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Solution to Linear Algebra Hoffman & Kunze Chapter 6.6

Exercise 6.6.1

Let $V$ be a finite-dimensional vector space and let $W_1$ be any subspace of $V$. Prove that there is a subspace $W_2$ of $V$ such that $V=W_1\oplus W_2$.

Solution: Since $V$ is finite-dimensional, so is $W_1$. Let $w_1,\dots,w_n$ be a basis of $W$. Then by Theorem 5 of Page 45, we can extend it to a basis $w_1,\dots,w_n$, $v_1,\dots,v_m$ of $V$.

Let $W_2=\mathrm{span}(v_1,\dots,v_m)$. Then it is clear that $V=W_1\oplus W_2$.

Exercise 6.6.2

Let $V$ be a finite-dimensional vector space and let $W_1,\dots,W_k$ be subspaces of $V$ such that
$$V=W_1+\cdots+W_k\quad\text{and}\quad\dim(V)=\dim(W_1)+\cdots+\dim(W_k).$$Prove that $V=W_1\oplus\cdots\oplus W_k.$

Solution: See Exercise 2C.16 of Linear Algebra Done Right.

Exercise 6.6.3

Find a projection $E$ which projects $\mathbb R^2$ onto the subspace spanned by $(1,-1)$ along the subspace spanned by $(1,2)$.

Solution: Consider the linear map $E:\R^2\to \R^2$ such that\[E(1,-1)=(1,-1),\quad E(1,2)=(0,0).\]In terms of the standard basis, we have\[E(1,0)=\frac{1}{3}(2,-2),\quad E(0,1)=\frac{1}{3}(-1,1).\]

Exercise 6.6.4

If $E_1$ and $E_2$ are projections onto independent subspaces, then $E_1+E_2$ is a projection. True or false?

Solution: False. Let $E_1$ be the projection in Exercise 6.6.3 and $E_2$ be the projection onto the subspace spanned by $(1,0)$ along the subspace spanned by $(0,1)$.

Then $$(E_1+E_2)(1,-1)=(1,-1)+(1,0)=(2,-1)$$and \begin{align*}(E_1+E_2)(2,-1)&=E_1(2,-1)+E_2(2,-1)\\&=\frac{1}{3}(5,-5)+(2,0)\ne (2,-1).\end{align*}Hence $(E_1+E_2)^2\ne E_1+E_2$, so $E_1+E_2$ is not a projection.

Exercise 6.6.5

If $E$ is a projection and $f$ is a polynomial, then $f(E)=aI+bE$. What are $a$ and $b$ in terms of the coefficents of $f$?

Solution: If $E$ is a projection, then $E^2=E$. Moreover, we have $$E^{k}==E$$for all $k>00$. Hence $a$ is the constant term of $f$ while $b$ is the sum of all other coefficients of $x^{k+1}$, for all $k\geq 0$.

Exercise 6.6.6

True or false? If a diagonalizable operator has only the characteristic values $0$ and $1$, it is a projection.

Solution: True. Since $T$ is diagonalizable and has only the characteristic values $0$ and $1$. By Theorem 6 of page 204, the minimal polynomial of $T$ is \[p=x(x-1).\]Namely, $$T(T-I)=0\Rightarrow T^2=T.$$Hence $T$ is a projection.

Exercise 6.6.7

Prove that if $E$ is the projection on $R$ along $N$, then $(I-E)$ is the projection on $N$ along $R$.

Solution: We have $V=R\oplus N$. By definition, we have\[Ev_R=v_R,\quad Ev_N=0\]for all $v_R\in R$ and $v_N\in N$. Therefore $$(I-E)v_R=v_R-v_R=0,$$$$(I-E)v_N=v_N-0=v_N.$$Moreover,$$(I-E)^2=I-2E+E^2=I-E.$$Hence $I-E$ is the projection on $N$ along $R$.

Exercise 6.6.8

Let $E_1,\dots, E_k$ be linear operators on the space $V$ such that $E_1+\cdots+E_k=I$.

(a) Prove that if $E_iE_j=0$ for $i\not=j$, then $E_i^2=E_i$ for each $i$.

(b) In the case $k=2$, prove the converse of (a). That is, if $E_1+E_2=I$ and $E_1^2=E_1$, $E_2^2=E_2$, then $E_1E_2=0$.

Solution: For part (a), we have\begin{align*}E_i^2&=E_i(I-E_1-\cdots-E_{i-1}-E_{i+1}-\cdots-E_{n})\\&=E_i-\sum_{j\ne i}E_{i}E_{j}=E_i,\end{align*}since $E_iE_j=0$ for $i\ne j$.

(b) We have $$E_1=E_1(E_1+E_2)=E_1^2+E_1E_2.$$Since $E_1=E_1^2$, we have $E_1E_2=0$. Similarly, one can show that $E_2E_1=0$.

