1. Solution: Let $u_1,u_2,\cdots,u_n$ be a basis of $U$. Thus $n=\dim U=\dim V$. Hence $u_1,u_2,\cdots,u_n$ is a linearly independent list of vectors in V with length $\dim V$. By 2.39, $u_1,u_2,\cdots,u_n$ is a basis of $V$. In particular, any vector in $V$ can be written as a linear combination of $u_1,u_2,\cdots,u_n$. As $u_i\in U$, it follows that $V\subset U$. This means that $U=V$.

2. Solution: The dimension of a subspace $U$ of $\R^2$ can only be 0,1,2. If $\dim U=0$, then $U=\{0\}$. If $\dim U=2$, then $U=\R^2$ by problem 1. If $\dim U=1$, then for any nonzero $x\in U$, it follows that \[U=\{kx:k\in\R\},\]which it is the line through $x$ and the origin.

3. Solution: It is similar to Problem 2. If $\dim U=2$, there exist two linearly independent $x,y\in\R^3$. Then \[U=\{k_1x+k_2y:k_1\in\R,k_2\in\R\},\]which it is the plane through $x$, $y$ and the origin.

4. Solution: (a) A basis of $U$ is $x-6$, $x^2-6x$, $x^3-6x^2$ and $x^4-6x^3$. Of course, $x-6$, $x^2-6x$, $x^3-6x^2$ and $x^4-6x^3$ is linearly independent since they has different degrees (It is easy to check). Moreover, if $p(6)=0$, then $p(x)$ is divided by $x-6$, hence \begin{align*}p(x)=&(x-6)(k_3x^3+k_2x^2+k_1x+k_0)\\=&k_3(x^4-6x^3)+k_2(x^3-6x^2)+k_1(x^2-6x)+k_0(x-6)\end{align*} is a linear combination of $x-6$, $x^2-6x$, $x^3-6x^2$ and $x^4-6x^3$.

(b) Of course, $1$, $x-6$, $x^2-6x$, $x^3-6x^2$ and $x^4-6x^3$ is a basis of $\ca P_4(\mb F)$.

(c) Let $W=\{c:c\in\mb F\}$, then $\ca P_4(\mb F)=U\oplus W$ by (b).

5. Solution: (a) For polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$, consider the condition when $f”(6)=0$. Then you will get a linear equation about $a,b,c,d,e$. Find a basis of the solution space of this linear equation. Then substitute it into $f(x)=ax^4+bx^3+cx^2+dx+e$, you will get a basis of $U$(why?). I skip the details and give a example of basis. $1$, $x$, $x^3-18x^2$, $x^4-12x^3$ is a basis of $U$.

(b) Of course, $1$, $x$, $x^2$, $x^3-18x^2$ and $x^4-12x^3$ is a basis of $\ca P_4(\mb R)$.

(c) Let $W=\{cx^2:c\in\mb R\}$, then $\ca P_4(\mb R)=U\oplus W$ by (b).

6. Solution: (a) For polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$, consider the condition when $f(2)=f(5)$. Then you will get a linear equation about $a,b,c,d,e$. The dimension of the solution space of this linear equation is $4$, so is $U$(why?). Thus we only have to give $4$ linearly independent polynomials in $\ca P_4(\mb F)$ such that each of them attain the same value at $x=2$ and $x=5$. A good example is $1$, $x^2-7x+10$, $x(x^2-7x+10)$ and $x^2(x^2-7x+10)$.

(b) Of course, $1$, $x$, $x^2-7x+10$, $x(x^2-7x+10)$ and $x^2(x^2-7x+10)$ is a basis of $\ca P_4(\mb F)$.

(c) Let $W=\{cx:c\in\mb F\}$, then $\ca P_4(\mb F)=U\oplus W$ by (b).

7. Solution: (a) For polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$, consider the condition when $f(2)=f(5)=f(6)$. Then you will get $2$ linear equations about $a,b,c,d,e$. The dimension of the solution space of this linear equation is $3$, so is $U$(why?). Thus we only have to give $3$ linearly independent polynomials in $\ca P_4(\mb F)$ such that each of them attain the same value at $x=2$, $x=5$ and $x=6$. A good example is $1$, $(x-2)(x-5)(x-6)$ and $x(x-2)(x-5)(x-6)$.

