Calculate the determinant of $A-5I$ Question
Let $ A = 
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
 A: Row-reducing the matrix $A-5I$ is quicker.
$$
\begin{gather}
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
6 & 2 & 8 & 9 & 10 \\
11 & 12 & 8 & 14 & 15 \\
16 & 17 & 18 & 14 & 20 \\
21 & 22 & 23 & 24 & 20
\end{pmatrix} \longrightarrow
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
23 & 6 & -1 & 2 & 0 \\
32 & 9 & 6 & -2 & 0 \\
37 & 14 & 11 & 8 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
60 & 5 & 10 & 0 & 0 \\
-75 & 30 & -5 & 0 & 0
\end{pmatrix} \longrightarrow 
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-70 & 20 & 0 & 0 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-110 & 0 & 0 & 0 & 0
\end{pmatrix}
\end{gather}
$$
This takes not more than $30$ multiplications in all. The determinant is clearly $$5\cdot (-1) \cdot (-5) \cdot (-25) \cdot (-110) = 68750.$$
I chose to row-reduce the matrix in this fashion to make use of the fact that every entry in the last column is a multiple of $5$. It made the calculations easier when doing it by hand.
A: The calculation of $\lambda_4 \lambda_5$ (which in this case is equivalent to the cubic term of the characteristic polynomial) is indeed given by the sum of principal $2\times2$ subdeterminants
$$\lambda_4 \lambda_5 = \sum_{i=2}^5 \sum_{j=1}^{i-1} \left|\begin{matrix}a_{ii}&a_{ij}\\a_{ji}&a_{jj}\end{matrix}\right|.$$
In this case we actually have the easy formula $a_{ij} = 5i+j-5$, so the above subdeterminant simplifies to $-5(i-j)^2$.  We can of course evaluate the double sum algebraically, but it's just as easy to regroup the sum according to the value of $i-j$:
$$\sum_{1\le j < i \le 5} (i-j)^2 = 4\cdot 1^2 + 3\cdot 2^2 + 2\cdot 3^2 + 1\cdot4^2 = 4 + 12 + 18 + 16 = 50.$$
Thus $\lambda_4 \lambda_5 = -5(50) = -250$ and $\det(A-5I) = (-5)^3 (-250-5\cdot 65+25) = 68750$.
A: Let the standard basis of $\mathbb R^5$ be denoted $\{e_1,e_2,\dotsc,e_5\}$. Let $\vec l$ denote the leftmost column of $A$, and let $\vec 1$ be the column vector of all $1$s. We have that for all $i \in \{1,\dotsc,5\}$: $$\begin{align}&Ae_i = \vec l+(i-1)\vec1 \\&\vec l = \sum_{i=1}^5 ie_i \\&\vec 1 = \sum_{i=1}^5 e_i\end{align}$$
Since every vector in the image (or range, if you call it that) of $A$ is a linear combination of $\vec 1$ and $\vec l$, it follows that the rank of $A$ is $2$. From which it follows that $3$ of $A$'s eigenvalues are $0$.
We now proceed to find the remaining eigenvalues. We focus on the subspace of $\mathbb R^5$ spanned by $\vec 1$ and $\vec l$. We see that $$A \vec 1 = 10\vec 1 + 5\vec l \\ A \vec l= 160\vec 1 + 55\vec l$$
Restrict $A$ to the subspace $\operatorname{span}\{\vec 1, \vec l\}$, and change to the basis $\{\vec 1, \vec l \}$. You get:$$B=\begin{bmatrix}10 & 160 \\ 5 & 55 \end{bmatrix}$$
Any eigenvalue of $B$ is an eigenvalue of $A$. The eigenvalues of $B$ are $λ_{12} = \frac52 (13 \pm \sqrt{209})
$.
We've now found all of $A$'s eigenvalues.
Finally, the determinant of $A$ is $\prod_{i=1}^5(\lambda_i - 5) = \frac52(11 + \sqrt{209})\frac52(11 - \sqrt{209})(-5)(-5)(-5)=68750$.
A: To determine $\det(A - 5I)$, it suffices to compute the eigenvalues of $A$.  To that end, we note that $A$ has rank $2$, which means that it has at most $2$ non-zero eigenvalues.
Let $\lambda_1,\lambda_2$ denote these eigenvalues.  We note that
$$
\lambda_1 + \lambda_2 = \operatorname{tr}(A) = 65\\
\lambda_1^2 +  \lambda_2^2 = \operatorname{tr}(A^2) = 4725
$$
Solving these equations for $\lambda_{1},\lambda_2$ yields
$$
\lambda_{1,2} = \frac 12\left(65 \pm 5 \sqrt{209}\right)
$$
We thereby compute
$$
\det(A - 5I) = \prod_{k=1}^5 (\lambda_k - 5) = \frac 14(55^2 - 25 \cdot 209) \cdot (-5)^3 = 68750
$$

Another strategy to calculate this determinant:
Note that $A$ has a rank factorization $A = UV$, where
$$
U = \pmatrix{0&1\\5&1\\10&1\\15&1\\20&1}, \quad 
V = \pmatrix{1 & 1 & 1 & 1 & 1\\
1 & 2 & 3 & 4 & 5}.
$$
The Sylvester determinant identity then yields
$$
\det(UV - 5 I) = 5^{3} \det(VU - 5 I) = (-5)^3 \det \pmatrix{45 & 5\\200 & 10} = 68750.
$$

If you want to find the eigenvalues:
Note that whenever matrices $U,V$ are such that $UV$ and $VU$ are well defined products, $UV$ and $VU$ will have the same non-zero eigenvalues. Thus, the non-zero eigenvalues of $A = UV$ are equal to the non-zero eigenvalues of
$$
B = VU = \pmatrix{1 & 1 & 1 & 1 & 1\\
1 & 2 & 3 & 4 & 5}
\pmatrix{0&1\\5&1\\10&1\\15&1\\20&1} = 
\pmatrix{50 & 5\\200 & 15}.
$$
The eigenvalues of this matrix (both of which are non-zero) are equal to the solutions of the characteristic equation
$$
\lambda^2 - \operatorname{tr}(B) \lambda + \det(B) = 0 \implies\\
\lambda^2 - 65 \lambda - 250 = 0  \implies \\
\lambda = \frac 12\left(65 \pm 5 \sqrt{209}\right).
$$
