If J is a 101×101 matrix with all entries equal to 1 and let I denote the identity matrix of order 101. Then what is the determinant of J-I?

Question

If $J$ is a $101\times 101$ matrix with all entries equal to $1$ and let $I$ denote the identity matrix of order $101$. Then what is the determinant of $J-I$ ?

Answer 1

Let $J$ be the $n\times n$ matrix with all entries equal to $1$ and let $I$ be the $n\times n$ identity matrix.

Furthermore, let $A=-I$ and let $u=v=\left(\begin{array}{c}1\\\vdots\\1\end{array}\right)$. Then $J-I=A+uv^T$, such that by this Lemma: $$\det(J-I)=\det\left(A+uv^T\right)=\left(1+v^TA^{-1}u\right)\det(A)$$

• $A=-I$, such that $\det(A)=(-1)^n$.
• $A^{-1}=-I$, such that $A^{-1}u=\left(\begin{array}{c}-1\\\vdots\\-1\end{array}\right)$ and $v^TA^{-1}u=-n$, thus

$$\det(J-I)=(1-n)(-1)^n$$ and for $n=101$ you get $$\det(J-I)=(1-101)(-1)^{101}=100.$$

Answer 2

The characteristic polynomial of $J$ is $$ \det(J-XI) $$ where $X$ is an indeterminate. However, $J$ has just two eigenvalues, namely $101$ and $0$, the latter with multiplicity $100$. Therefore the characteristic polynomial is $$ (0-X)^{100}(101-X) $$ and evaluating this polynomial at $X=1$ gives the answer.

More generally, if $J$ has order $n$, the characteristic polynomial is $$ (0-X)^{n-1}(n-X) $$

Note. The algebraic multiplicity of the eigenvalue $n$ is at least $1$, whereas the geometric multiplicity of the eigenvalue $0$ is exactly $n-1$, so also the algebraic multiplicity of the eigenvalue $0$ must be $n-1$ (as it cannot be larger than $n-1$ because of the other eigenvalue).

Answer 3

Try for $3\times 3$ matrix and an identity matrix $I_3$ you get the determinant of $J-I$ as $2$ then what will the detreminant of $101\times 101$ matrix its $100$. Note diagonal elements become $0$. Hope its clear.