Assignments

0

Easy

1

1. ** (in-class)
   1. 3/5
   2. 4/5
   3. failed
2. Not attempted
3. **
   1. 5/5
   2. 4/5
4. **
   1. hard, didn’t think of unit vectors
   2. hard
5. *** hard, just algebra though
6. *** way to easy with our current tools

2

1. ** remembered concept, formal proof failed
2. ** 3/5, easy with our concepts and $A=v{\lambda }^{\top }$
3. **
   1. irrelevant for exam
   2. easy
4. **
   1. easy with linearity axioms
   2. also easy
5. • Calculations basically trivial with RREF
6. **
   1. triv
   2. easy
   3. triv
   4. trivial with determinants
7. • easy, just rewriting

3

1. 1. triv
   2. triv
2. • easy
3. • decent (forgot to write out full final for (a))
4. • decent, formality hard though

5. ---

   1. decent, slight mistake but found before checking sol
   2. easy using induction
   3. easy, but mistake, nilpotency degree at most $k$, not $=k$.
   4. easy
   5. not solved, formalism way too hard but easy conceptually
6. not attempted yet

4

2. • easy, formalism
3. **
   1. extremely easy
   2. easy
   3. easy
4. ** trivial using RREF
5. ** pretty ok, induction proof

6. ---

   1. easy using RREF
   2. easy using determinant
   3. by contradiction, is ok but formalism
   4. …

5

2. *** impossible for me, even to understand revisit
3. • easy with RREF
4. • Calculation mistake in RREF, like an idiot
5. **
   1. easy but calc mistake that I caught
   2. easy to do, hard to prove/formalise
6. ** provable, but the solution is way more elegant

6.) Show that given $v_1,v_2,v_3$ independent, we have $w_1=v_1+v_2$, $w_2=-v_1+v_2$ and $w_3=v_1+v_2+v_3$ independent.

6

1. • very good training exercise CR-Decomp computation
2. 1. very easy
   2. hard, especially formalism wise To prove something is a basis, we show it’s independent and spans the space.
3. **
   1. easy
   2. doable, formalism again, just yapping to prove something
4. • Made an error. To show the dimenion of $\text{Span}(u,v,w)$ we only have to show that they are independent and then say $3$. If they are not independent, then remove the dependent ones and the elements left are the dimension.
5. **
   1. Easy
   2. Easier if you use orthogonal subspaces and the fact that $H_d=N(d)$
   3. formalism but easy
6. ** Cool exercise, good for subspace proofs and nice proof technique, hard

7

1. • I’m really bad at calculations
   1. Redo the RREF, good exercise
   2. Good nullspace finding exercise
   3. Good recap for the dimensions
   4. easy, good reminder that $R(A)=R(R)$ while the column space is not preserved by RREF.
2. ** pretty ok, some mistakes
   1. some formalism, where we choose two names for the same vector to show that there is a vector such that the $Ax=w$ thus $w\in C(A)$ even though we defined $x=w$.
3. • easy enough, good practice for dimensions
4. 1. easy
   2. hard to find counterexample, skill issue
5. *** TODO
   1. easy
   2. again
   3. again
   4. again
6. **
   1. $N(A)=N(2A)$ for square $A$ makes sense
   2. quite easy that for square $C(A)=C(2A)$
   3. easy enough, we just solve the RREF and see that all three rows are independent as we have $\text{rank}(A)=3$, thus they only intsersect at $0$
   4. easy, but I miscalculated twice

8

1. • Very good Least Squares practice!
2. **
   1. figured this out by myself, only missing the assumption from Lemma 6.12 that the first two columns are independent.
   2. very good least squares practice. Because $A$ has independent columns, the solution is unique!
3. • some easy linear transformation proofs, formalism wise interesting.
4. ** interesting, use property that $u_1,\dots ,u_n$ are independent to argue that the only combination giving $0$ is $\lambda =0$ and thus $\lambda =\mu =0$, thus they are all independent.
5. **
   1. the determinant only exists for square matrices! This was harder than I thought, maybe redo
   2. Copy proof into cheatsheet, important
6. *** TODO

