0

While self studying Linear Algebra from Hoffman Kunze, I have some questions in proof of a lemma in section - Simultaneous Triangulation, Simultaneous Diagonalization.

As I have asked 4 questions here so I will give bounty of 50 points( under the criteria of rewarding existing answer) to anyone who answers all 4 questions as it will take a good amount of time.

Adding Image of lemma( Page 207 of the book): enter image description here

This lemma is used in the proof :

enter image description here

Questions: (1) Why assumptions that $\mathcal{F}$ contains only finite number of operators holds? Author then assume the maximal set to contain finite operators , why can't maximal set have infinite operators?

(2) How $V_{1}$ is larger than $W$?

(3) How does $T(T_{1} - c_{1} I) \beta \epsilon$ W is equivalent to statement $T\beta \in V_1$, for all $\beta \in V_1$ and all $T \in \mathcal{F}$?

(4) How is $V_{2} $ invariant under $\mathcal{F}$?

Kindly tell how should I reason these!!

alphaomega
  • 2,296

1 Answers1

1

(1)

By taking a maximal linearly independent subset of $\mathcal{F}$ means that every other function in $\mathcal{F}$ can be expressed as a linear combination of those functions in the maximal set, and of course condition (b) is invariant under linear combinations (if $f_1$, $f_2$ satisfy (b) then their linear combination satisfies (b)). Also -for some reason that I cannot see at the moment, and I hope it's not the main question- this maximal set can be considered finite.

(2)

I suppose you meant "$V_1$ larger than $W$". Now if $x\in W$, then $T_1(x)\in W$ as $W$ is invariant under $\mathcal{F}$. Hence $T_1(x)-c_1I(x)=T_1(x)-c_1x$ is a linear combination of elements of $W$, and thus is in $W$. So every $x\in W$ is in $V_1$.

(3)

I would suggest you to take a closer look. It doesn't say that the statements are equivalent. He proves that If $T$ commutes with $T_1$ (and that's the case for every operator in $\mathcal{F}$, by definition) then $T(\beta)$ is in $V_1$ for any $\beta \in V_1$. In other words $V_1$ is invariant under $\mathcal{F}$ (keep that in mind for your fourth question)

(4)

$V_2$ was defined for $T_2$ as $V_1$ was defined for $T_1$ with the additional property that ot is a subspace of $V_2$. With the same reasoning as in (3) (and also using (3)), we can show that for $T(\beta)\in V_2$ for any $\beta \in V_2$, for all $T\in \mathcal{F}$. In other words (again) $V_2$ is invariant under $\mathcal{F}.$

Edit: Question (1)

It's finite because $\dim V=n<+\infty$, for some $n\in \mathbb{N}$. Every linear operator $T$ can be represented by one $n\times n$ matrix. Every euch matrix is a linear combination of (the linear independent) matrices of the form:

  • A_{ij} = $a_{ij}=1$ and every other $a_{i' j'}=0$, $i'j'\neq ij$, $ \ $ where $i,j,i',j'\in \{1,2,\ldots , n\}$.

Of course the set of all such matrices is finite.

alphaomega
  • 2,296