Prove $B^{-1}=I+B-B^2$

Question

I've been trying to prove this for so many hours but nothing seems to work. Probably I am just missing something. Anyone has any idea about it?

The question is to prove: $B^{-1}=I+B-B^2$ Any suggestions are accepted!

I should mention that I have the values for $B$ and $B^{-1}$ but as it is asking for proof, I guess I am not allowed to use them.

Edit: as requested B is a 3x3 matrix and has an inverse ad det(B)!=0

B= \begin{bmatrix}1&3&2\\0&-1&4\\0&0&1\end{bmatrix}

Answer 1

In general $$(I+B-B^2)(B) = B+B^2-B^3$$ and there is no reason to believe that it is the identity matrix.

Answer 2

The relation is satisfied iff $B^3-B^2-B+I=0$. In other words you are asking if $B$ satisfies a certain polynomial relation $P(B)=0$ for some polynomial $P(t) \in \mathbb{Z}[t]$ (in your case $P(t)=t^3-t^2-t+1$). This only happens if the minimal polynomial of $B$ divides $P(t)$. So, in general: every time you have this type of question you should compute the minimal polynomial $m_B(t)$ and test with the Euclidean division algorithm if $m_B(t)$ divides $P(t)$.

Answer 3

$B = $\begin{bmatrix}1&3&2\\0&-1&4\\0&0&1\end{bmatrix}

$B^2 = $ \begin{bmatrix}1&0&16\\0&1&0\\0&0&1\end{bmatrix}

$B^3 = $ \begin{bmatrix}1&3&18\\0&-1&4\\0&0&1\end{bmatrix}

Now it's easy to verify that $B + B^2 - B^3 = I$

Answer 4

It's easy to see that the eigenvalues of $B$ are $1$, $1$ and $-1$. So the characteristic polynomial is $$p(\lambda)=(\lambda-1)^2(\lambda+1)=\lambda^3-\lambda^2-\lambda+1$$ And by the Cayley-Hamilton theorem, $p(B)=0$.