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$$A=\begin{bmatrix} 2&3&4\\ 3&-5&5\\ 4&5&0\end{bmatrix}$$ Find a unitary matrix $Q$ such that $A = QHQ^{H}$, where $H$ is Hessenberg.

I am having a little trouble finding my $Q$. I know that the first row and column in $Q$ look like $[1\quad 0 \quad 0]$, and the rest of the matrix (what's left is a $2\times 2$ matrix) is what my teacher called $F$. I know that $$F = I - \frac{2vv^{T}}{v^{T}v},$$ but I'm not sure how to come up with the $v$-vector in order to find $F$, so that I can find my $Q$. Then to find $H$, I know $H = Q^{H}AQ$, which once I find $Q$, seems to be easy to find, especially since all the eigenvalues of $A$ are real and don't have any imaginary parts, since $A$ is symmetric. And since $A$ is symmetric, then $H$ will be tridiagonal.

Any help would be greatly appreciated.

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Take the first column of your matrix, and ignore the first entry. Call that v, form F, embed this into an identity matrix with one extra row and column, and see what happens. Iterate on this procedure.


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