Show that matrix is invertible

Question

I have the following exercise:

Show that $\left(E_n + M\right)^{−1} = E_n − \frac{M}{\gamma + 1}$ with $M^2 = \gamma M$ and $\gamma \neq -1$ holds true.

What would be the determinant of $\left(E_n + M\right)^{−1}$?

Answer 1

HINT

Assuming $E_n$ denotes the $n \times n$ identity matrix, we have $E_n^2 = E_n$. Now note that $$ \begin{split} (E_n + M)&\left(E_n - \frac{1}{\gamma+1}M\right)\\ &= E_n^2 + ME_n - \frac{1}{\gamma + 1}E_n M - \frac{1}{\gamma + 1} M^2 \\ &= E_n + M - \frac{1}{\gamma + 1} M - \frac{\gamma}{\gamma + 1} M \\ &= E_n + \left(1 - \frac{1}{\gamma + 1} - \frac{\gamma}{\gamma + 1} \right) M \\ &= E_n \end{split} $$

Answer 2

For the determinant part:

If $\gamma = 0$ then $M^2 = 0$, and it follows that all the eigenvalues of $M$ are zero. Since the eigenvalues of $E_n + M$ are $\alpha + 1$ where $\alpha$ is an eigenvalue of $M$, this implies all the eigenvalues of $E_n + M$ are $1$, and the determinant, the product of the eigenvalues, is $1.$

If $\gamma \neq 0$ we will show that there is a non-singular matrix $P$ such that $$M = P\begin{bmatrix} \gamma E_r & 0 \\ 0 & 0 \end{bmatrix}P^{-1},\label{e:1}\tag{*}$$ where $E_r$ is the identity matrix of order $r$ where $r = \text{rank}(M)$.

So, $$E_n + M = P \begin{bmatrix} (1+\gamma) E_r & 0 \\ 0 & E_{n-r}\end{bmatrix} P^{-1},$$

and hence, $$\det(E_n + M) = (1+\gamma)^r,$$ that is, $$ \det(E_n + M)^{-1} = \frac{1}{(1+\gamma)^r}.$$

The proof of $\eqref{e:1}$ follows.

Note that in this case $\text{rank}(M^2) = \text{rank}(M)$.

For a matrix $X$ let $\mathcal{C}(X)$ denotes its column space and $\mathcal{N}(X)$ its null space.

Then since for any two matrices $A$ and $B$ we have $\text{rank}(AB) = \text{rank}(B) - \text{dim}\left(\mathcal{C}(B) \bigcap \mathcal{N}(A)\right)$, plugging $A=B=M$ in the formula gives us $\text{dim}\left(\mathcal{C}(M) \bigcap \mathcal{N}(M)\right) = 0.$

From the rank plus nullity formula we have $\dim(\mathcal{C}(M))+\dim(\mathcal{N}(M))= n$, from which it follows $\mathcal{C}(M) \oplus \mathcal{N}(M) = \mathbb{R}^n$ here $\oplus$ denotes direct sum.

Let $r = \text{rank}(M)$, then there is a basis of $\mathbb{R}^n$ : $x_1,x_2,\dots,x_n$ such that $x_1,\dots, x_r$ form a basis of $\mathcal{C}(M)$ and $x_{r+1},\dots,x_n$ form a basis of $\mathcal{N}(M)$.

For $1 \leq i \leq r$, since $x_i \in \mathcal{C}(M)$ we have $x_i = My_i$ for some $y_i$, hence $Mx_i = M^2 y_i = \gamma M y_i = \gamma x_i$, and for $i > r$ clearly $M x_i = 0$.

Let $P$ be the matrix with columns $x_1,x_2,\dots, x_n$, then $P$ is invertible, and $$MP = M \begin{bmatrix} x_1 & x_2 & \dots & x_r & x_{r+1} & \dots x_n \end{bmatrix} = \begin{bmatrix} Mx_1 & Mx_2 & \dots & Mx_r & Mx_{r+1} & \dots Mx_n \end{bmatrix} = \gamma \begin{bmatrix} x_1 & x_2 & \dots & x_r & 0 & \dots & 0 \end{bmatrix}.$$ Since $P^{-1}x_i = e_i$ where $e_i$ is the vector with $1$ at the $i$th place and zero elsewhere we have, $P^{-1}MP = \begin{bmatrix} \gamma E_r & 0 \\ 0 & 0 \end{bmatrix}.$