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.

## 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).

## 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.