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

• I
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?

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fresh_42
Mentor
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\}##.

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\}##.

fresh_42
Mentor
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:
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

fresh_42
Mentor
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.

arpon
mathwonk