Linear-Algebra A Matrixs Rank and Calculating Inverse Matrices-2, guaranteed and verified 100% Pass

1


A Matrix’s Rank and Calculating Inverse Matrices


Def. If 𝐴 ∈ 𝑀𝑚×𝑛 (ℝ), we define the rank of 𝑨, denoted 𝑅𝑎𝑛𝑘(𝐴), to be the
rank of the linear transformation associated with 𝐴, 𝐿𝐴 , 𝐿𝐴 : ℝ𝑛 → ℝ𝑚 .


If 𝑚 = 𝑛, then notice that an 𝑛 × 𝑛 matrix is invertible if and only if its rank is 𝑛.
This follows from an earlier theorem about linear transformations. This is
because any matrix 𝐴 is the matrix representation of a linear transformation. In
fact we have:


Theorem: Let 𝑇: 𝑉 → 𝑊 be a linear transformation between finite dimensional
vector spaces, and let 𝐵1 and 𝐵2 be ordered bases for 𝑉 and 𝑊 respectively.
𝐵
Then 𝑅𝑎𝑛𝑘(𝑇) = 𝑅𝑎𝑛𝑘([𝑇]𝐵21 ).


Theorem: Let 𝐴 be an 𝑚 × 𝑛 matrix. If 𝑃 is an 𝑚 × 𝑚 matrix and 𝑄 is an 𝑛 × 𝑛
matrix, both invertible, then
a. 𝑅𝑎𝑛𝑘(𝐴𝑄) = 𝑅𝑎𝑛𝑘(𝐴)
b. 𝑅𝑎𝑛𝑘(𝑃𝐴) = 𝑅𝑎𝑛𝑘(𝐴)
c. 𝑅𝑎𝑛𝑘(𝑃𝐴𝑄) = 𝑅𝑎𝑛𝑘(𝐴).


Proof: a. 𝑅(𝐿𝐴𝑄 ) = 𝑅(𝐿𝐴 𝐿𝑄 ) = 𝐿𝐴 𝐿𝑄 (ℝ𝑛 )

= 𝐿𝐴 (𝐿𝑄 (ℝ𝑛 )) = 𝐿𝐴 (ℝ𝑛 ) (since 𝐿𝑄 is onto)

= 𝑅 (𝐿𝐴 ).
Thus we have:

𝑅𝑎𝑛𝑘(𝐴𝑄 ) = dim (𝑅(𝐿𝐴𝑄 )) = dim(𝑅(𝐿𝐴 )) = 𝑅𝑎𝑛𝑘(𝐴).

, 2


b. Similarly, we have:

𝑅(𝐿𝑃𝐴 ) = 𝑅(𝐿𝑃 (𝐿𝐴 )) = 𝐿𝑃 (𝐿𝐴 (ℝ𝑛 )).
But 𝐿𝐴 (ℝ𝑛 ) is a subspace of 𝑅 𝑚 .
Since 𝑃 is invertible we have:

dim (𝐿𝑃 (𝐿𝐴 (ℝ𝑛 )) = dim(𝐿𝐴 (ℝ𝑛 )) = 𝑅𝑎𝑛𝑘(𝐴).
So the 𝑅𝑎𝑛𝑘(𝑃𝐴) = 𝑅𝑎𝑛𝑘(𝐴).


c. Follows from parts a and b.


Corollary: Elementary row and column operations on a matrix are rank
preserving.


Proof: Every elementary row or column operation can be viewed as a
multiplicaion of a matrix by an invertible matrix on the left (elementary row
operations) or the right (elementary column operations).


Theorem: The rank of any matrix equals the maximum number of its linearly
independent columns. Thus the rank of a matrix is the dimension of the subspace
generated by its columns.


Proof: Let 𝐴 ∈ 𝑀𝑚×𝑛 (ℝ).

𝑅𝑎𝑛𝑘(𝐴) = 𝑅𝑎𝑛𝑘(𝐿𝐴 ) = dim(𝑅(𝐿𝐴 )).
Let 𝐵 be the standard ordered basis for ℝ𝑛 .