Tutor profile: Adriana C.

Prove of give a counterexample: If 2 is an eigenvalue of $$A$$, then 5 is an eigenvalue of $$A^{2} + I$$

A good way to begin proofs is to identify your "givens" and what you "want to show." GIVEN that 2 is an eigenvalue of $$A$$ and using the definition of eigenvalue, we know that there must exist a vector $$\textbf{x}$$ such that $$A\textbf{x} = 2\textbf{x}$$ Now, we WANT TO SHOW that 5 is an eigenvalue of $$A^{2} + I$$. This implies that there should exist a vector $$\textbf{x}$$ such that $$(A^{2} + I)\textbf{x} = 5\textbf{x}$$ Working with the left side, we observe that: Left Hand Side: $$(A^{2} + I)\textbf{x}\\$$ $$= A^{2} \textbf{x}\ + \textbf{x}\\$$ $$= A A \textbf{x}\ + \textbf{x}\\$$ Now using the GIVEN fact that $$A\textbf{x} = 2\textbf{x}$$ we can rewrite this as: $$= A 2 \textbf{x}\ + \textbf{x}\\$$ Noting that scalars commute and using the fact from our GIVENS $$= 2A \textbf{x}\ + \textbf{x} \implies 2 \cdot 2x + \textbf{x}\\$$ $$= 4\textbf{x} + \textbf{x} = 5\textbf{x}$$ (Right Hand Side) Therefore, by definition, 5 is an eigenvalue of $$A^{2} + I $$ Any questions?

Solve the equation $$sin\theta = 1.2$$

Alright, so we are dealing with a trigonometric equation right? Notice that you are being asked to solve for $$\theta$$ an angle that makes this statement $$sin\theta = 1.2$$ true. Do you see anything wrong with that logic? What is the range of the sine function? Exactly, there's no solution. You can also think about it this way: $$\theta = sin^{-1}(1.2)$$, recall that the domain of the inverse sine function is...right, $$[-1, 1]$$ and $$1.2 \notin [-1, 1]$$ so we arrive at the same conclusion--there is no solution. Does this concept make sense now? Want to see an easy way to remember the domain/range restrictions for all trig function and their inverses?

Determine whether the geometric series $$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{4^{n}}{5^{n+1}} $$ is convergent or divergent. If it is convergent, find its sum.

First off, can you recall the definition of a geometric series...try it, before you peak. $$\textbf{Defn:} $$ A geometric series is a series of the form $$\sum_{n=1}^{\infty} ar^{n-1}$$ Now, let us use some rewriting magic on our given geometric series: Given $$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{4^{n}}{5^{n+1}} \\ $$ $$= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{4 \cdot 4^{n-1}}{5^{2} \cdot 5^{n-1}}\\ $$ $$=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{4 \cdot 4^{n-1}}{5^{2} \cdot 5^{n-1}}\\ $$ $$=\sum_{n=1}^{\infty} \frac{4}{25} (-\frac{4}{5})^{n-1} \\ $$ Observe that $$ a = \frac{4}{25} $$ and $$ \textbf{r} = -\frac{4}{5}\\$$ Recall that a geometric series converges to $$\frac{a}{r-1}$$ if $$ -1 < \textbf{r} < 1 \\$$ by a theorem. Since $$\textbf{r} = -\frac{4}{5} \in (-1,1)$$, our geometric series $$ \sum_{n=1}^{\infty} (-1)^{n-1} \frac{4^{n}}{5^{n+1}} $$ converges to: $$\frac {\frac{4}{25}}{1 - (-\frac{4}{5})} = \frac{\frac{4}{25}}{\frac{9}{5}} = \frac{4}{45}. $$ How are you feeling--ready for another one or do you have follow up questions?