maths 1B last minute questions (1 Viewer)

InteGrand · Nov 1, 2016

Drsoccerball said:

$\noindent a) Suppose $\lambda$ is an eigenvalue for $P$ with eigenvector $\mathbf{v}$, so $P(\mathbf{v}) = \lambda \mathbf{v}$. Apply $P$ to both sides: $P(P(\mathbf{v})) = P(\lambda \mathbf{v}) \Rightarrow P(\mathbf{v}) = \lambda P( \mathbf{v} ) \Rightarrow \lambda \mathbf{v} = \lambda^2 \mathbf{v}$ (we used idempotency and linearity here). Since $\mathbf{v}$ is not the zero vector (as it is an eigenvector), we have $\lambda = \lambda^{2}$, so $\lambda$ is either $0$ or $1$.$

$\noindent b) Let $P$ be idemptotent and not the zero or identity map. Then (since $P$ is not the identity) there is a vector $\mathbf{w}\in V$ such that $P(\mathbf{w}) \ne \mathbf{w}$. So we can write $P(\mathbf{w}) = \mathbf{w} + \mathbf{y}$, where $\mathbf{y}\neq \mathbf{0}$. Apply $P$ to both sides, using idempotency on LHS and linearity on RHS, so $P(\mathbf{w}) = P(\mathbf{w}) + P(\mathbf{y}) \Rightarrow P(\mathbf{y}) = \mathbf{0} = 0\mathbf{y}$. Since $\mathbf{y}$ is non-zero, this shows $0$ is an eigenvalue.$

$\noindent Also note $P(P(\mathbf{x})) = P(\mathbf{x})$ by idempotency, and there is an $\mathbf{x} \in V$ such that $P(\mathbf{x}) \neq \mathbf{0}$, since $P$ is not the zero map. So $1$ is an eigenvalue (with this vector $P(\mathbf{x})$ a corresponding eigenvector).$

InteGrand · Nov 1, 2016

Drsoccerball said:

Which parts do you want help with? All of them? What progress have you made so far? Here's how to do the first part.

$\noindent The result (namely that any eigenvalue of $T$ must be $\pm 1$) actually will hold for any linear map with $T\circ T = I$, where $I$ is the identity map (i.e. any linear map that, when applied twice to an input, leaves the original input unchanged. The transpose map is clearly such a map.) To see this, suppose $T(C) = \lambda C$ where $C$ is non-zero, then applying $T$ to both sides and using $T(T(C)) = C$ on the LHS gives us $C = T(\lambda C) = \lambda T(C) = \lambda^{2}C$, i.e. $C = \lambda^{2}C$. So since $C$ is non-zero, we must have $\lambda^{2} = 1 \Rightarrow \lambda = \pm 1$.$

Drsoccerball · Nov 1, 2016

InteGrand said:
Which parts do you want help with? All of them? What progress have you made so far? Here's how to do the first part.

$\noindent The result (namely that any eigenvalue of $T$ must be $\pm 1$) actually will for any linear map with $T\circ T = I$, where $I$ is the identity map (i.e. any linear map that, when applied twice to an input, leaves the original input unchanged. The transpose map is clearly such a map.) To see this, suppose $T(C) = \lambda C$ where $C$ is non-zero, then applying $T$ to both sides and using $T(T(C)) = C$ on the LHS gives us $C = T(\lambda C) = \lambda T(C) = \lambda^{2}C$, i.e. $C = \lambda^{2}C$. So since $C$ is non-zero, we must have $\lambda^{2} = 1 \Rightarrow \lambda = \pm 1$.$

All of them ahahah

InteGrand · Nov 1, 2016

Drsoccerball said:
All of them ahahah

Drsoccerball said:

$\noindent b) For simplicity we'll refer to the basis matrices as $P, Q, R, S$ in that order. So just transpose them to find that $T(P) = P, T(Q) = R, T(R) = Q$ and $T(S) = S$. It follows that the desired matrix is: $B = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$. (You should probably revise your notes/textbook on matrices of linear maps if you're not sure how this follows.)$

$\noindent c) The rank is $4$ (e.g. by inspecting the matrix, or since clearly $T$ is onto).$

$\noindent d) From a), only $\pm 1$ can be eigenvalues. To see why these are both eigenvalues, note that $T(M) = M$, where $M$ is any non-zero symmetric matrix ($2\times 2$ of course), and $T(N) = -N$, where $N$ is any non-zero skew-symmetric $2\times 2$ matrix.$

$\noindent e) The eigenspace $E_{1}$ is the space of all symmetric $2\times 2$ matrices and $E_{-1}$ is the space of all skew-symmetric $2\times 2$ matrices. I'll let you try finding bases for these. Hint: find conditions on $a,b,c,d$ such that $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is symmetric (to find basis for $E_1$), and then so that it is skew-symmetric (to find basis for $E_{-1}$).$

Drsoccerball · Nov 1, 2016

The matrix the answers is

1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
?

InteGrand · Nov 1, 2016

Drsoccerball said:
The matrix the answers is

1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1
?

Oh yeah sorry, I got a bit complacent/was on autopilot and miscalculated the transposes of P and Q. It should be T(P) = P and T(S) = S, of course, and those answers are right. Thanks!

Drsoccerball · Nov 1, 2016

Sorry I can't find it anywhere in my notes... Why does the transpose of each matrix make up the rows and columns of B ?

