Are the columns space and row space same for idempotent matrix?

  • #1
236
16
Suppose, ##A## is an idempotent matrix, i.e, ##A^2=A##.
For idempotent matrix, the eigenvalues are ##1## and ##0##.
Here, the eigenspace corresponding to eigenvalue ##1## is the column space, and the eigenspace corresponding to eigenvalue ##0## is the null space.
But eigenspaces for distinct eigenvalues of a matrix have intersection ##\{0\}##.
So, null space and column space are complementary for idempotent matrix. That means the row space and column space are the same for idempotent matrix.
Is this argument correct?
 

Answers and Replies

  • #2
13,554
10,661
There's a contradiction in your conclusion. If you have an eigenvalue ##0##, then there is a vector ##v \neq 0## with ##A.v=0##, i.e. the eigenspace of eigenvalue ##0## cannot be ##\{0\}##.
 
  • #3
236
16
There's a contradiction in your conclusion. If you have an eigenvalue ##0##, then there is a vector ##v \neq 0## with ##A.v=0##, i.e. the eigenspace of eigenvalue ##0## cannot be ##\{0\}##.
I have not said that the eigenspace of eigenvalue ##0## is ##\{0\}##.
 
  • #4
13,554
10,661
Yep, I translated it wrongly. I'm used to the term kernel for what you call nullspace. Sorry.

Before I also misinterpret column or row spaces I simply ask: What about ##\begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}##?
 
Last edited:
  • #5
236
16
Yep, I translated it wrongly. I'm used to the term kernel for what you call nullspace. Sorry.

Before I also misinterpret column or row spaces I simply ask: What about ##\begin{bmatrix}1 & 0 \\ 1 & 0 \end{bmatrix}##?
The column space and row space are not the same in the example. But I cannot figure out what was wrong in my argument.
The row space of a matrix is the vector space spanned by the row vectors of the matrix and the column space of a matrix is the vector space spanned by the column vectors of the matrix
 
  • #6
13,554
10,661
You must not handle subspaces like sets. The error is the usage of "complementary". As in the example you can have ##V=V_1 \oplus V_2## and ##V=V_1 \oplus V_3## without ##V_2## and ##V_3## even being contained in one another. Not to speak of being equal. Think of two lines through the origin. Together with, say the ##x-##axis each of them span the plane, although they only have ##0## in common.

You may conclude that ##V/V_1 \cong V_2 \cong V_3## are isomorphic, but not equal. To conclude equality, one always has to show a real inclusion ##V_2 \subseteq V_3##, too, when dealing with vector spaces.
 
  • Like
Likes arpon
  • #7
mathwonk
Science Advisor
Homework Helper
11,039
1,230
you seem to have argued that because two subspaces are complementary to the same subspace that they are equal. but "complementary" (in the algebraic sense) simply means as you say, their intersection is zero (and their dimensions are complementary). Can you give an example of two different subspaces of the x,y plane that both intersect the x axis at only the origin? (it would be different if both subspaces were orthogonal complements of the same subspace.)
 
  • Like
Likes arpon

Related Threads on Are the columns space and row space same for idempotent matrix?

  • Last Post
Replies
1
Views
5K
Replies
1
Views
602
  • Last Post
Replies
6
Views
2K
Replies
4
Views
2K
Replies
7
Views
7K
Replies
1
Views
2K
  • Last Post
2
Replies
26
Views
5K
Replies
2
Views
617
Replies
4
Views
4K
Top