---
# System prepended metadata

title: Midterm 2 Grading Scheme
tags: [MAT224-Organize]

---

---
tags: MAT224-Organize
---
# Midterm 2 Grading Scheme

[**GO BACK TO INFORMATION PAGE**](http://224.y7y.cc)


Use navigation on right bar for your convenience, or use it with links in this page.

Problem|Note
-|-
[Problem 1](#Problem-1)|20 marks
[Problem 2](#Problem-2)|20 marks
[Problem 3](#Problem-3)|10 marks
[Problem 4](#Problem-4)|20 marks
[Problem 5](#Problem-5)|8 marks
[Problem 6](#Problem-6)|12 marks
$$
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\newcommand{\Rref}{\mathrm{rref}}
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\newcommand{\Rank}{\mathrm{rank}}
\newcommand{\rank}{\Rank}
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\def\em\@tt#1\em{\@tt|\em}
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\begin{tabular}{#1}
\hline
#2
\\\hline
\end{tabular}\@gobble
}
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\@ifnextchar,{\@tt #1 & #2\\\hline\expandafter\expandafter\expandafter\@gobble\expandafter\@gobble\@gobble\@ttt}{
\@ifnextchar.{\cm\@tt|l #1 & #2\@ttt}{\@tt|l #1 & #2\@ttt}
}
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\@ifnextchar,{\@tut #1 & #2\\\expandafter\expandafter\expandafter\@gobble\expandafter\@gobble\@gobble\@tutt}{
\@ifnextchar.{\begin{pmatrix}#1 &#2\end{pmatrix}\@gobble}{\@tut #1 & #2\@tutt}
}
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\def\bm{\@tuut\@gobble\@tuutt}
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\@ifnextchar,{\@tuut #1 & #2\cr\expandafter\expandafter\expandafter\@gobble\expandafter\@gobble\@gobble\@tuutt}{
\@ifnextchar.{\bordermatrix{#1 &#2\cr}\@gobble}{\@tuut #1 & #2\@tuutt}
}
}
$$
# Problem 1
[Go top](#Midterm-2-Grading-Scheme)

**Solution**:FFTFF

**Grading Policy:**
- For True Questions, True= 4pt, False=0pt
- For False Questions, True = 0pt, False + incorrect Counter example = 2pt, False+ correct Counter Example = 4pt.
- Problem 1.4: If student say need $\ker\perp\im$, it still consider correct even if they do not have a counter example 
- Problem 1.3: If student say false but his counter example saitiesfies all condition, partial marks: 2pt.
:

# Problem 2
[Go top](#Midterm-2-Grading-Scheme)

**Solution**:	$$
	\dim(\ker (T^2))=6
	$$	
$$
	\dim(\ker (T^4))=10
	$$	
$$
	\dim\left(\mathrm{Im} (T^2)\cap\ker (T^3)\right)=6
	$$
$$
	\dim(\mathrm{Im} (T^7)) = 0
	$$
$$
	\dim\left(\ker(T+I)\right) =0
	$$

**Grading Policy:**
- Each answer is 2 marks. No partial credits.



# Problem 3
[Go top](#Midterm-2-Grading-Scheme)

**Navigation**

Problem|Note
-|-
[Problem 3 a)](#Problem-3a)|5 marks
[Problem 3 b)](#Problem-3b)|5 marks
[Problem 3 c)](#Problem-3c)|10 marks

## Problem 3a
[go back](#Problem-3)

**Solution**

We want to show if there exists $a_1,a_2,a_3,a_4,a_5\in\mathbb{C}$ such that
$$
a_1T^2(\vv)+ a_2T(\vv)+ a_3\vv+ a_4T(\ww)+ a_5\ww=\0
$$
Then $a_1=a_2=a_3=a_4=a_5=0$. (**3 marks**)

Apply $T^2$ we have

$$
a_3T^2(\vv)=\0
$$
Since $\vv\neq\0$ Therefore $a_3=0$.

So we left 

$$
a_1T^2(\vv)+ a_2T(\vv)+ a_4T(\ww)+ a_5\ww=\0
$$
Apply $T$ we have

$$
a_2T^2(\vv)+a_5T(\ww)=\0
$$
Since $\{T(w),T^2(v)\}$ is linearly independent, we have $a_2=a_5=0$
So we left
$$
a_1T^2(\vv)+a_4T(\ww)=\0
$$
Since $\{T(w),T^2(v)\}$ is linearly independent, we have
$a_1=a_4=0$.


