# Determinant of Large Matrix

$$A=\left[\begin{array}{ccccc}{-2} & {-1} & {} & {\cdots} & {-1} \\ {-1} & {-2} & {-1} & {\cdots} & {-1} \\ {} & {} & {\ddots} & {} & {} \\ {-1} & {\cdots} & {-1} & {-2} & {-1} \\ {-1} & {\cdots} & {} & {-1} & {-2}\end{array}\right] \in \mathbb{R}^{53 \times 53}$$

So we want to find determinant of this big matrix. I tried for some cases I got the pattern like for even dimension determinant is $$n+1$$ and and for odd dimension it is $$-n-1$$ so answer should be $$-54$$ ;I guess. But what is formal method to do this calculation ; idea I have in mind is to find eigenvalue and then product will give me determinant.

• First, find the eigenvalues of the matrix $J$, consisting only of $1$s (check out here: math.stackexchange.com/questions/217521/…). Then your matrix is $-J - I$. Working out the determinant should be easy enough from there. – Theo Bendit Oct 3 '19 at 7:18
• $A = -I -J$ where $I$ is the identity matrix and $J$ is the matrix full of $1$’s. $J$ has a one-dimensional eigenspace with eigenvalue n, and an $(n-1)$-dimensional eigenspace with eigenvalue $0$. So $A$ has a one-dimensional eigenspace with eigenvalue $-n-1$, and an $(n-1)$-dimensional eigenspace with eigenvalue $-1$. Therefore the determinant of $A$ is $(-n-1)(-1)^{n-1}$ – Joppy Oct 3 '19 at 7:22

We can work with general dimension $$n$$. The wanted determinant is $$(-1)^n$$ times the determinant of $$B=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & { 2}\end{array}\right].$$ Do the combination $$C_n\leftarrow \sum_{i=1}^nC_i$$. Then $$\det(B)=\det(C)$$, where $$C=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 2+n-1} \\ { 1} & { 2} & { 1} & {\cdots} & { 2+n-1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & {2+n- 1} \\ { 1} & {\cdots} & {} & { 1} & { 2+n-1}\end{array}\right],$$ hence $$\det(A)=(-1)^n(n-1)\det \left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & {1}\end{array}\right].$$ Finally, do the substitutions $$C_i\leftarrow C_i-C_n$$, $$1\leqslant i\leqslant n$$ to get that the last determinant is one.
Notice matrix $$A = -(I_{53}+ uu^T)$$ where $$u$$ is the $$53 \times 1$$ matrix full of $$1$$.
\begin{align}\det A &= (-1)^{53} \det(I_{53} + uu^T) = (-1)^{53}\det(I_{53})(1 + u^T I_{53}^{-1} u)\\ &= -(1 + 53) = -54\end{align}
An alternative approach is treat any $$53\times 1$$ matrix as a vector in $$\mathbb{R}^{53}$$.
$$e_1 = \frac{1}{\sqrt{53}} u$$ becomes an unit vector in $$\mathbb{R}^{53}$$. Extend $$e_1$$ to an orthronormal basis $$e_1, \ldots, e_{53}$$ of $$\mathbb{R}^{53}$$. i.e. take another $$52$$ vectors so that $$e_i \cdot e_j = e_i^T e_j = \begin{cases} 1, & i = j\\ 0, & \text{ otherwise }\end{cases}$$ We will have $$\displaystyle\;I_{53} = \sum_{k=1}^{53} e_i e_i^T\;$$ and $$uu^T = 53 e_1 e_1^T$$. This leads to $$A = -54 e_1 e_1^T - \sum_{i=2}^{53} e_i e_i^T$$ In this basis, $$A$$ is diagonal with diagonal entry $$-54,-1,-1,\ldots,-1$$. This means $$A$$ has a simple eigenvalue $$-54$$ and an eigenvalue $$-1$$ with multiplicity $$52$$. From this, we obtain (again) $$\det A = (-54)(-1)^{52} = -54$$