If A is linear transformation of rank one, then there exists a unique scalar $\alpha$ such that $A^2= \alpha A$
b) If $\alpha\neq1$, then 1- A is invertible
Please, help

Question

If A is linear transformation of rank one, then there exists a unique scalar $\alpha$ such that $A^2= \alpha A$
b) If $\alpha\neq1$, then 1- A is invertible
Please, help

loser66 · Answer

Please, I do not know how to start. I have a formula: Ax = [x,y0]x0 but I don't know how to use it. hihihi... stupid me!!

kinggeorge · Answer

Well, if it has rank 1, then all the rows are scalar multiples of the first row. So if we apply the linear transformation to a vector $(x_1,...,x_n)^T$, then we get$$A(x_1,...,x_n)^T=(a,a_2\cdot a,...,a_n\cdot a)^T$$for some constants $a_2,...,a_n$. Applying the transformation again, we get$$A(a,a_2\cdot a,...,a_n\cdot a)^T=(b,a_2\cdot b,...,a_n\cdot b)^T.$$Let $b=\alpha a$. Then we have the relations$$A^2(x_1,...,x_n)^T=(\alpha a,a_2\cdot \alpha a,...,a_n\cdot \alpha a)^T=\alpha(a,a_2\cdot a,...,a_n\cdot a)^T.$$This last bit is exactly $A(x_1,...,x_n)^T$, so $A^2=\alpha A$ for some constant $\alpha$.

kinggeorge · Answer

Typo. The last line should read: This last bit is exactly $\alpha A(x_1,...,x_n)^T$, so $A^2=\alpha A$ for some constant $\alpha$.

loser66 · Answer

I am sorry, I don't get the first equation
A is linear transformation, let say defined by [A] = (a_ij) how you get the right hand side?

kinggeorge · Answer

Let $a_{11},...,a_{1n}$ be the first row of the matrix. Then we can say that $$a_{11}x_1+...+a_{1n}x_n=a.$$Since the rows are all multiples of the first row, let the $i$-th row be $a_i$ times the first row. So the $i$-th row is $a_ia_{11},...,a_ia_{1n}$. Thus, $$a_ia_{11}x_1+...+a_ia_{1n}x_n=a_ia$$

loser66 · Answer

Thanks so much, I got it now.
Please, give me a hint for b)

kinggeorge · Answer

I'm not actually sure about part b myself right now.

loser66 · Answer

No worry, I still have time for this problem. :)

kinggeorge · Answer

Have you done anything with eigenvalues/eigenvectors?

kinggeorge · Answer

In any case, since $A^2=\alpha A$, it must be that $A\vec{x}=\alpha\vec{x}$ for any vector $\vec{x}$. Also, we know that $I-A$ is NOT invertible if and only if $\det(I-A)=0$ if and only if there is some non-zero vector $\vec{x}$ such that $$\vec{0}=(I-A)\vec{x}=I\vec{x}-A\vec{x}=\vec{x}-\alpha\vec{x}=(1-\alpha)\vec{x}.$$Since $\vec{x}$ is non-zero, the rightmost term is zero if and only if $\alpha=1$. So it follows that $I-A$ is not invertible if and only if $\alpha=1$, and therefore if $\alpha\neq1$, then $I-A$ must be invertible.