9

1. **
   1. use Lemma and full rank factorisation
   2. Just expand using CR
   3. More difficult, first show that it holds for full col-rank, then full row-rank then for any $A=CR$ decomposition.
2. ** very good practice for certificate of no solutions, prove indirectly
3. • interesting application, redo at the end
4. • difficult for me, uses intersection of $N(A{)}^{\perp }\cap N(A)$ and $N(A)=N(A^{\top }A)$ for the proof.
5. ** very interesting, might redo
   1. Uses the fact that for $x\in C(A)$, $A{A}^{†}$ is the identity as $A{A}^{†}$ projects onto $C(A)$ and $x$ already in $C(A)$
   2. Similar for the other side, where $A^{†}A$ is the projection onto $C(A^{\top })$ and thus for an $x\in C(A^{\top })$, $A^{†}A$ is the identity.
6. ** nice exercise on Projection matrices and pseudoinverses
   1. show symmetry by just inserting everything
   2. we use the fact that $R(R)=R(A)$ since it’s by RREF and thus $R^{†}R$ is just a projection onto $R(A)$.

10

1. • good practice for G-S!
   1. made 10 mistakes
   2. was ok
   3. upper-triangular
   4. Not true, think of $-I$ as a counterexample!
2. **
   1. easy
   2. easy
   3. easy
3. • easy, uses kronecker delta and $Q{e}_i$ to get $q_i$ $i$-th column of $q$
4. ** absolutely cursed function argument

5. ---

   1. we just need to apply the Certificate creatively
   2. good exercise for subspaces and applying stuff hard, had to cheat

6. ---

   1. straightforward application of the determinant formula as sum of products
   2. quite easy, just use property we just proved
   3. possible but the formalism is insane

11

1. • good determinant practice
   1. Easy, but I miscalculated
   2. Also easy
2. **
   1. concept easy, formalism: Observe that there is a unique non-zero entry in each of it’s first $m$ columns. Thus every permutation $\sigma$ that contributes to the determinant of $M^{\prime }$ in the formula $\det M^{\prime }={\sum }_{\sigma \in {\Pi }_n}\text{sign}(\sigma ){\prod }_{i=1}^n{M}_{i,\sigma (i)}^{\prime }$must select these non-zero entries, i.e. $\sigma (i)=i$ for all $i\in [m]$. The formula then simplifies to $\det M^{\prime }={\sum }_{\sigma \in {\Pi }_{n-m}}\text{sign}(\sigma )\prod {i=1}^{n-m}{C}_{i,\sigma (i)}=\det C$as the non-zero entries in the first $m$ columns are all one.
   2. easy
3. • Complex numbers, quite easy with the formulas
4. ** easy, multilinearity exercise
5. **
   1. easy EW proof
   2. easy but I had a slight miscalculation
6. • easy concept hard formalism once more

12

1. • Decent Eigenvector practice
   1. Ez
   2. Ez
   3. Interesting application of similar matrices
2. **
   1. Good proof maybe put on sheet
   2. important proof of complete set of EV conditions Remember, a complete set of eigenvectors is linearly independent and thus forms a basis of ${\mathbb{R}}^{n\times n}$.
   3. Reuses previous results
   4. reuses previous results
3. ** Good training for a calculation question if it comes up
4. **
   1. Good determinant trick with multiplication from both sides. As $\det (S)\cdot \det (S^{-1})=1$ it doesn’t change anything.
   2. Very good diagonalisation exercise. For ${\Lambda }_A,{\Lambda }_B$ diagonal we have ${\Lambda }_A{\Lambda }_B={\Lambda }_B{\Lambda }_A$ thus $AB=(V{\Lambda }_A{V}^{-1})(V{\Lambda }_B{V}^{-1})$ can be swapped around.
   3. Hard, good formalism template for the Diagonalisation
   4. hard
5. **
   1. Nice to know $I-n{n}^{\top }$ is the projection onto a plane in $3$d with $n$ orthogonal to the plane.

6. ---

   1. easy
   2. easy
   3. Nice diagonalisation calculation
7. ** easy, just find EWs, check that $1$ is in them and then find EV for $\lambda =1$

13

1. • easy, old exam question
2. ** Prove $B$ is PSD then easy, just write out the full sum and then replace and factorise
3. ** Literally impossible, even the solution skips more steps than an elevator.
4. ** Quite ok, just apply the Rayleigh and remember arithmetic
5. ** interesting, we pray it’s not at the exam
6. ** same as 5.