Vtyjkyuk

find the singular value of the $ n times n $ real symmetrics matrices .

$$A=left(beginmatrix
1 & 1 & ... & 1\
1 & -1 & ... & 0 \
.& .& . & .\
.& . & . & .\
. & .& & . \1 & 0 & ... & -1
endmatrixright)$$

What are the eigenvalue of the matrix $A $ ?
this is orginal question photo .

My attempts: Here A is symmetrics, as i know that if A is a real symmetrics matrics ,then the singular value are absolute value of the eigenvalue of A..

here matrix is $n times n $, so im very confuse how to find the eigenvalue..

Any hints/solution will be aprreciated

Thanks in advance

Assuming the pattern suggested by Theo Bendit, let $x$ be a vector and calculating $Ax$ gives:
$$
Ax=left(beginmatrix
1 & 1 & ... & 1\
1 & -1 & ... & 0 \
.& .& . & .\
.& . & . & .\
. & .& & . \1 & 0 & ... & -1
endmatrixright)left(beginmatrix
x_1\
x_2 \
.\
.\
.\
x_n
endmatrixright)
$$
This gives
$$
Ax=left(beginmatrix
x_1+x_2+..+x_n\
x_1-x_2 \
x_1-x_3\
.\
.\
x_1-x_n
endmatrixright)
$$
Now to find eigenvalues, consider $Ax-lambda x$
$$
Ax-lambda x = left(beginmatrix
x_1+x_2+..+x_n - lambda x_1\
x_1-(lambda+1)x_2 \
x_1-(lambda+1)x_3\
.\
.\
x_1-(lambda+1)x_n
endmatrixright)qquad(1)
$$
Now if we make $lambda = -1$ and $x_1 = 0$, we get first element of $Ax-lambda x$ as $x_2+x_3+...+x_n$ and remaining all elements as zeros. To make first element zero, we can do it in n-1 ways varying $x_2, x_3,..$etc in such a way that corresponding $x$ eigenvectors are independent and $sum_i=2^n x_i = 0$.

With this we get $n-1$ eigenvalues as $-1$.

To find other eigenvalue, equate (1) to zero. We get from 1st element:
$$
x_1+x_2+...+x_n - lambda x_1 = 0
$$
$$
x_2+x_3+...+x_n = (lambda - 1) x_1 qquad(2)
$$
Adding all the remaining $n-1$ equations:
$$
(n-1)x_1 = (lambda+1)(x_2+x_3+...+x_n)
$$
From (2)
$$
(n-1)x_1 = (lambda+1)(lambda - 1) x_1
$$
since we have already considered $x_1 = 0$ case, we get
$$
lambda ^2 = n
$$
or $lambda = sqrt n$. Note that $lambda$ cannot be equal to $- sqrt n$ as it will mean $n<1$

So we have $n$ eigenvales as $sqrt n, -1, -1, ... $(n-1)times

Hint: Assuming the pattern is as Theo Bendit suggested, you can take $n-2$ eigenvectors in the form $u_2 - u_j$ (where $u_j$ is the vector with $1$ in position $j$ and $0$ elsewhere), while the other two eigenvectors are of the form
$$ pmatrixacr 1cr .cr .cr .cr 1cr$$