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Tutor profile: Diego C.

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Diego C.
Maths PhD graduate
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Questions

Subject: Linear Algebra

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Question:

Consider a linear transformation $$T: \mathbb{R}^n \to \mathbb{R}^n$$ for which $$Ker(T) = \{\vec{0}\}$$. If $$\{\vec{x_1}, \vec{x_2}, \dots \vec{x_k}]$$ is a linearly independent subset of $$\mathbb{R}^n$$, show that $$\{T(\vec{x_1}), T(\vec{x_2}), \dots T(\vec{x_k})]$$ is also linearly independent.

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Diego C.
Answer:

First, recall that $$\{T(\vec{x_1}), T(\vec{x_2}), \dots T(\vec{x_k})$$ is linearly independent if the the expression $$a_1T(\vec{x_1} ) + a_2T(\vec{x_2} ) + \dots a_kT(\vec{x_k}) = \vec{0}$$ is only satisfied by $$a_1 = a_2 = \dots a_k = 0$$. Since T is a linear transformation, the above linear combination can be written as: $$T(a_1\vec{x_1} + a_2\vec{x_2} + \dots + a_k\vec{x_k}) = \vec{0}$$. This means that the combination $$a_1\vec{x_1} + a_2\vec{x_2} + \dots + a_k\vec{x_k} \in Ker(T)$$, but we have been told that $$Ker(T) = \{\vec{0}\}$$ so this implies that $$a_1\vec{x_1} + a_2\vec{x_2} + \dots + a_k\vec{x_k} = \vec{0}$$. However, since we know that $$\{\vec{x_1}, \vec{x_2}, \dots \vec{x_k}]$$ is a linearly independent subset this means that this expression is true only if all coefficients $$a_i$$ are zero. In this way we conclude that $$\{T(\vec{x_1}), T(\vec{x_2}), \dots T(\vec{x_k})$$ is linearly independent.

Subject: Calculus

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Question:

Find the minima and maxima of the function $$f(x) = x^2 - 4x +3$$.

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Diego C.
Answer:

First, we find the critical points by differentiating $$f(x)$$ and equating it to zero: $$f'(x) = 2x - 4 = 0$$ The solution to this equation is: $$x = \tfrac42 = 2$$, so the critical point has x-coordinate $$x=2$$. In order to find the value of the y-coordinate we simply substitute this into the original function. This is: $$f(x =2) = (2)^2 - 4(2) + 3 = -1$$. Then, the critical point is $$(2,-1)$$. Finally, we need to figure out whether it is a maximum or minimum. This can be done easily by obtaining the second derivate of $$f(x)$$: $$f''(x) = 2$$. Since $$f''(x)$$ is positive everywhere (as it is a constant), then our critical point is a minimum.

Subject: Algebra

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Question:

Find the values for which the inequality $$x^2 -4x +3 \lt 0$$ is satisfied.

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Diego C.
Answer:

The quadratic expression can be factorised as: $$x^2 -4x +3 = (x-1)(x-3) \lt 0$$. From here, we can see that the two factors, $$(x-1)$$ and $$(x-3)$$, must have opposite signs. There are two possible cases: $$\textbf{Case 1}$$. $$x-1>0$$ and $$x-3<0$$, which means that $$x>1$$ and $$x<3$$. These two conditions are satisfied by the values in the interval $$(1,3)$$. $$\textbf{Case 2}$$. $$x-1<0$$ and $$x-3>0$$, which means that $$x<1$$ and $$x>3$$. However, these conditions cannot be satisfied at the same time, so this case does not give further solutions. In conclusion, the values solving the inequality are the ones in the interval $$(1,3)$$.

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