MATH2601 Higher Linear Algebra (2 Viewers)

davidgoes4wce · Apr 25, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

I don't know much about Graph Theory and Group Theory but are they two different topics? Or Different names but same subjects?

I can't be bothered Googling it

InteGrand · Apr 25, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

davidgoes4wce said:
I don't know much about Graph Theory and Group Theory but are they two different topics? Or Different names but Two dame subjects?

I can't be bothered Googling it

Two different topics.

leehuan · May 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

I was wondering if given the trace and the determinant of a matrix could you write down a unique matrix satisfying these conditions, or a simple formula for the family of matrices satisfying it?

Mostly asking for the 2x2 case

InteGrand · May 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

leehuan said:
I was wondering if given the trace and the determinant of a matrix could you write down a unique matrix satisfying these conditions, or a simple formula for the family of matrices satisfying it?

Mostly asking for the 2x2 case

No, the value for the trace and determinant of a real or complex matrix does not uniquely specify the matrix.

$\noindent For $2\times 2$, let $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and suppose the trace is given to be $\alpha$ and the determinant $\beta$. This is equivalent to $a + d = \alpha$ and $ad - bc = \beta$. So these are the conditions that make $A$ have given trace $\alpha$ and determinant $\beta$.$

(Note that for a 2x2 complex matrix, the trace and determinant will uniquely specify the eigenvalues of the matrix though.)

leehuan · May 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

The intuition of the question was this

$A=\begin{bmatrix}3&1\\-2&0\end{bmatrix}\\ \text{Find all diagonalisable matrices }B\text{ such that }B^2=A$

I then realised that B may share the same eigenvectors as A, and have eigenvalues equal to the square root of those of A. But I'm not sure where to proceed from there.

InteGrand · May 8, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

leehuan said:
The intuition of the question was this

$A=\begin{bmatrix}3&1\\-2&0\end{bmatrix}\\ \text{Find all diagonalisable matrices }B\text{ such that }B^2=A$

I then realised that B may share the same eigenvectors as A, and have eigenvalues equal to the square root of those of A. But I'm not sure where to proceed from there.

$\noindent Here's some hints. If $B$ is a diagonalisable matrix such that $B^{2} = A$, then write $B = PDP^{-1}$ (so $P$ is an invertible matrix whose columns are eigenvectors of $B$, with corresponding eigenvalues in the diagonal entries of the diagonal matrix $D$). Then $B^{2} = PD^{2}P^{-1} = A$. So the eigenvalues of $A$ are the squares of those of $B$ and the corresponding eigenvectors are the eigenvectors from $B$.$

leehuan · May 22, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

$\\\text{Let }T:V\to V\text{ be linear, }\\W_1\text{ a subspace of }V\\ W_2\text{ a subspace of }W_1$

$\text{Suppose }W_1\text{ is invariant under }T.\\ \text{Must }W_2\text{ be invariant under }T\text{ and why/why not?}$

InteGrand · May 22, 2017

Re: MATH2601 Linear Algebra/Group Theory Questions

leehuan said:
$\\\text{Let }T:V\to V\text{ be linear, }\\W_1\text{ a subspace of }V\\ W_2\text{ a subspace of }W_1$

$\text{Suppose }W_1\text{ is invariant under }T.\\ \text{Must }W_2\text{ be invariant under }T\text{ and why/why not?}$

No it need not be. Say V = R^2 and W1 = R^2 (= V) and W2 be the line (t, 0) (the x-axis). Take T to be a rotation map by 90 degrees counter-clockwise about the origin say. Then T is a linear map from V to V, so T(V) = T(W1) is a subspace of W1 = V = R^2, and W2 is a subspace of W1 (which is a subspace of V), but clearly W2 is not invariant under T (e.g. the point (1, 0) in W1 does not get mapped to a point in W2 by T; it gets mapped to (0, 1)).

leehuan · Jun 7, 2017

$V\text{ is a finite dimensional V.S. over }\mathbb{C}\text{ and }T:V\to V\text{ is linear}$

$\\\text{Suppose }T^2 = T = T^*\text{ (i.e. idempotent and self-adjoint)}\\ \text{Prove that there exists a subspace of }W\text{ such that }\\ T(\textbf{v}) = \text{proj}_{W}\textbf{v}$

My approach thus far: Write $\textbf{v} = \textbf{x}+\textbf{y}, \quad \textbf{x}\in W\text{ and }\textbf{y} \in W^\perp$

$<\textbf{x} \mid \textbf{y} > 0 \implies \textbf{x} \perp \textbf{v} - \textbf{x}$

Is this a dead end? Because I don't see how I can use what I know about T here

InteGrand · Jun 7, 2017

leehuan said:
$V\text{ is a finite dimensional V.S. over }\mathbb{C}\text{ and }T:V\to V\text{ is linear}$

$\\\text{Suppose }T^2 = T = T^*\text{ (i.e. idempotent and self-adjoint)}\\ \text{Prove that there exists a subspace of }W\text{ such that }\\ T(\textbf{v}) = \text{proj}_{W}\textbf{v}$

My approach thus far: Write $\textbf{v} = \textbf{x}+\textbf{y}, \quad \textbf{x}\in W\text{ and }\textbf{y} \in W^\perp$

$<\textbf{x} \mid \textbf{y} > 0 \implies \textbf{x} \perp \textbf{v} - \textbf{x}$

Is this a dead end? Because I don't see how I can use what I know about T here

Claim: W := im(T) is such a subspace.