**Grading Policy:**

- Set up what to prove correctly: at least **3 marks**
- Minor Issue in set up but set up almost correctly **2 marks**


## Problem 3b
[go back](#Problem-3)
**Solution**

We want to show $T(W)\subset W$.

Since
$$
T(\{T^2(\vv),T(\vv),\vv,T(\ww),\ww\})=\{T^3(\vv),T^2(\vv),T(\vv),T^2(\ww),T(\ww)\}=\{\0,T^2(\vv),T(\vv),T(\ww)\}\subset W
$$
Therefore 
$$
T(W) = T(\spann\{T^2(\vv),T(\vv),\vv,T(\ww),\ww\})=\spann\{\0,T^2(\vv),T(\vv),T(\ww)\}\subset \spann W=W.
$$

**Grading Scheme**
- Set up what to prove correctly: at least **3 marks**
- Minor Issue in set up but set up almost correctly **2 marks**
- If they calculated out
$$
T|_W\mat{T^2(v)& T(v)& v& T(w)& w}=$$$$\mat{T^2(v)& T(v)& v& T(w)& w}\mat{0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&0&0}
$$
And then conclude it is invariant subspace, that is also correct.



## Problem 3c
[go back](#Problem-3)
**Solution**
$$
T|_W\mat{T^2(v)& T(v)& v& T(w)& w}=$$$$\mat{T^2(v)& T(v)& v& T(w)& w}\mat{0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&0&0}
$$

Or they can just write 
$$
\mat{0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&0&0}
$$

**Grading Policy:**


- Minor number issue **-1 marks**
- **If they have fatal issue, like say T=matrix, take 4 points off**




# Problem 4
[Go top](#Midterm-2-Grading-Scheme)

**Navigation**

Problem|Note
-|-
[Problem 4 a)](#Problem-4a)|5 marks
[Problem 4 b)](#Problem-4b)|8 marks
[Problem 4 c)](#Problem-4c)|8 marks
[Problem 4 d)](#Problem-4d)|5 marks

## Problem 4a
[go back](#Problem-4)

**Solution**
$\dim(\ker T)=2$, $\dim(\ker T^2)=4$, $\dim(\ker T^3)=6$

**Grading Policy**
- Each correct Answer is 2 points

## Problem 4b
[go back](#Problem-4)

**Solution**
$$
\mat{0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&0&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&0&0&0&0&0}
$$

**Marking Scheme**
- If they did wrong because the last question wrong, give full credits if their answer is correct based on last question.
- Minor number issue **-1 marks**
- **If they have fatal issue, like say T=matrix, take 2 points off**
- If they did not write down JCF but draw correct tableau, **3 marks**





# Problem 5
[Go top](#Midterm-2-Grading-Scheme)

**Navigation**

Problem|Note
-|-
[Problem 5 a)](#Problem-5a)|7 marks
[Problem 5 b)](#Problem-5b)|7 marks
[Problem 5 c)](#Problem-5c)|8 marks
[Problem 5 d)](#Problem-5d)|8 marks

## Problem 5a
[go back](#Problem-5)
**Solution**
We want to show there exists $n$ such that $(S\circ T)^n=0$, here $\map 0VV$ means the zero operator.

Since $T\circ S$ is nilpotent, there exists some integer $m$ such that $(T\circ S)^m=0$. Therefore let $n=m+1$ then
$$
(S\circ T)^n=(S\circ T)^{m+1}=S\circ(T\circ S)^m\circ T=S\circ 0\circ T=0
$$

Therefore $S\circ T$ is a nilpotent operator.


**Grading Policy:**
- Please note there are other methods, they could use that the characteristic polynomial of ST and TS only differ by a factor of $\lambda$ and that operator is nilpotent if and only if eigenvalue are all 0.
- If they set up (Wish to show...)correctly, at least give **3 marks**: 
    - If they interpretes definition correctly, by saying we know that $(T\circ S)^m=0$ for some $m$: **2marks**. 
- Minor issue or minor issue in set up **-1 mark**



## Problem 5b
[go back](#Problem-5)
**Solution**
We would like to show $T(\im(T))\subset\im(T)$, that is to show for any $\vv\in\im(T)$, $T\vv\in\im(T)$.

