Vtyjkyuk

Can we find the entry $s_23$ of $S=H^3$ where $H=pmatrix2&-1&0\3&1&2\-1&1&1
$ without finding $S$. I know that $s_23$ is given by multiplying the second row of $H^2$ and the third coloumn of $H$ but it is dull. Thank you for any hints!

If you don't want to calculate $H^3$ or $H^2$, you can take the vector $v=beginpmatrix0&0&1endpmatrix^T$ and calculate $w=H cdot H cdot H cdot v = H cdot H cdot beginpmatrix0&2&1endpmatrix^T$. The entry $s_23$ will be the second coordinate of $w$.

You need 18 multiplications and 12 additions to calculate this. Calculating $H^2$ would need 27 multiplications and 18 additions. I don't know whether this is still too 'dull' for you ;)

This works since $s_23$ is the $e_2$-component of the image of $e_3$ under $S$.

Edit: Even less work is needed by a comment of celtschk. We may extract the $e_2$-component of $H H beginpmatrix0&2&1endpmatrix^T$ by multiplying with $beginpmatrix0&1&0endpmatrix$ from the left. Therefore, we can calculate $$s_23 = beginpmatrix0&1&0endpmatrix HH beginpmatrix0&2&1endpmatrix^T = beginpmatrix3&1&1endpmatrix H beginpmatrix0\2\1endpmatrix$$ with 12 multiplications and 8 additions.

The characteristic polynomial of $H$ is $$X^3-4X^2+6X-3.$$

Therefore $$H^3=4H^2-6H+3I=H(4H-6I)+3I.$$

The element $(H^3)_23$ is then the product of the second row of $H$ times the third column of $4H-6I$ (plus nothing since $3I$ has element $(2,3)$ equal to $0$).

We need $18$ products and $12$ additions to get the polynomial, plus the expanding of terms. After that we need 1 addition (from the -6I), 3 multiplications (from the 4H), plus 3 multiplications and 2 additions from the second row-third column product.

But hey, this way is not dull!