2. Let a be m × N matrix, B is n × M matrix, and m < n, ab = I is known, where I is an m-order identity matrix. It is proved that the column vector group of B is linear

2. Let a be m × N matrix, B is n × M matrix, and m < n, ab = I is known, where I is an m-order identity matrix. It is proved that the column vector group of B is linear

It is proved that firstly, there is R (b) > = R (AB) = R (I) = M
B has only m columns, so r (b)

Let a be an a x m matrix, B be an M x n matrix, n be less than m, and E be an n-intermediate unit matrix. If AB = e, it is proved that the column vectors of B are linearly independent

Counter evidence:
If the column vectors of B are linearly correlated, then R (b) has n = R (E) = R (AB) < = R (b), that is, N, so r (b) = n
That is, column B vector is linearly independent

Let a be mxn matrix and B be NXS matrix. It is proved that if AB = 0, then R (a) + R (b)

There is also a prompt. \ x0d please see the picture: \ x0d
\X0d \ x0d if you are satisfied, please adopt ^ ^

Let a be nxm matrix and B be mxn matrix, where n

n = r(En) = r(AB)

It is known that non-zero vectors a and B satisfy that (a-2b) is perpendicular to a and (b-2a) is perpendicular to B, then the included angle of a and B is Given that non-zero vectors a and B satisfy (a-2b) perpendicular to a and (b-2a) perpendicular to B, what is the angle between a and B

As you said, if a (a-2b) = 0 and B (b-2a) = 0, then aa-2ab = 0, then bb-2ab = 0,
Subtract the above two formulas to get AA BB = 0, so the length of a = the length of B, that is, AA = BB. Add the above formula to get - > A - > A + - > b - > b = 4 - > b - > A, then the left formula is equal to 2 - > A - > A, then a length a = 2A length B length * [cos angle], get cos angle = 0.5, so the angle is 60 degrees

Given that a and B are non-zero vectors, and (a-2b) ⊥ a, (b-2a) ⊥ B, the included angle between a and B is ()

（a-2b）⊥a
a^2-2ab=0
（b-2a）⊥b
b^2-2ab=0 a^2=b^2 |a|=|b|
a^2-2ab=0
|a|^2-2*|a|*|b|*cos=0
1-2cos=0
cos=1/2 =60°