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What does the phrase "no pain, no gain" mean?

Ask by Pena Reyes. in the United States
Dec 17,2024

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"No pain, no gain" means that success often requires hard work and perseverance through challenges.

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The phrase "no pain, no gain" suggests that in order to achieve success or improve oneself, one must endure hardship or discomfort. It's like saying that the path to reaching your goals often requires hard work, perseverance, and overcoming challenges—essentially, embracing the struggle to reap the rewards later! This motto finds its roots in fitness culture, where pushing through tough workouts leads to physical results. However, it’s versatile and applicable beyond the gym—whether in academics, career growth, or personal development, enduring a little hardship can pave the way to impressive achievements.

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Numéro d'étudiant : La qualité de la rédaction sera prise en compte. Exercice 1. Soit \[ \mathcal{B}=\left\{\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\right\} \] la base canonique de \( \operatorname{Mat}_{2}(\mathbb{R}) \) et soit \( f: \operatorname{Mat}_{2}(\mathbb{R}) \rightarrow \operatorname{Mat}_{2}(\mathbb{R}) \) l'endomorphisme de \( \operatorname{Mat}_{2}(\mathbb{R}) \) tel que, en base canonique, \[ f\left(\left[\begin{array}{ll} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array}\right]\right)=\left(\left[\begin{array}{cc} x_{1}+2 x_{3} & 2 x_{1}-x_{2}+4 x_{3}-2 x_{4} \\ -x_{3} & -2 x_{3}+x_{4} \end{array}\right]\right) \] (a) Montrer que \[ A=\mu_{\mathcal{B}, \mathcal{B}}(f)=\left(\begin{array}{cccc} 1 & 0 & 2 & 0 \\ 2 & -1 & 4 & -2 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right) \] où \( \mu_{\mathcal{B}, \mathcal{B}}(f) \) est la matrice associée à \( f \) dans la base canonique. (b) Déterminer le polynôme caractéristique \( \chi_{f}(x) \). (c) Déterminer les valeurs propres de \( f \), leurs multiplicités algébriques et montrer que l'endomorphisme \( f \) est diagonalisable. (d) Déterminer une base \( \mathcal{B}^{\prime} \) de \( \operatorname{Mat}_{2}(\mathbb{R}) \) formée de vecteurs propres de \( \operatorname{Mat}_{2}(\mathbb{R}) \), la matrice de changement de base \( P:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right) \) et la matrice diagonale \( D:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f) \) telles que \[ \mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f)=\left(\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right)\right)^{-1} \mu_{\mathcal{B}, \mathcal{B}}(f) \mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right) \] Autrement dit, \[ D=P^{-1} A P \] où \( A=\mu_{\mathcal{B}, \mathcal{B}}(f) \).
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Numéro d'étudiant : 22007890 La qualité de la rédaction sera prise en compte. Exercice 1. Soit \[ \mathcal{B}=\left\{\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\right\} \] la base canonique de \( \operatorname{Mat}_{2}(\mathbb{R}) \) et soit \( f: \operatorname{Mat}_{2}(\mathbb{R}) \rightarrow \operatorname{Mat}_{2}(\mathbb{R}) \) l'endomorphisme de \( \operatorname{Mat}_{2}(\mathbb{R}) \) tel que, en base canonique, \[ f\left(\left[\begin{array}{ll} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array}\right]\right)=\left(\left[\begin{array}{cc} x_{1}+2 x_{3} & 2 x_{1}-x_{2}+4 x_{3}-2 x_{4} \\ -x_{3} & -2 x_{3}+x_{4} \end{array}\right]\right) \] (a) Montrer que \[ A=\mu_{\mathcal{B}, \mathcal{B}}(f)=\left(\begin{array}{cccc} 1 & 0 & 2 & 0 \\ 2 & -1 & 4 & -2 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right) \] où \( \mu_{\mathcal{B}, \mathcal{B}}(f) \) est la matrice associée à \( f \) dans la base canonique. \( ~ \) trer que l'endomorphisme \( f \) est diagonalisable. Déterminer une base \( \mathcal{B}^{\prime} \) de \( \operatorname{Mat}_{2}(\mathbb{R}) \) formée de vecteurs propres de \( \operatorname{Mat}_{2}(\mathbb{R}) \), la matrice de changement de base \( P:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\mathrm{Mat}_{2}(\mathbb{R})}\right) \) et la matrice diagonale \( D:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f) \) telles que \[ \mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f)=\left(\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right)\right)^{-1} \mu_{\mathcal{B}, \mathcal{B}}(f) \mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right) \] Autrement dit, \[ D=P^{-1} A P \] où \( A=\mu_{\mathcal{B}, \mathcal{B}}(f) \).
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