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where is an invertible matrix. Then, we have:
(3.51)
To choose , note that:
(3.52)
From Theorem 3.1 , the first columns of must be linearly independent of each other and of the other columns. Now, is chosen such that:
(3.53)
Regarding (3.45 ), can be expressed as:
(3.54)
where has columns. Then, equation (3.45 ) leads to:
(3.55)
Hence, and is a free matrix. According to (3.46 ), we have:
(3.56)
(3.57)
Defining , where has rows, we obtain:
(3.58)
Since is required to be a stable matrix, the pair must be detectable [35].
From (3.35 ) and (3.36 ), we have:
(3.59)
Assuming that has full column rank, there exists a matrix such that:
(3.60)
Left‐multiplying both sides of the equation (3.59 ) by , and then using (3.60 ), the input vector can be estimated based on the state‐vector estimate as:
(3.61)
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