Can all square matrices be converted to Hessenberg form?

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I wanted to know if there are any constraints on the type of square matrices that can be converted to Hessenberg form using householder reduction?

Or can the householder reduction be used to convert all $\mathbf{A\epsilon R^{\mathit{n}x\mathit{n}}}, n>2$ to hessenberg form.

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$n-2$ Householder reductions can be used to convert all $A \in \mathbb{R}^{n \times n}$ to Hessenberg form.

Let say you are creating a lower Hessenberg matrix,the main idea is that if you perform the Householder reduction from left to right, you are first creating zero below a certain number by using a rotation. As you move on to the next column, the zeros that you created in the earlier column doesn't get affected. There is no restriction on the matrix $A$, we can always find $Q$ such that

$$Q^TAQ=H$$