InteGrand · Nov 1, 2016

Drsoccerball said:
Sorry I can't find it anywhere in my notes... Why does the transpose of each matrix make up the rows and columns of B ?

$\noindent It's because the map $T$ is the transpose map. Remember, the matrix of a linear map $T$ with respect to a basis $\mathcal{B} = \left\{\vec{b}_1, \vec{b}_{2},\ldots , \vec{b}_n\right\}$ in the domain and $\mathcal{C}$ in the codomain (for our case $\mathcal{B}$ and $\mathcal{C}$ were the same) is found as follows: column $j$ ($j=1,2,\ldots n$) of the desired matrix is \emph{the coordinate vector of $T\left(\vec{b}_j\right)$ with respect to the basis $\mathcal{C}$}. In other words, evaluate $T\left(\vec{b}_1\right)$, find the coordinates of this wrt the specified codomain basis, and this is our column $1$ of the desired matrix of $T$. Then repeat for the other columns using the other domain basis vectors.$

$\noindent So for our example, for first column, we compute that $T(P) = P$, which has coordinates $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ wrt the codomain basis (which is also the domain basis here) $ \mathcal{B} = \left\{P,Q,R,S\right\}$. So this is column $1$ of our desired matrix.$

$\noindent Now apply $T$ to the second domain basis vector: $T(Q) = R$, and $R$ has coordinates $\begin{bmatrix}0\\0\\1\\0\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our second column.$

$\noindent Now apply $T$ to the third domain basis vector: $T(R) = Q$, and $Q$ has coordinates $\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our third column.$

$\noindent Finally, apply $T$ to the fourth domain basis vector: $T(S) = S$, and $S$ has coordinates $\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$ wrt codomain basis $\mathcal{B}$, so this is our fourth column.$

Now that we've found all the columns of the desired matrix, just put them together to get said matrix.

Drsoccerball · Nov 1, 2016

Okay that makes sense.. so it was just a coincidence that each quarter of the matrix was made of each basis matrix's transpose ?

InteGrand · Nov 1, 2016

Drsoccerball said:
Okay that makes sense.. so it was just a coincidence that each quarter of the matrix was made of each basis matrix's transpose ?

Pretty much, the main thing to focus on in terms of how to get that matrix is the columns (as these came from the procedure described above).

Drsoccerball · Nov 1, 2016

What is normal approximation and what is the formula ?

InteGrand · Nov 1, 2016

Drsoccerball said:
What is normal approximation and what is the formula ?

$\noindent We can approximate some distributions with the Normal distribution sometimes (using Central Limit Theorem). E.g. we can do a normal approximation to the Binomial distribution. If $X \sim \mathrm{Bin}(n,p)$ and $n$ is large, then approximately, $X$ is normal with mean $np$ and variance $np(1-p)$. There's several websites with more info, as well as YouTube videos.$

Drsoccerball · Nov 1, 2016

InteGrand said:
$\noindent We can approximate some distributions with the Normal distribution sometimes (using Central Limit Theorem). E.g. we can do a normal approximation to the Binomial distribution. If $X \sim \mathrm{Bin}(n,p)$ and $n$ is large, then approximately, $X$ is normal with mean $np$ and variance $np(1-p)$. There's several websites with more info, as well as YouTube videos.$

I was always seeing Bin had no idea what it meant ahahah thanks Integrand!

Drsoccerball · Nov 1, 2016

Drsoccerball said:
I was always seeing Bin had no idea what it meant ahahah thanks Integrand!

There's another formula called the normal distribution probability density function is that pretty much the same thing?

InteGrand · Nov 1, 2016

Drsoccerball said:
There's another formula called the normal distribution probability density function is that pretty much the same thing?

Not really. That's just the probability density function (PDF) of the normal distribution. All continuous distributions have their PDF that characterises them. (For discrete distributions, the analogue to the PDF is often called a 'probability mass function' (PMF).)

Drsoccerball · Nov 2, 2016

Question : Is E(1/(x+1)) the same thing as 1/(E(x) + 1) ? How about E(1/(x^2 - x +1)) = 1/(E(x^2) - E(x) + 1) ?

InteGrand · Nov 2, 2016

Drsoccerball said:
Question : Is E(1/(x+1)) the same thing as 1/(E(x) + 1) ? How about E(1/(x^2 - x +1)) = 1/(E(x^2) - E(x) + 1) ?

To both: in general no.

Drsoccerball · Nov 2, 2016

is there any other way to find variance besides the E(x^2) -(E(x))^2 way ? Like lets say you have Y = 2X + 3

InteGrand · Nov 2, 2016

Drsoccerball said:
is there any other way to find variance besides the E(x^2) -(E(x))^2 way ? Like lets say you have Y = 2X + 3

$\noindent Yeah we can. Recall the property of variance $\mathrm{Var}\left(aX + b\right) = a^2 \, \mathrm{Var}(X)$, for any constants $a,b$. So if we're given the distribution of $X$ in your example, we can calculate its variance and use that to find $Y$'s variance, as we will have $Y$'s variance given by $\mathrm{Var}(Y) = \mathrm{Var}(2X + 3) = 2^2 \, \mathrm{Var}(X) = 4\,\mathrm{Var}(X)$.$

Drsoccerball · Nov 2, 2016

How would you do it without the formula ? find E(y^2) and E(y) ? How do you even find E(y^2)

y = 2X + 3

y^2 = 4X^2 + 12X + 9

E(y^2) = E(4X^2 + 12X + 9) ?

maths 1B last minute questions (1 Viewer)

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