Exercise 6.6.9

Let $V$ be a real vector space and $E$ an idempotent linear operator on $V$, i.e., a projection. Prove that $(I+E)$ is invertible. Find $(I+E)^{-1}$.

Solution: Since $E$ is idempotent, we have\[E^2-E-2I=-2I,\]i.e.\[(I+E)(2I-E)=2I.\]Hence $(I+E)^{-1}$ is equal to $\frac{1}{2}(2I-E)$.

Exercise 6.6.10

Let $F$ be a subfield of the complex numbers (or, a field of characteristic zero). Let $V$ be a finite-dimensional vector space over $F$. Suppose that $E_1,\dots,E_k$ are projections of $V$ and that $E_1+\cdots E_k=I$. Prove that $E_iE_j=0$ for $i\not=j$ (Hint: use the trace function and ask yourself what the trace of a porjection is.)

Solution: Let $W_i$ be the image $W_i$ of the projection $E_i$

Since $E_i$ is a projection, its minimal polynomial is a monic divisor of $x(x-1)$. Since $x(x-1)$ has no multiple roots, by Theorem 6 of page 204, $E_i$ is diagonalizable. Moreover, the eigenvalues are zero or one. Therefore the trace $\mathrm{tr}(E_i)$ is exactly the dimension of $W_i$.

Clearly, we have $W_1+W_2+\cdots+W_k\subset V$. But $E_1+\cdots+E_k=I$, for any $v\in V$, we have\[v=E_1v+\cdots+E_kv\in W_1+W_2+\cdots+W_k.\]Hence $W_1+W_2+\cdots+W_k=V$. Moreover, taking the trace on $E_1+\cdots+E_k=I$, we get that (trace function is linear)\[\dim W_1+\cdots+\dim W_k=\dim V.\]Therefore, we have $V=W_1\oplus\cdots\oplus W_k$ by Exercise 6.6.2.

Taking $w_i\in W_i$, we have\[w_i=Iw_i=\sum_{j=1}^k E_jw_i=w_i+\sum_{j\ne i}^kE_jw_i.\]Since $V=W_1\oplus\cdots\oplus W_k$, we have $E_jw_i=0$ for $j\ne i$. Hence $E_jW_i=0$ for all $j\ne i$. Therefore $$E_jE_iv\in E_j W_i=0$$ for all $v\in V$ and $j\ne i$. Namely, $E_jE_i=0$ for $j\ne i$.

Exercise 6.6.11

Let $V$ be a vector space, let $W_1,\dots,W_k$ be subspace of $V$, and let
$$V_j=W_1+\cdots W_{j-1}+W_{j+1}+\cdots+W_k.$$Suppose that $V=W_1\oplus\cdots\oplus W_k$. Prove that the dual space $V^*$ has the direct-sum decomposition $V^*=V_1^0\oplus\cdots\oplus V_k^0$.

Solution: We use the transpose from Chapter 3.7. We also use the notation from Theorem 9 of page 212.

By Theorem 9 of page 212, we have projections $E_i$ such that $E_iE_j=0$ for $i\ne j$ and $I=E_1+\dots+E_k$. Define the map $E_i^t:V^*\to V^*$ by\[(E^t_if)(v)=f(E_iv),\]for all $f\in V^*$ and $v\in V$. If $i\ne j$, we have\begin{align*}(E_i^tE_j^tf)v=(E_j^tf)(E_iv)=f(E_jE_iv)=f(0)=0.\end{align*}Hence $E_j^tE_i^t=0$ for $j\ne i$. Similarly, one shows that $E_i^t$ is an idempotent.

We also have\begin{align*}((E_1^t+\cdots+E_k^t)f)v&=f(E_1v)+\cdots+f(E_kv)\\&=f(E_1v+\cdots+E_kv)=f((E_1+\cdots+E_k)v)\\&=f(Iv)=f(v).\end{align*}Hence $E_1^t+\cdots+E_k^t=I$.

Let $S_i$ be the image of $E_i^t$, then by Theorem 9 of page 212, we have\[V^*=S_1\oplus \cdots\oplus S_k.\]Therefore, it suffices to show that $S_i=V_i^0$.

Let $v_i\in V_i$, then $E_iv_i=0$ (page 211). Hence for all $f\in V^*$, we have\[(E_i^tf)v_i=f(E_iv_i)=f(0)=0.\]Hence $E_i^tf\in V_i^0$. Namely $S_i\subset V_i^0$.

Conversely, for $v\in V$, we write $v=w_i+v_i$ where $w_i\in W_i$ and $v_i\in V_i$. Moreover, $E_iv=w_i$. If $f\in V_i^0$, then $$f(v)=f(w_i+v_i)=f(w_i),$$$$(E_i^tf)v=f(E_iv)=f(w_i).$$Therefore, we have $E_i^tf=f$ for $f\in V_i^0$. This implies that $V_i^0\subset S_i$.

We conclude that $S_i=V_i^0$ and we are done.


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