(b) Of course, $1$, $x$, $x^2$, $(x-2)(x-5)(x-6)$ and $x(x-2)(x-5)(x-6)$ is a basis of $\ca P_4(\mb F)$.

(c) Let $W=\{cx+dx^2:c\in\mb F,d\in\mb F\}$, then $\ca P_4(\mb F)=U\oplus W$ by (b).

8. Solution: (a) For polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$, consider the condition when $\int_{-1}^1 f=0$. Then you will get a linear equation about $a,b,c,d,e$, which is $a/5+c/3+e=0$. Find a basis of the solution space of this linear equation. Then substitute it into $f(x)=ax^4+bx^3+cx^2+dx+e$, you will get a basis of $U$(why?). I skip the details and give a example of basis. $x$, $3x^2-1$, $x^3$ and $5x^4-1$ is a basis of $U$.

(b) Of course, $1$, $x$, $3x^2-1$, $x^3$ and $5x^4-1$ is a basis of $\ca P_4(\mb R)$.

(c) Let $W=\{c:c\in\mb R\}$, then $\ca P_4(\mb R)=U\oplus W$ by (b).

9. Solution: Note that \[v_2-v_1=(v_2+w)-(v_1+w),\]it follows that $v_2-v_1\in \mathrm{span}(v_1+w,\cdots,v_m+w)$. Similarly, $v_i-v_1\in \mathrm{span}(v_1+w,\cdots,v_m+w)$ for all $2\leqslant i\leqslant m$.

Actually, $v_2-v_1$, $\cdots$, $v_m-v_1$ is linearly independent since $v_1$, $\cdots$, $v_m$ is linearly independent in $V$. (It is easy to prove, see examples in Exercise 2.A and 2.B). By 2.33, it follows that \[\dim\mathrm{span}(v_1+w,\cdots,v_m+w)\geqslant m-1.\]

10. Solution: Because $p_0$ has degree $0$, we have $\mathrm{span}(p_0)=\mathrm{span}(1)$. If we assume that \[ \mathrm{span}(p_0,p_1,\cdots,p_i)=\mathrm{span}(1,x,\cdots,x^i). \] Then by assumption, it is trivial that \[ \mathrm{span}(p_0,p_1,\cdots,p_i,p_{i+1})\subset \mathrm{span}(1,x,\cdots,x^i,x^{i+1}). \]On the other hand, $p_{i+1}$ has degree $i+1$, hence it can be written as \[p_{i+1}=a_{i+1}x^{i+1}+f_{i+1}(x),\]where $a_{i+1}\ne0$ and $\deg f_{i+1}(x)\leqslant i$. Then \[x^{i+1}=\frac{1}{a_{i+1}}(p_{i+1}-f_{i+1}(x))\in \mathrm{span}(1,x,\cdots,x^i,p_{i+1}).\]Note that $\mathrm{span}(p_0,p_1,\cdots,p_i)=\mathrm{span}(1,x,\cdots,x^i)$, we conclude \[\mathrm{span}(1,x,\cdots,x^i,p_{i+1})=\mathrm{span}(p_0,p_1,\cdots,p_i,p_{i+1}).\]Thus \[x^{i+1}\in \mathrm{span}(p_0,p_1,\cdots,p_i,p_{i+1}),\]then \[\mathrm{span}(1,x,\cdots,x^i,x^{i+1})\subset \mathrm{span}(p_0,p_1,\cdots,p_i,p_{i+1}).\] By induction, we have \[ \mathrm{span}(p_0,p_1,\cdots,p_i)=\mathrm{span}(1,x,\cdots,x^i). \]for all $0\leqslant i\leqslant m$. In particular, \[ \mathrm{span}(p_0,p_1,\cdots,p_m)=\mathrm{span}(1,x,\cdots,x^m) \]means $p_0$, $p_1$, $\cdots$, $p_m$ is a basis of $\ca P(\mb F)$. Because $p_0$, $p_1$, $\cdots$, $p_m$ is a spanning list of $\ca P(\mb F)$ with the same length as the dimension of $\ca P_m(\mb F)$ (2.42).

11. Solution: By 2.43, we have \[\dim(U\cap W)=\dim U+\dim W-\dim(U+W)=8-\dim(\mb R^8)=0.\]Hence $U \cap W=\{0\}$, combining with $U+W=\mb R^8$, it follows that $\mb R^8=U\oplus W$.

