How do you compute eigenvalues/vectors of big $n\times n$ matrix? The product of the non-zero eigenvalues of the matrix is ____.
$$\pmatrix{1&0&0&0&1\\0&1&1&1&0\\0&1&1&1&0\\0&1&1&1&0\\1&0&0&0&1}$$

My attempt:
Well, answer is $6$. It's a big matrix (but not more), so finding eigenvalues by characteristics equation will be lengthy process. I'm trying for any short trick to find eigenvalues of a big $n \times n$ matrix.
This explained "eigenvalues by inspection". I'm not getting properly.

Can you explain, eigenvalues and eigenvectors by inspection for this matrix in steps please?

Can you explain in steps.
 A: You can easily see that the rank of the matrix is $2$: a row reduction brings it in the form
$$
\begin{pmatrix}
1 & 0 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$
Thus you know that the $0$ eigenvalue has geometric multiplicity $5-2=3$. On the other hand, the vectors
$$
\begin{pmatrix}
1 \\ 0 \\ 0 \\ 0 \\ 1
\end{pmatrix}
\qquad\text{and}\qquad
\begin{pmatrix}
0 \\ 1 \\ 1 \\ 1 \\ 0
\end{pmatrix}
$$
are clearly eigenvectors relative to $2$ and $3$ respectively.
Denote by $m(\lambda)$ and $d(\lambda)$ the algebraic and geometric multiplicities of $\lambda$ as eigenvalue of the matrix. Then you have
\begin{gather}
d(0)=3\le m(0) \\[6px]
m(2)\ge 1,\quad m(3)\ge 1\\[6px]
m(2)+m(3)+m(0)\le5
\end{gather}
and so you can conclude that
$$
m(2)=1,\quad m(3)=1,\quad m(0)=3
$$
and that there are no more eigenvalues.
Thus the product of the nonzero eigenvalues is $6$.
A: You can do it by guess-and-check in this case, although in general of course this won't work. The rank is $2$, so there are only two non-zero eigenvalues. 
One sees by guess-and-check the eigenvectors are $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}$ with eigenvalues $2$ and $3$. (When guessing and checking, it's not a bad idea to start with a vector of all $1$s. If you do that in this case, it won't work, but you'll quickly see why these two vectors are the right guess.)
A: Given that we are not asked to compute the eigenvalues but only the product: 
Another way is to use the property $\sum \prod \{\lambda_k\} =\sum |A_k|$ where $\{\lambda_k\}$ means a $k-$sized subset of eigenvectors (the sums on the left is over all such subsets), and $|A_k|$ are the $k-$ principal minors. (The more familiar properties $\sum \lambda_i=tr(A)$ and $\prod \lambda_i=|A|$ are particular cases of this fact, related to the coefficients of the characteristic polynomial - see eg)
In our case (two non-zero eigenvalues) we must sum over all the $2-$principal minors, which can be done quickly by inspection. 
Starting for the upper-left element, $a_{11}$ gives three $2-$minors with value $1$, (the other is zero). $a_{22} \cdots a_{44}$ give one non-zero minor each, also with value $1$. The sum then is $6$.
A: Hint the product of eigenvalues is  the determinant of the matrix $A$ here the determinant of your matrix can be solved by creating only a diagonal matrix with elements only $1$ with some operations on determinant . if you want to directly solve check it out here.
How to find the determinant of this $5 \times 5$ matrix?. 1 more trick sum of eigenvalues is trace of matrix. By observation given by Others we can see 3 eigenvalues are $0$ so sum of others should be $5$ so it can be eithe $1,4$ or $2,3$ so we just need to check whether they are eigenvalues or not .
A: Since the coefficients $c_0,\dots,c_n$ of the characteristic polynomial $\chi_A$ of $A$ are given by the symmetric functions $f_k(x_1,\dots,x_n)=\sum_{1\le i_1<\dots<i_k\le n}x_{i_1}\cdots x_{i_k}$ applied to its eigenvalues $\lambda_1,\dots,\lambda_n$ (specifically, $c_k=(-1)^{(n-k)}f_{n-k}(\lambda_1,\dots,\lambda_n)$), one can see that the lowest non-zero coefficient $c_i$ is exactly $(-1)^{n-i}$ times the product of the non-zero eigenvalues.
The reason is that if there are (counted with multiplicity) $j$ non-zero eigenvalues $\lambda_{i_1},\dots,\lambda_{i_j}$, then each monomial in $f_k$, where $k>j$, must contain at least one factor which evaluates to $0$ at $(\lambda_1,\dots,\lambda_n)$, while $f_j$ contains the (unique!) monomial $x_{i_1}\cdots x_{i_j}$ which evaluates to $\lambda_{i_1}\cdots\lambda_{i_j}\neq 0$. 
For the given matrix $A$, $\chi_a(x)=x^5-5x^4+6x^3$, giving the solution $6$.
A: You can find the characteristic polynomial, and then the last nonzero coefficient is $\pm$ the product of the non-zero eigenvalues, depending on the degree. 
For example, the characteristic polynomial for above is:
$$p(\lambda)=\det(\lambda I-A) = \lambda^5-5\lambda^4+6\lambda^3,$$ the product of the non-zero eigenvalues would be $6$. 
This is only true, in general, if you want to count with duplication, so if there is a repeated non-zero eigenvalue, then you include those in the product multiple times.
You can get the non-repeated product of the non-zero eigenvalues by computing the GCD of the two polynomails, $p(x),p'(x)$, and dividing that from $p(x)$. So:
$$q(x)=\frac{p(x)}{\gcd(p(x),p'(x))}$$
is a polynomial with the same roots as $p(x)$, but no repeated roots. Then the product of the non-zero roots, without multiplicity, will be $\pm$ the last non-zero coefficient of $q(x)$.
A: By inspection (because of all the zeros in the matrix) you can see that the span $W_2$ of the first and fifth standard basis vectors is an invariant subspace, as is the span $W_3$ of the remaining three standard basis vectors. You therefore have a decomposition $\Bbb R^5=W_2\oplus W_3$ a direct sum of two invariant subspaces (of dimension $2$ and $3$), and you can find the eigenvalues of the restrictions to those subspaces separately, and then combine them.
For $W_2$ you linear operator is a rank $1$ matrix with trace $2$, so you get eigenvalues $0$ and $2$. For $W_3$ the action is given by the $3\times 3$ sub-matrix at the center. It is rank $1$ with trace $3$, so the eigenvalues are $0$ (double) and $3$. So your nonzero (simple) eigenvalues are $2,3$ and $2\times3=6$.
This uses that fact that a rank$~1$ matrix of size $n>1$ has an eigenvalue$~0$ of multiplicity at least $n-1$ (the dimension of its kernel), and a final eigenvalue equal to its trace (which is the sum of the eigenvalues). And that a nonzero matrix with all entries equal has rank$~1$. These facts are obvious, whence "by inspection". I did not need the eigenvectors of the rank$~1$ matrices themselves, but clearly any (nonzero) column of such a matrix is an eigenvector. In this case, taking into account the initial decomposition into invariant subspaces, this gives you as eigenvectors the first two columns of your original matrix.

Since they are asking for the product of nonzero eigenvalues and your matrix is diagonalisable (since symmetric), you could alternatively interpret that number as the determinant of the action of this matrix on its image subspace. That image has (by inspection) the first two columns of you matrix as basis, and on that basis the restricted action is given by the matrix $({2\atop0}~{0\atop3})$ which has determinant $6$.
