Find a real matrix \(B\) such that \[ B^4= \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&-1&1 \end{pmatrix}. \]
Solution
Put \[ a\coloneqq\sqrt[4]{2}, \qquad C\coloneqq \begin{pmatrix} a&0&0\\ 0&a&0\\ 0&1-a&1 \end{pmatrix}. \]
1. One has\[ C^2= \begin{pmatrix} a^2&0&0\\ 0&a^2&0\\ 0&1-a^2&1 \end{pmatrix}. \]
The only nondiagonal entry that requires computation is the \((3,2)\) entry. It is \[ (1-a)a+(1-a)=1-a^2. \] The remaining entries follow immediately from the triangular form of \(C\).
2. The fourth power of \(C\) is the prescribed matrix:\[ C^4= \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&-1&1 \end{pmatrix}. \]
Squaring the matrix in step 1 gives \[ C^4= \begin{pmatrix} a^4&0&0\\ 0&a^4&0\\ 0&1-a^4&1 \end{pmatrix}. \] Since \(a^4=2\), this is exactly the displayed matrix.
3. Therefore one valid choice is\[ \boxed{ B= \begin{pmatrix} \sqrt[4]{2}&0&0\\ 0&\sqrt[4]{2}&0\\ 0&1-\sqrt[4]{2}&1 \end{pmatrix}. } \]
This matrix is \(C\), and step 2 proves that its fourth power is the required matrix.
4. Q.E.D.
Step 3 gives the requested real matrix.