- #1
TheSodesa
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- 7
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.