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Chapter 2 Exercise C


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 $a/5+d/3+e=0$. 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\[\dim U_1\oplus \cdots \oplus U_m=\dim(W+U_m)=\dim W+\dim U_m=\dim U_1+\cdots+\dim U_{m-1}+\dim U_m.\]


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.

Linearity

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This Post Has 14 Comments

  1. can someone explain those “why?”s in question 5,6,7,8 or at least direct me to some resources pls?

  2. Regarding 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!

    1. This 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 $$w=k_1v_1+\cdots+k_mv_m.$$

    1. The 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 $m-1$

      1. Oh yes, I get it now. Thanks for clarification.

  3. #14
    Hi. Your solution of excercise #14 is using dimension of V, but the excercise do not mention nothing about dimV.

    1. Yes, you are correct. Thank you.

  4. You don’t need to use induction for 16; more insight ensues if you use another method

  5. #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

    1. it is aslo true if $U_1 = \{ 0 \}$ or $U_2 = \{ 0 \}$.

  6. The 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.

    1. It is correct since U_j is obviously one-dimensional.

  7. The solution to 16 is incomplete because the text is wrong. It’s an equality, not an inequality.

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