finite dimensional vector space and which of the following are true

Question

Let $V$ be a finite dimensional real vector space and let $A:V\to V$ be a linear map such that $A^2=A$. Assume that $A\ne0$ and that $A\ne I$. Which of the following statements are true?

a. $ker(A)\ne0$

b. $V=ker(A)\oplus R(A)$

c. The map $I+A$ is invertible

(c) is true since eigen value of $A$ cann't be $-1$. (a) and (b) are too true??. Not sure about them

If $\lambda$ is an eigen value of $A$ then from $A^2=A$ we will have $\lambda^2=\lambda$ implying either $\lambda=0$ or$1$. — tattwamasi amrutam, Jan 25 '14 at 15:00
Yes, but you must have some reason to think they are true. What is that reason? — Daniel Fischer, Jan 25 '14 at 15:07
@DanielFischer: Can we conclude from $A^2=A$ that $rank(A^2)=rank(A)$? — Mikasa, Jan 25 '14 at 15:11
@DanielFischer because other than $zero$ matrix there are matrices which stisfy $A^2=A$.. — tattwamasi amrutam, Jan 25 '14 at 15:12
@TattwamasiAmrutam You already know that $A$ can only have $0$ and $1$ as eigenvalues. What are the corresponding eigenspaces? — Daniel Fischer, Jan 25 '14 at 15:18
Is it possible to conclude eigen spaces from eigenvalues?? @DanielFischer — tattwamasi amrutam, Jan 25 '14 at 15:20
Not really, but from $A^2 = A$ we can obtain the eigenspaces here. — Daniel Fischer, Jan 25 '14 at 15:21
The eigenspace for the eigenvalue $0$ (if that is an eigenvalue) is always the kernel, $\ker A$. The eigenspace for the eigenvalue $1$ is $E(1) = { v : A(v) = v}$. So it is contained in $R(A)$ (well, all eigenspaces are contained in $R(A)$, whatever $A$). But since $A^2 = A$, we have $R(A) = E(1)$. — Daniel Fischer, Jan 25 '14 at 15:28
A cannot have both eigenvalues as 1 ,as that would imply A is identity.if it has a 0 as as an eigenvalue then ker(A)! = 0. — Prayagdeep Parija, Jan 25 '14 at 17:21

Bobby · Answer 1 · 2014-01-26T17:59:58.407

1

The answer is (b). First, We show that $\rm{Ker}(A)\cap R(A)=0$. Let $x\in \rm{Ker}(A)\cap R(A)$. There exists $y \in V$ such that $x=A(y)$. Thus $A(x)=A^2(y)=A(y)=0$ and so $x=0$.

Now, we show that $\rm{Ker}(A)\oplus R(A)=V$. Assume that $v\in V$. Consider element $v-A(v)+A(v)$. It is not hard to see that $v-A(v)\in \rm{Ker(A)}$, as desired.

edited Jan 26 '14 at 17:59

answered Jan 25 '14 at 15:19

Bobby

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finite dimensional vector space and which of the following are true

1 Answers1