 8-\lambda & 0 \cr 0 & -8 & 0 & 8-\lambda \cr }$

$\implies -\lambda * [ (8- \lambda)\det \pmatrix{ 8-\lambda & 0 & -8 \cr 0 & 8-\lambda & 0 \cr -8 & 0 & 8-\lambda \cr }-8 \det \pmatrix{ 0 & 8-\lambda & -8 \cr -8 & 0 & 0 \cr 0 & -8 & 8-\lambda \cr }]$

etc.

There's got to be an easier way?

Answer 1

Hint: if $A$ and $B$ are square of the same order, $$\det\begin{pmatrix} A & B \\ B & A\end{pmatrix}=\det(A-B)\det(A+B)$$ So setting $$A=\begin{pmatrix}8-\lambda & 0 \\ 0 & 8-\lambda\end{pmatrix}$$

and

$$B=\begin{pmatrix}-8 & 0\\0&-8\end{pmatrix}$$

shows that the determinant of your $4\times 4$ matrix is $(16-\lambda)^2\lambda^2$, and therefore the determinant of your original matrix is $$\boxed{-\lambda^3(16-\lambda)^2}$$

Answer 2

Here's a quick way to do the problem.

We can write this as a block matrix in the following form: $$ B = \pmatrix{ 0&0&0\\ 0&8I&-8I\\ 0&-8I&8I } = 8\pmatrix{ 0&0&0\\ 0&I&-I\\ 0&-I&I } $$ $B$ is a block diagonal matrix; one block is $0$ (corresponding to an eigenvalue of $0$ with corresponding eigenvector $[1,0,0,0,0]^T$), and the other block is $8 \pmatrix{I&-I\\-I&I}$. It suffices to find the eigenvalues of $A = \pmatrix{I&-I\\-I&I}$ and multiply the result by $8$.

In fact, if you are aware of tensor products, you may note that $A = \pmatrix{1&-1\\-1&1} \otimes I$ (in fact, I will be indirectly exploiting this).

Note that the matrix $\pmatrix{1&-1\\-1&1}$ has eigenvalues $0$,$2$ and corresponding eigenvectors $[1,1]^T$ and $[1,-1]^T$.

Now, suppose that $v$ is an eigenvector of $I$ (in other words, let $v$ be any non-zero $2 \times 2$ vector). We find that $$ A \pmatrix{v\\v} = 0 $$ and $$ A \pmatrix{v\\-v} = \pmatrix{2v\\-2v} $$ We may thus establish that $A$ has eigenvectors $0,0,2,2$ As corresponding eigenvectors we may take, for example, $$ v_{0,1} = \pmatrix{1\\0\\1\\0}; v_{0,2} = \pmatrix{0\\1\\0\\1}; v_{2,1} = \pmatrix{1\\0\\-1\\0}; v_{2,2} = \pmatrix{0\\1\\0\\-1} $$

Multiplying the eigenvectors of $A$ by $8$, we note that the matrix $B$ has eigenvectors $0,0,0,16,16$. We may take the eigenvectors $$ \pmatrix{1\\0\\0\\0\\0}; \pmatrix{0\\1\\0\\1\\0}; \pmatrix{0\\0\\1\\0\\1}; \pmatrix{0\\1\\0\\-1\\0}; \pmatrix{0\\0\\1\\0\\-1} $$

Answer 3

For simple matrices, you can often find the eigenvalues and eigenvectors by observation. Once you guess an eigenvalue, its easy to find the eigenvector by solving the linear system $(A-\lambda I)x=0$. Here, you already know that the matrix is rank deficient, since one column is zero. (The corresponding eigenvector is $[1~0~0~0~0]^T$.) So $\lambda=0$ is one eigenvalue.

Also, if you can guess the eigenvector (up to a constant) by looking at the structure of $(A-\lambda I)$, you can find the corresponding eigenvalue. Here, one possibility is $k[0~1~1~1~1~1]^T$, since adding up the columns of $A$ gives you a vector of zeros. The corresponding eigenvalue is zero. So 0 has multiplicity 2.

Another way to get zeros is by subtracting the 4th column from the 2nd column (with $\lambda=16$). The eigenvector is $k[0~1~0~-1~0~0]^T$. $k$ can be found by normalization as $1/\sqrt{2}$. We similarly get your second eigenvector by considering the 3rd and 5th columns.

Now that we know 4 of the 5 eigenvalues and eigenvectors, the last is easy to find using the eigenvalue decomposition. i.e. $\lambda_5 v_5 v_5^T = A-\sum_{i=1}^4 \lambda_i v_i v_i^T$

Answer 4

The first row and column are all zeros so we may consider the $4 \times 4$ submatrix of $B$ given my removing the first row and column and remembering that $0$ is an eigenvalue.

The third row is the same as the first and the fourth row is the same as the second. So 0 is an eigenvalue with multiplicity at least 2.

The structure of the matrix reveals that if $[v_1 , v_2 , v_3 , v_4]$ is an eigenvector then $[v_4 , v_1 , v_2 , v_3]$ is also an eigenvector with the same eigenvalue! Since the all ones vector has already been accounted for (from the null space), the last two eigenvectors must correspond to the same eigenvalue.

The trace of the matrix is 32 so the last two eigenvalues must be 16 (since the trace is equal to the sum of the eigenvalues). At this point direct computation gives an eigenvector of $\lambda = 16$ and by the above observation, the other eigenvector.

Answer 5

Please refer to Finding Eigenvectors: Fast & Nontraditional way or the arXiv preprint for fast and Nontraditional approach without using the Gaussian-Jordan elimination process.

When the matrix is diagonalizable (There is a way to check that) and has a spectrum of two, there is no need to evaluate eigenvectors at all since they already appear as nonzero column vectors of certain matrices that we would like to call The eigenmatrix. We have given a general theory for diagonalizable and nondiagonalizable matrices as well.

I answered the exact same question there under examples. Please note that we did not use the elimination (Gaussian) process there since the eigenvectors are already visible as nonzero columns of certain matrices. Here is a reproduction of the answer, which is in Wikipedia. The eigenvectors are the nonzero columns of the matrix

$$\kappa(B) = \pmatrix{ \textbf{-16} & 0 & 0 & 0 & 0 \cr 0 & \textbf{-8} & 0 & -8 & 0 \cr 0 & 0 & \textbf{-8} & 0 & -8 \cr 0 & -8 & 0 & \textbf{8} & 0 \cr 0 & 0 & -8 & 0 & \textbf{8} \cr }$$

So, the associated eigenvectors of the eigenvalue 0 are the first three column vectors and that of 16 are the last two.

You may also rescale them to get nicer eigenvectors though it is not necessary. Please note that we can subtract the eigenvalue from diagonal entry of any column, if the resulting column is nonzero. If it is zero, simply pick another column to start with.

I hope you enjoy the idea. You have access to the part of the preprint under the same link.