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I found a true/false exam. Here is the exercise : (other question are here)

1) Let $X=\{x_1,x_2,x_3,x_4\}$ a set. Let $S:Mat(3,2,\mathbb R)\to \mathcal F(X,\mathbb R)$ and $T:\mathcal F(X,\mathbb R)\to Mat(3,2,\mathbb R)$ linear application where $\mathcal F(X,\mathbb R)=\{f:X\to \mathbb R\mid f\text{ is a function}\}$. It's possible to have $Spec(T\circ S)=\{-5,-2,3\}$ where $Spec(T)=\{\text{eigenvalue of T}\}$.

2) Let $V$ a $\mathbb C-$vector space. Let $S,T\in \mathcal L(V)$. If $Spec(S)=\{-1,\frac{-1}{4},\frac{3}{4}\}$ and $Spec(T)=\{\frac{1}{2},1\}$. There is $\lambda \in Spec(T\circ S)$ s.t. $0<|\lambda |<1$.


My Answers:

1) Unfortunately I have no idea.

2) Unfortunately no idea.

Any help would be appreciated.

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  • $\begingroup$ (1) Look at the dimensions. $\operatorname{Mat}(3,2,\mathbb{R})$ has dimension $6$, while $\mathcal{F}(X,\mathbb{R})$ has dimension $4$. Then $S$ must send some non-zero vectors to $0$. $T$ has no choice but to send zero to zero. Therefore, $T\circ S$ sends some non-zero vectors to $0$. $\endgroup$ – user578878 Jul 27 '18 at 11:43
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I assume that $M(3,2,\mathbb{R})$ are the $3\times 2$ real matrices?

Notice that $\mathcal{F}(X,\mathbb{R})\cong \mathbb{R}^4$. Hence $T\circ S$ is an endomorphism of a $6$-dimensional space factoring through a $4$-dimensional space, hence $\dim\ker(T\circ S)\geq 2$. It follows that $0$ must be an eigenvalue of $T\circ S$.

For the second one, recall that if $A$ is a square matrix, then $\det(A)=\prod_i\lambda_i$ where the $\lambda_i$ are the eigenvalues of $A$. Now assuming that $S,T$ are linear operators on a finite-dimensional space $V$, we have that $\det(T\circ S)=\det(T)\det(S)$. Hence $|\det(T\circ S)|<1$ On the other hand, this should be equal to the product of all eigenvalues of $T\circ S$. You can easily proceed by contradiction.

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  • $\begingroup$ $dim(V)<\infty$ is not given. $\endgroup$ – user578878 Jul 27 '18 at 11:55
  • $\begingroup$ For 2), we need multiplicity to compue the determinant. i.e. $\det\begin{pmatrix}2&0\\0&2\end{pmatrix}=4\neq 2$.. $\endgroup$ – MSE Jul 27 '18 at 12:06
  • $\begingroup$ @MSE There is no problem in finite dimensions. No matter the multiplicity, what is important is that the product has absolute values $<1$. Sure, those equal signs that they wrote are not exactly true, but not a big deal. The question for you is, is $dim(V)<\infty$ given? $\endgroup$ – user578878 Jul 27 '18 at 12:08
  • $\begingroup$ @nextpuzzle: yes we suppose $\dim(V)<\infty $. $\endgroup$ – MSE Jul 27 '18 at 12:14
  • $\begingroup$ Yes, sorry I forgot the multiplicity there, but it doesn't matter. I'm not sure whether it's true for infinite dimensional spaces. I don't see an immediate reason why the point spectrum of $T\circ S$ couldn't be empty for example. $\endgroup$ – Mathematician 42 Jul 27 '18 at 12:23

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