Proof "(A^ta) = R (A) $

First, it needs to be clear that the entry point for this proof is that the solution of the equation group contains the linked relationship.

1) Proof R (ATA) ≤r (a) R (A^ta) \leq R (a)

When ax=0, atax=0 ax = 0 o'clock, A^tax = 0 is bound to be, namely: the solution of atax=0 A^tax = 0, which contains the solution of ax=0 ax = 0, which shows that the base solution of atax=0 A^tax = 0 contains the basic solution system of ax=0 ax = 0 。

Therefore: N−r (ATA) ≥n−r (a) n-r (A^ta) \geq n-r (a), i.e.: R (ATA) ≤r (a) R (A^ta) \leq R (a)

This can also be directly obtained using R (AB) ≤min (R (a), R (b)) R (AB) \leq min (R (a), R (b)).

This proof uses two angles of row rank and column rank, and uses the method of linear table out and rank comparison to prove the process is worth thinking about.

2) Proof R (ATA) ≥r (a) R (A^ta) \geq R (a)

by atax=0 a^tax = 0, you can get: xtatax=0 x^ta^tax = 0, transform get: (AX) tax=0 (ax) ^tax = 0, visible vector AX = 0.
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