Give an expression for an inverse of a matrix, i.e., $B^{-1}$, using only $I, B, B^2$.

This seems like it shouldn't be too difficult but I am unable to find anything close.


closed as unclear what you're asking by Dietrich Burde, Yanior Weg, Lord Shark the Unknown, Lee David Chung Lin, max_zorn May 7 at 4:36

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  • $\begingroup$ Do you mean the inverse of $B^{-1}$? That's clearly $B$. $\endgroup$ – Jerry Chang May 5 at 13:25
  • $\begingroup$ No, I am asking for $B^{-1}=...$, not $(B^{-1})^{-1}=...$. Sorry for the confusion. $\endgroup$ – Montes May 5 at 13:27
  • $\begingroup$ See this question. If $B^3=0$ we only need $I,B,B^2$. But your question is not clear. $\endgroup$ – Dietrich Burde May 5 at 13:32
  • $\begingroup$ If the matrix is $3\times3$, then, given the Cayley-Hamilton theorem, we have that $B$ satisfy its own characteristic equation $B^3 + a_1 B^2 + a_2 B + a_3 I = 0$, so that $B^3$ can be expressed in terms of $B^2, B, I$. $\endgroup$ – enzotib May 5 at 13:41

If you want to express $M^{-1}$ by a linear combination of $I_n$, $M$ and $M^2$, this work if $n\leq 3$ using Cayley–Hamilton theorem. But its not possible if $n\geq 4$. Take $$M=\left(\begin{array}{cccc} 1&1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 0&0&0&1\\ \end{array}\right)$$

Then $$M^2=\left(\begin{array}{cccc} 1&2&1&0\\ 0&1&2&1\\ 0&0&1&2\\ 0&0&0&1\\ \end{array}\right)$$

So $Vect(I_4,M,M^2)\subset\{A\in M_4(\Bbb{R})~|~A_{1,4}=0\}$.

But $$M^{-1}=\left(\begin{array}{cccc} 1&-1&1&-1\\ 0&1&-1&1\\ 0&0&1&-1\\ 0&0&0&1\\ \end{array}\right)$$

So $M^{-1}$ is not a linear combination of $I_4,M$ and $M^2$.


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