12. Solution: By 2.43, we have \[\dim(U\cap W)=\dim U+\dim W-\dim(U+W)=10-\dim(U+W).\]Note that $U+W$ is a subspace of $\mb R^9$, it follows that $\dim(U+W)\leqslant 9$ (by 2.38). Hence $\dim(U\cap W)\geqslant 1$, i.e. $U \cap W\ne\{0\}$.

13. Solution: By 2.43, we have \[\dim(U\cap W)=\dim U+\dim W-\dim(U+W)=8-\dim(U+W)\geqslant 8-\dim(\mb C^6)=2.\]Hence there exists $e_1, e_2\in U\cap W$ such that $e_1$ and $e_2$ are linearly independent. Then neither of $e_1$ or $e_2$ is a scalar multiple of the other.

14. Solution: Choose a basis $\ca W_i$ of $U_i$, then by definition of direct sum, $U_1+\cdots+U_m$ can be spanned by the union of $\ca W_1$, $\cdots$, $\ca W_m$. From 2.31, we conclude \[\dim(U_1+\cdots+U_m)\leqslant \dim U_1+\cdots+\dim U_m,\]since the cardinality of the gather of $\ca W_1$, $\cdots$, $\ca W_m$ is no more than $\dim U_1+\cdots+\dim U_m$. In particular, $U_1+\cdots+U_m$ is finite-dimensional.

15. Solution: Let $(v_1,\cdots,v_n)$ be a basis of $V$. For each $j$, let $U_j$ equal $\mathrm{span}(v_j)$; in other words, $U_j=\{av_j:a\in\mathbb F\}$. It is easy to see that $\dim U_j=1$ for all $j=1,\cdots,n$. Because $(v_1,\cdots,v_n)$ is a basis of $V$, each vector in V can be written uniquely in the form \[a_1v_1+\cdots+a_nv_n,\]where $a_1$, $\cdots$, $a_n\in\mathbb F$. By definition of direct sum, this means that $V=U_1\oplus \cdots \oplus U_n$.

16. Solution: Since $U_1+\cdots+U_m$ is a direct sum, it follows that\[U_1\oplus \cdots \oplus U_m=U_1+\cdots+U_m.\]Hence $U_1\oplus \cdots \oplus U_m$ is finite dimensional by Problem 14. Now we use induction on $m$ to show\[\dim U_1\oplus \cdots \oplus U_m= \dim U_1+\cdots+\dim U_m.\]By 2.43, for $m=2$, we have\[\dim(U_1+U_2)=\dim U_1+\dim U_2-\dim(U_1\cap U_2)=\dim U_1+\dim U_2\]since $U_1\cap U_2=0$ as $U_1+U_2$ is a direct sum.

Suppose the equality is true for $m-1$. Now consider the case $m$, if $U_1+\cdots+U_m$ is a direct sum, then the only way to write $0$ as a sum $u_1+\cdots+u_m$, where each $u_j$ is in $U_j$, is by taking each $u_j$ equal to $0$. Therefore the only way to write $0$ as a sum $u_1+\cdots+u_{m-1}$, where each $u_j$ is in $U_j$, is by taking each $u_j$ equal to $0$. It follows that $U_1+\cdots+U_{m-1}$ is a direct sum, hence\[\dim U_1\oplus \cdots \oplus U_{m-1}= \dim U_1+\cdots+\dim U_{m-1}.\]On the other hand, let $W=U_1\oplus \cdots \oplus U_{m-1}$, then $U_1\oplus \cdots \oplus U_m=W+U_m$. Suppose $0=x+y$, where $x=x_1+\cdots+x_{m-1}\in W$ and $y\in u_m$, where each $x_j$ is in $U_j$, it follows from 1.44 that $x_i=0$ and $y=0$. Hence $W+U_m$ is a direct sum again by 1.44. Therefore by the inductive assumption, we have\begin{align*}&\ \dim U_1\oplus \cdots \oplus U_m\\ =&\ \dim(W+U_m)=\dim W+\dim U_m\\ =&\ \dim U_1+\cdots+\dim U_{m-1}+\dim U_m.\end{align*}

17. Solution: To give a counterexample, let $V=\mb R^2$, and let \[U_1=\{(x,0):x\in\R\},\]\[U_2=\{(0,y):y\in\R\},\]\[U_3=\{(x,x):x\in\R\}.\]Then $U_1+U_2+U_3=\R^2$, so $\dim(U_1+U_2+U_3)=2$. However, \[\dim U_1=\dim U_2=\dim U_3=1\]and \[\dim(U_1\cap U_2)=\dim (U_2\cap U_3)=\dim (U_3\cap U_1)=\dim (U_1\cap U_2\cap U_3)=0.\]Thus in this case our guess would reduce to the formula $2=3$, which is obviously false.

