Exact shift

Let’s see what happens when $\mu$ is equal to an eigenvalue.

The first step is:

$$U_{k+1}{R}_{k+1}=T_k-\mu I$$

The matrix is singular.

Let’s assume that $T_k$ is unreduced; that is, there is no zero on the sub-diagonal. $t_{i+1,i}\ne 0$.

In the second step, the last row of $R_{k+1}$ is 0:

$$T_{k+1}=R_{k+1}{U}_{k+1}+\mu I$$

The last row is exactly $[0,\dots ,0,\mu ]$. Matrix $T_{k+1}$ is exactly in block upper triangular form. We recover the exact eigenvalue as expected.

In the next step, we can deflate and work with a smaller matrix.