To prove, choose $\ww=\vv$, therefore we find some $\ww\in V$ that $T\vv=T\ww$, therefore $T\vv\in\im(T)$. (One can also say $T\vv\in\im(T)$ is clearly true.)

**Marking Scheme**
<!--- They can also use the theorem in slides, if they say since $T$ is commutative with $T$, any image or kernel of the operator commute with $T$ is invariant subspace, it is fine. -->
- Set up what to prove correctly: at least **3 marks**
- Minor Issue in set up but set up almost correctly **-1 marks**

## Problem 5c
[go back](#Problem-5)

**Solution**
We wish to show, **if** there exists an inverse $$\map{(T\circ S)^{-1}}WW$$ such that $(T\circ S)^{-1}(T\circ S)=I_W$ and $(T\circ S)\circ(T\circ S)^{-1}=I_W$, \textbf{THEN} for any $\ww\in W$, $S(\ww)=\0$(injective) and that for any $\ww\in W$, there exists $\vv\in V$ so that $T(\vv)=\ww$ (surjective). 


Indeed, if $S(\ww)=\0$, then $$
S\circ (T\circ S)^{-1}(\ww)=(T\circ S)^{-1}\0=\0
$$
This implies
$$
\ww=I_W(\ww)=T\circ S\circ (T\circ S)^{-1}(\ww) = T\0=\0
$$
So $S$ is an injective. Further, to show $T$ is a surjective, for any $\ww\in W$ Let 
$$
\vv=S\circ(T\circ S)^{-1}(\ww)
$$
Then 
$$
T(\vv)=T\circ S\circ(T\circ S)^{-1}(\ww)=I_W(\ww)=\ww.
$$
Therefore $T$ is a surjective.

**Marking Scheme**
<!--- They can also use the theorem in slides, if they say since $T$ is commutative with $T$, any image or kernel of the operator commute with $T$ is invariant subspace, it is fine. -->
- If they directly use invertible as isomorphism, it is fine
- **They can also use left factor of surjective is surjective, right factor of injective is injective** But they have to first say: because $T\circ S$ is invertible, then it is both injective and surjective
- Set up what to prove correctly: at least **4 marks**
    - Set up one direction correctly: **2 marks**
- Minor Issue in set up but set up almost correctly **-1 marks**

## Problem 5d
[go back](#Problem-5)

**Solution**
($\Leftarrow$ ) We wish to show if $\mathrm{Im}(T) \subseteq \ker(S)$ then $S\circ T=0$, which is to show $S\circ T(\vv)=\0$ for any $\vv\in V$. Indeed, $T(\vv)\in \im(T)\subset \ker(S)$, so $T(\vv)\in\ker(S)$ so $S\circ T(\vv)=S(T(\vv))=0$.



($\Rightarrow$) We wish to show if $S\circ T=0$, then $\mathrm{Im}(T) \subseteq \ker(S)$, this is to show for any $\ww\in W$ such that $\ww=T(\vv)$ for some $\vv\in V$, we have $S(\vv)=0$. Indeed, if $\ww=T(\vv)$, 
$$
S(\ww)=S(T(\vv))=S\circ T(\vv)=0(\vec{v})=\0
$$

- Set up what to prove correctly: at least **4 marks**
    - Set up one direction correctly: **2 marks**
- Minor Issue in set up but set up almost correctly **-1 marks**

# Problem 6
[Go top](#Midterm-2-Grading-Scheme)
**10 marks**

The Jordan Canonical Form is tehrefore

$$
\mat{0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0\\0&0&0&0&0&0}
$$
or
$$
\mat{0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&0&0\\0&0&0&0&0&1\\0&0&0&0&0&0}
$$

**Marking Scheme**

- Minor number issue **-1 marks**
- If Among them there is only one correct: Give them 4 points.
- Among them there is two correct: Give them 8 points.

- Did one correct but missing the other: 6 points.
- Did all two correct: 10 points.

- For only drawing Young Tableau, only substract 1 point based on it