## Yannis V.

26 Jun 2021Hello! This is my first contribution on this great site. I think I found one more solution to problem #9.

We can identify the following subspaces of V (using the fact that w \in V)

U = span(v_1, ..., v_m)

U1 = span(v_1 + w, ... , v_m+ w)

U2 = span(w)

Note that U is a subset of U1+U2, since for each u in U:

u = a_1*v_1 + ... + a_m * v+m

= a_1*(v_1+w) + ... + a_m*(v_m+w) - (a_1+...+a_m) * w

We can then see that U is a subspace of U1 + U2.

Then m = dim(U) <= dim(U1+U2) [by 2.38]

= dimU1 + dimU2 - dim(U1 \intersection U2) [by 2.43]

= m - 1

## Yannis V.

26 Jun 2021(typo in the last line: should be <= m -1)

## sam

6 Jan 2021For questions 2 and 3, to complete the questions don't we have to show that those subspaces are precisely $R^{2}$ and $R^{3}$, respectively? So for example for question 2, we show that those are precisely the subspaces of $R^{2}$ by showing {0}, $R^{2}$, and $U = \{kx, k \in R\}$ is a direct sum equal to $R^{2}$?

## LERONG YANG

13 Aug 2020For #17, the counterexample is good but text " U1+U2+U3=R2” is inaccurate.

R2 is plane obviously, and U1, U2, U3 are three axises (Lines)， we better say dim(U1+U2+U3)=dim(R2)

## Marie

4 Jul 2020I do not understand how to approach question 5. I started by letting p(x)=a+bx+cx^2+dx^3+ex^4 then came with an expression for p''(6) that is 2c+36d+432e=0. I do not know where to go from here on. Can someone explain what to do next? thanks.

## Linearity

4 Jul 2020Compute the second derivative $$p''(x)=12ex^2+6dx+2c.$$Plug in $6$ and use $p''(6)$ to get it.

## Obi

20 Jul 2020That is what she did - she computed the second derivative, plugged in 6 and got a linear equation that links c, d and e. The question was about what to do next.

I'm also not sure

## Lucida

22 Jul 2020I think at this point of the book, we shouldn't introduce the idea of linear equation system to solve problem #5,#6 and #7. Analogous to examples in the book, I think what the author might except ideas like this:

For #5, it's easy to find 4 linearly independent elements: 1,p,(p-6)^3,(p-6)^4, and since the whole space has its dimension equal to 5, the dimension of this subspace must be less than 5, so these 4 linearly independent elements is a basis of given subspace.

For #6, we can find 1,(x-7/2)^2,x(x-7/2)^2,x^2(x-7/2)^2, and the remaining part is analogous to #5.

For #7, we can find 1,(x-2)(x-5)(x-6),x(x-2)(x-5)(x-6), and since the subspace defined in #7 is a subspace of that one defined in #6, so its dimension should be less than 4, so since we have 3 linearly independent elements, it's done.

## sam

7 Jan 2021I don't understand why the subspace must be less than 5, can you show me why the p''(6) = 0 property of the subspace makes this true? By equation 2.38 a subspace U of a vector space V has dimension s.t. dim(U) <= dim(V).

## Karam

8 May 2021I think an easier solution to problem #4 would be as follows:

To prove linear independence, we can write each of the polynomials (t-6),(t^2-6t),(t^3-6t^2),(t^4-6t^3) with respect to the standard basis of P4 (i.e. as column vectors in R5). This will give you a 5-by-4 matrix and thereby using Gaussian elimination will ascertain linear independence.