Proof: Exercise.

leehuan · Jun 7, 2017

InteGrand said:
Claim: W := im(T) is such a subspace.

Proof: Exercise.

Where does the inspiration come from that it just happens to be the image that satisfy this criteria

InteGrand · Jun 7, 2017

leehuan said:
Where does the inspiration come from that it just happens to be the image that satisfy this criteria

$\noindent Well for one thing, if $T(\mathbf{v}) = \mathrm{proj}_{W}(\mathbf{v})$ for all $\mathbf{v} \in V$, we need that $T(\mathbf{v})\in W$ for all $\mathbf{v} \in V$ (because by definition $\mathrm{proj}_{W}(\mathbf{v})\in W$). Now as we know, a subspace of $V$ that contains $T(\mathbf{v})$ for all $\mathbf{v} \in V$ is $\mathrm{im}(T)$ (in fact of course any subspace with this property must contain the image, i.e. the image of $T$ is the ``smallest'' subspace with this property). So it would make sense to try the image of $T$. (And if there was any other subspace $W$ that would work, since $W$ would have to contain $\mathrm{im}(T)$, it would have to be the case that $\mathrm{im}(T)$ works too. So just try $\mathrm{im}(T)$.)$

leehuan · Jun 12, 2017

He-Mann · Jun 12, 2017

InteGrand · Jun 12, 2017

leehuan said:
$A=\begin{pmatrix}3 &-1&2\\ -1&3&2\\ 2&2&0\end{pmatrix}$

$\text{Found in part iv): }A=QDQ^T\text{ where }\\ Q=\begin{pmatrix}\frac{1}{\sqrt3} & \frac{1}{\sqrt2} & -\frac{1}{\sqrt5}\\ \frac{1}{\sqrt3} & -\frac{1}{\sqrt2} & -\frac{1}{\sqrt 6}\\ \frac{1}{\sqrt 3}& 0 & \frac{2}{\sqrt6}\end{pmatrix}\\ D = \begin{pmatrix}4 &0&0\\ 0&4&0\\ 0&0&-2\end{pmatrix}$

$\text{v) Write down an expression for a matrix }B\text{ such that }B^2=A$

$\noindent If $A = PDP^{-1}$ in general ($D$ diagonal), then defining $B= PD^{\frac{1}{2}}P^{-1}$, we have $B^{2} = A$. Here $D^{\frac{1}{2}}$ is a square root of $D$, which is a diagonal matrix with diagonal entries all being square roots of the entries in $D$. In your example, you can do this if you are willing to accept complex entries (a $\sqrt{2}i$ for example). You can find out more about square roots of matrices here:$

https://en.wikipedia.org/wiki/Square_root_of_a_matrix#By_diagonalization .

boredofstudiesuser1 · Jun 12, 2017

InteGrand said:
$\noindent If $A = PDP^{-1}$ in general ($D$ diagonal), then defining $B= PD^{\frac{1}{2}}P^{-1}$, we have $B^{2} = A$. Here $D^{\frac{1}{2}}$ is a square root of $D$, which is a diagonal matrix with diagonal entries all being square roots of the entries in $D$. In your example, you can do this if you are willing to accept complex entries (a $\sqrt{2}i$ for example). You can find out more about square roots of matrices here:$

https://en.wikipedia.org/wiki/Square_root_of_a_matrix

Woah, how do you do it so fast?

He-Mann · Jun 12, 2017

boredofstudiesuser1 said:
Woah, how do you do it so fast?

He's a machine!

leehuan · Jun 12, 2017

I feel bad lol. I had the same idea as InteGrand, I just mucked up my matlab input when I went to check my answer
_______________

$\text{Is this identity true?}\\ \text{proj}_W\textbf{v} = \textbf{v} - \text{proj}_{W^\perp}\textbf{v}$

InteGrand · Jun 12, 2017

leehuan said:
I feel bad lol. I had the same idea as InteGrand, I just mucked up my matlab input when I went to check my answer
_______________

$\text{Is this identity true?}\\ \text{proj}_W\textbf{v} = \textbf{v} - \text{proj}_{W^\perp}\textbf{v}$

Yes

boredofstudiesuser1 · Jun 12, 2017

leehuan said:
I feel bad lol. I had the same idea as InteGrand, I just mucked up my matlab input when I went to check my answer
_______________

$\text{Is this identity true?}\\ \text{proj}_W\textbf{v} = \textbf{v} - \text{proj}_{W^\perp}\textbf{v}$

It's ok, we'll call you a machine too if it makes you feel better.

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