Does Every Matrix with a Right Inverse Span Rm?

  • Thread starter Thread starter Bob
  • Start date Start date
  • Tags Tags
    Inverse
Click For Summary
An m*n matrix A has a right inverse if there exists an n*m matrix C such that AC equals the identity matrix Im, indicating that the column vectors of A span Rm. If A has a right inverse, it implies that rank(A) equals m, confirming that the columns are linearly independent and span the entire space Rm. A matrix cannot have a right inverse if n < m, as this would lead to a situation where the reduced row echelon form of AC contains at least one all-zero row. Conversely, if n >= m, A can have a right inverse, provided that its rank is m. Overall, the existence of a right inverse is directly linked to the spanning properties of the column vectors of A.
Bob
Messages
28
Reaction score
0
An m*n matrix A is said to have a right inverse if there exists an n*m matrix C such that AC=Im. A is said to have a left inverse if there exists an n*m matrix D such that DA=In.

(a) If A has a right inverse, show that the column vectors of A span Rm.
(b) Is it possible for an m*n matrix to have a right inverse if n<m? n>=m? Explain.

? How to do that? I dont' know how to start? what if A doesn't span Rm?
 
Physics news on Phys.org
Note that the jth column of AC is equal to Ac_j where c_j[/itex] is the jth column of C. <br /> <br /> Think of the Linear transformation L_A:\mathbb{R}^n \rightarrow \mathbb{R}^m defined as left multiplication by A. Another way to phrase the above statement is that the jth column of AC is equal to L_A(c_j)<br /> <br /> Note that the columns of AC are linearly independent.<br /> <br /> What does this tell you about rank(A)?
 
Last edited:
It's min{m,n}

thanks
 
Bob said:
It's min{m,n}

thanks
It is unclear to me based on your result whether you have actually arrived at a correct solution. To say that rank(A) = min{m,n}, while technically not wrong, leads me to suspect that you don't fully understand the problem, but before I make any accusations perhaps you could show me your work stating clearly your line of reasoning.
 
AC=Im

rank(A)=m

column vectors of A span Rm

if n>m, column vectors are linear dependent.
if n<m, AC does not equal Im, the reduced row echelon form of AC has at least one all zero row.

:cool:
 
Last edited:
What you have written is correct, but you left out many of the details of the proof. In you're actual writeup you should fill in the gaps by either proving these statements or citing pre-existing theorems that allow you to make these jumps.

For example what are the justifications for going from AC = I am to rank(A) = m, and then how does this lead you to conclude that the column vectors of A span Rm?

Why if n<m does A not have a right inverse? What you wrote is true but is not an explanation.

aside:
If A has a right inverse, saying rank(A) = min{m,n} is the same as saying rank(A) = m. Do you see why?

regards,
nocturnal
 
Last edited:
:bugeye:
If A has a right inverse

AC=(e_1 ,e_2 ,...,e_m)is consistent

Ac_1 = e_1
Ac_2 = e_2
...
Ac_m = e_m

an arbitrary vector in R_m can be written as

\alpha _1 e_1 + \alpha _2 e_2 + ... + \alpha _m e_m

(Therom:A liner system Ax=b is consistent if and only if b is in the column space of A)

since e_1,e_2,...,e_m are in the column space of A,
\alpha _1 e_1 + \alpha _2 e_2 + ... + \alpha _m e_m is in the column space of A.

the column vectors of A span R_m
 
Last edited:

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K