Review

Zero matrix

𝐀 = [^{4\ 5\ 6} _{5\ 6\ 7}]; 𝟎 = [^{_{0\ 0}} _{^{0\ 0} _{0\ 0}}]₃ₓ₂

𝐀𝟎 = [^{0\ 0} _{0\ 0}] = 𝟎₂ₓ₂

Non-homogeneous system has no zero solution

𝐀𝐁 ≠ 0 means A, B both cannot be zero.

Diagonal matrix and identical matrix

𝐁 is a diagonal matrix.

𝐀𝐁 is scaling each column of 𝐀 by the value of element on the diagonal times.

I = []

Vectors multiplication

Rank relation

𝐀 = [^{1\ 2} _{0\ 3}]; 𝐁 = [^{2\ 3} _{0\ 4}]

𝐀𝐁 = [^{2\ 11} _{0\ 12}]

r(𝐀) = 2; r(𝐁) = 2; r(𝐀𝐁) = 2

If 𝐁 = [^{1\ 0} _{0\ 0}], r(𝐁) = 1, then 𝐀𝐁 = [^{1\ 0}_{0\ 0}]; r(𝐀𝐁) = 1.

Further letting 𝐀 = [^{1\ 2} _{0\ 0}], r(𝐀) = 1, then 𝐀𝐁 = [^{1\ 0}_{0\ 0}]; r(𝐀𝐁) = 1.

In addition, if 𝐀 = [^{1\ 2} _{0\ 0}], 𝐁 = [^{0\ -2} _{0\ 1}], then 𝐀𝐁 = [^{0\ 0}_{0\ 0}], r(𝐀𝐁) = 0

Conclusion: the rank of the product matrix is not greater than the rank of any multiplier.

r(𝐀𝐁) ≤ min{ r(𝐀), r(𝐁)}