- #1

TheSodesa

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## Homework Statement

Let ##A## and ##B## be square matrices, such that ##AB = \alpha BA##. Investigate, with which value of ##\alpha \in \mathbb{R}## the subspace ##N(B)## is ##A##-invariant.

## Homework Equations

If ##S## is a subspace and ##A \in \mathbb{C}^{n \times n}##, we define multiplying ##S## by ##A## as follows:

\begin{equation}

AS = \{ A \vec{x}: \vec{x} \in S \}

\end{equation}

The subspace ##S## is said to be ##A##-invariant, if

\begin{equation}

AS \subset S

\end{equation}

The null space of a matrix ##A \in \mathbb{C}^{n \times n}## is

\begin{equation}

N(A) = \{ \vec{x} \in \mathbb{C}^{n}: A \vec{x} = \vec{0} \}

\end{equation}

## The Attempt at a Solution

Alright, I have no idea what I'm doing and I came close to failing (or failed, the exam results haven't come back yet) my matrix algebra course, but here goes:

Let us assume that ##N(B)## is indeed ##A##-invariant, and that ##AB = \alpha BA##. Then according to ##(2)##:

[tex]

A N(B) \subset N(B)

[/tex]

Also, solving for ##\alpha## in the given equation gives us (assuming ##A## and ##B## are non-singular):

\begin{align*}

&& AB &= \alpha BA\\

&&\iff\\

&& ABA^{-1} &=\alpha B\\

& &\iff\\

&& AB A^{-1} B^{-1} &= \alpha I,

\end{align*}

Looking at the middle equation, ##\alpha B## seems to similar with ##B##. I wonder if this is something I can use to my advantage. Regardless, I'm not sure how to progress from here on out. Like I said, my understanding of the theory is on a very shaky foundation, so I really could use some help.