Because the length of any linearly independent list is at most the length of any spanning list, we have dim(U) is at least 4. And because U is a subspace of P4(F), we have dim(U) is at most 5. Thus 4<=dim(U)<=5. Clearly dim(U) cannot equal 5 since U is a subspace of P4(F); that is, if dim(U)=dim(P4(F))=5, then U=P4(F). However, U does not equal P4(F). Thus dim(U)=4, which means that we have a linearly independent list in U with length dim(U) and hence this list must be a basis of U. Therefore, the list (t-6),(t^2-6t),(t^3-6t^2),(t^4-6t^3) is a basis of U.

## Lex

7 Jun 2020#10 is has a much simpler solution.

Proof:

Suppose p0,p1,..,pm is linearly dependent. Then by linear dependence lemma, some pj = c0p0 + ... + c_{j-1}p_{j-1}. But the left side is degree j and the right side is degree j-1, contradiction. Therefore p0,p1,...,pm is linearly independent, and has the right length to be a basis for Pm(F), therefore it is a basis for Pm(F).

## Linearity

9 Jun 2020Yes, nice solution.

## Subhasish Mukherjee

21 May 2020Is there an intuitive reason to think 17 won't work?

## Benjamin Favre

29 Apr 2020Hi,

In exercise #4 shouldn't we choose the vectors of the basis in P4(F) since U is defined by U = {p belongs to P4(F) : p(6) = 0} ? The basis given in the solution has vectors not in P4(F), so it can not be a basis of U. Or am I missing something ?

A basis would be : x^3-(x^4)/6, x^2-(x^4)/(6^2), x-(x^4)/(6^3) and 1-(x^4)/(6^4). All those vectors belong to P4(F) and are 0 for x=6, then they belong to U.

Hope someone responds, I'am trying to learn linear algebra and needs to understand if I'm doing a mistake !

Thanks

## Linearity

29 Apr 2020Vectors in a basis do not have to be of degree 4.

## Johnson

12 Apr 2020#8 should be a/5 + c/3 + e = 0 right?

## Linearity

29 Apr 2020Yes, thanks!

## Ruwimal Pathiraja

2 Apr 2020I think there is an easier solution for problem 10 that uses previous results of the chapter. Please correct me if I am wrong.

We first show that the list p_0, p_1, ..., p_m is linearly independent. Indeed if there exist scalars a0, a1,...,am in F such that a0*p_0 + a1*p_1 + .... + am*p_m = 0, then since the degree of p_j is j for all j, equating degrees of the LHS and RHS yields that all ai must be 0.

Now since (1, z, z^2, ...., z^m) is trivially a basis for Pm(F), dim(Pm(F)) = m +1. Then since the length of p_0,...,p_m is also m + 1, it must be a basis for Pm(F), by 2.39.

## Random Name

28 Feb 2020It seems to me that problem 16 has an easier solution.

U_1\oplus \cdots \oplus U_m is finite dimensional by 14.

Let u_1 be a basis of U_1,..., u_m be a basis of U_m.

So if some u\inU_1\oplus \cdots \oplus U_m, then u=(some linear comb. of u_1)+\cdots + (some linear comb. of u_m), so u_1,\cdots,u_m spans U_1\oplus \cdots \oplus U_m. Since we are dealing with a direct sum, u_1,\cdots,u_m is also linearly independent, so it is a basis of U_1\oplus \cdots \oplus U_m.

Clearly, dimension of a basis is a number of it's elements, so \dim U_1\oplus \cdots \oplus U_m= \dim U_1+\cdots+\dim U_m.

## Siyu Wang

4 Feb 2020#6/7/8 Those constructions are very elegant.

## Henrik Young

28 Jan 2020#17

I don't quite get the idea how we get the counterexample to prove it is wrong, can anyone give me an advice?Any information is appreciated.

## Shubham Aich

10 Sep 2019can someone explain those "why?"s in question 5,6,7,8 or at least direct me to some resources pls?

## Kalle

5 Sep 2019Regarding exercise 9. I have a feeling that whenever dim = m-1, w = -v_i for an i = 1, ..., m. However, fail to prove it. Any advice is appreciated!

## Linearity

9 Sep 2019This is wrong. If the dimension is $m-1$, then $v_1+w$, $\cdots$, $v_m+w$ is linearly dependent. Hence there exist not all zero numbers $a_1,\dots,a_m$ such that $$a_1(v_1+w)+\cdots+a_m(v_m+w)=0.$$Hence $$a_1v_1+\cdots+a_mv_m+(a_1+\cdots+a_m)w=0.$$Clearly, we must have $a_1+\cdots+a_m\ne 0$. Otherwise, by the linear independence of $v_1,\dots,v_m$, we have $a_1=\cdots=a_m$. Therefore, we have$$w=-\frac{a_1v_1+\cdots+a_mv_m}{a_1+\cdots+a_m}.$$In other words, the dimension is equal to $m-1$ if and only if there exist $k_1+\cdots+k_m=-1$ such that $$w=k_1v_1+\cdots+k_mv_m.$$

## Brian Lee

26 Dec 2019Yes but the only if direction doesn't hold true. If $$w \in \mathrm{span}(v_1+w,...,v_m+w)$$ then for any $v \in \mathrm{span}(v_1+w,...,v_m+w)$ we have that $$v=a_1(v_1+w)+...+a_m(v_m+w)$$ which gives $$v=a_1v_1+...+a_mv_m+(a_1+...+a_m)w$$ $$=a_1v_1+...+a_mv_m+(a_1+...+a_m)(b_1v_1+...+b_mv_m)$$ $$ \in \mathrm{span}(v_1,...,v_m)$$ Since $v_1,...,v_m$ is linearly independent and spans $V$ we thus get that

$\dim{\mathrm{span}(v_1+w,...,v_m+w)} = m$ not $m-1$. So actually the span can never have dimension $m-1$.

## Linearity

26 Dec 2019Unfortunately, your argument is wrong. See the following counterexample. Take $m=2$ and let $w=-(v_1+v_2)/2$, then

$$\mathrm{span}(v_1+w,v_2+w)=\mathrm{span}(v_1-v_2),$$which is one-dimensional.

## Dwer

10 Apr 2020It seems that "w∈span(v1+w,…,vm+w)" in your argument doesn't hold in the first place. We'd only know that w∈V, but V is neither equal to span(v1+w,…,vm+w) nor span(v1,…,vm)

## Alex P

29 Jul 2019I think number 9 is wrong.

It is not a proof that dim>=m-1. It just shows an example that dim can be equal m-1.

There is a solution to this problem here: https://math.berkeley.edu/~olya/110_review_solutions.pdf

## Linearity

29 Jul 2019The proof is correct. I showed this span has a linearly independent of length $m-1$. (By 2.33) Every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space. Therefore, a basis has length no shorter than $m-1$

## Alex P

29 Jul 2019Oh yes, I get it now. Thanks for clarification.

## Diego Ramos

15 Jul 2017#14

Hi. Your solution of excercise #14 is using dimension of V, but the excercise do not mention nothing about dimV.

## Mohammad Rashidi

28 Jul 2017Yes, you are correct. Thank you.

## MetaMusic

18 Jun 2017You don't need to use induction for 16; more insight ensues if you use another method

## Max Zarazov

10 May 2017#10

no two polynomials are of the same degree => linearly independent

there are m+1 of those => size is the same as dimension of P_m

linearly independent + right size => basis

#14

induction proof is also simple

#17

it's only true iff the following holds:

(U1 + U2) ∩ U3 = (U1 ∩ U2) + (U1 ∩ U3)

which is true iff U1 ⊆ U3 and U2 ⊆ U3

## ZC

12 Jul 2019it is aslo true if $U_1 = \{ 0 \}$ or $U_2 = \{ 0 \}$.

## Eric

15 Dec 2016The solution to 15 seems incorrect. We are supposed to prove that ONE-DIMENSIONAL subspaces exist, and you don't seem to mention the dimension. I think you are supposed to apply 2.34 recursively: start with some one-dimensional subspace of V called U1. We know from 2.34 that there is some corresponding subspace to U1, let's call it W1, such that U1 ++ W1 = V (where ++ is direct sum).

Then, we replace V with W1 and do the same thing. Pick a one-dimensional subspace of W1 called U2 and apply 2.34 to get some corresponding subspace of W1 and call it W2. U2 ++ W2 = W1. Now we have U1 ++ U2 ++ W2 = U.

Repeat this process until you have Un-1 ++ Wn-1. Wn-1 will be our Un.

## Mohammad Rashidi

14 Jan 2017It is correct since U_j is obviously one-dimensional.

## Nuno Alvares Pereira

6 Sep 2016The solution to 16 is incomplete because the text is wrong. It's an equality, not an inequality.