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## Solution to Linear Algebra Hoffman & Kunze Chapter 3.3

#### Exercise 3.3.1

Let $V$ be the set of complex numbers and let $F$ be the field of real numbers. With the usual operations, $V$ is a vector space over $F$. Describe explicitly an isomorphism of this space onto $\mathbb R^2.$

Solution: The natural isomorphism from $V$ to $\mathbb R^2$ is given by $a+bi\mapsto(a,b)$. Since $i$ acts like a placeholder for addition in $\mathbb C$, $$(a+bi)+(c+di)=(a+c)+(b+d)i\mapsto(a+c,b+d)=(a,b)+(c,d).$$ And $$c(a+bi)=ca+cbi\mapsto(ca,cb)=c(a,b).$$ Thus this is a linear transformation. The inverse is clearly $(a,b)\mapsto a+bi$. Thus the two spaces are isomorphic as vector spaces over $\mathbb R$.

#### Exercise 3.3.2

Let $V$ be a vector space over the field of complex numbers, and suppose there is an isomorphism $T$ of $V$ into $\Bbb C^3$. Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be vectors in $V$ such that
$$T\alpha_1=(1,0,i),\quad T\alpha_2=(-2,1+i,0),$$ $$T\alpha_3=(-1,1,1),\quad T\alpha_4=(\sqrt{2},i,3).$$(a) Is $\alpha_1$ in the subspace spanned by $\alpha_2$ and $\alpha_3$?
(b) Let $W_1$ be the subspace spanned by $\alpha_1$ and $\alpha_2$, and let $W_2$ be the subspace spanned by $\alpha_3$ and $\alpha_4$. What is the intersection of $W_1$ and $W_2$?
(c) Find a basis for the subspace of $V$ spanned by the four vectors $\alpha_j$.

Solution:

(a) Since $T$ is an isomorphism, it suffices to determine whether $T\alpha_1$ is contained in the subspace spanned by $T\alpha_2$ and $T\alpha_3$. In other words we need to determine if there is a solution to
$$\left[\begin{array}{cc}-2&-1\\1+i&1\\0&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}1\\0\\i\end{array}\right].$$
To do this we row-reduce the augmented matrix
$$\left[\begin{array}{cc|c}-2&-1&1\\1+i&1&0\\0&1&i\end{array}\right] \rightarrow\left[\begin{array}{cc|c}1&1/2&-1/2\\1+i&1&0\\0&1&i\end{array}\right] \rightarrow\left[\begin{array}{cc|c}1&1/2&-1/2\\0&1&i\\1+i&1&0\end{array}\right]$$$$\rightarrow\left[\begin{array}{cc|c}1&1/2&-1/2\\0&1&i\\0&\frac{1-i}{2}&\frac{1+i}{2}\end{array}\right] \rightarrow\left[\begin{array}{cc|c}1&0&\frac{-1-i}{2}\\0&1&i\\0&0&0\end{array}\right]$$The zero row on the left of the dividing line has zero also on the right. This means the system has a solution. Therefore we can conclude that $\alpha_1$ is in the subspace generated by $\alpha_2$ and $\alpha_3$.

(b) Since $T\alpha_1$ and $T\alpha_2$ are linearly independent, and $T\alpha_3$ and $T\alpha_4$ are linearly independent, $\dim(W_1)=\dim(W_2)=2$. We row-reduce the matrix whose columns are the $T\alpha_i$:
$$\left[\begin{array}{cccc}1&-2&-1&\sqrt{2}\\0&1+i&1&i\\i&0&1&3\end{array}\right]$$which yields
$$\left[\begin{array}{cccc}1&0&-i&0\\0&1&\frac{1-i}{2}&0\\0&0&0&1\end{array}\right],$$from which we deduce that $T\alpha_1,T\alpha_2,T\alpha_3,T\alpha_4$ generate a space of dimension three, thus $\dim(W_1+W_2)=3$. Since $\dim(W_1)=\dim(W_2)=2$ it follows from Theorem 6, page 46 that $\dim(W_1\cap W_2)=1$.

Now $AX=0$ $\Leftrightarrow$ $RX=0$ where $R$ is the row reduced echelon form of $A$. This follows from the fact that $R=PA$; multiply both sides of $AX=0$ on the left by $P$. Solving for $X$ in $RX=0$ gives the general solution is of the form $(ic,\frac{i-1}{2}c,c,0)$. Letting $c=2$ gives
$$2iT\alpha_1+(i-1)T\alpha_2+2T\alpha_3=0$$which implies
$$T\alpha_3=-iT\alpha_1+\frac{1-i}{2}T\alpha_2$$ which implies $T\alpha_3\in TW_1$. Thus $\alpha_3\in W_1$. Thus $\alpha_3\in W_1\cap W_2$. Since $\dim(W_1\cap W_2)=1$ it follows that $W_1\cap W_2=\mathbb C\alpha_3.$

(c) We have determined in part (b) that the $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ span a space of dimension three, and that $\alpha_3$ is in the space generated by $\alpha_1$ and $\alpha_2$. Thus $\{\alpha_1,\alpha_2,\alpha_4\}$ give a basis for the subspace spanned by $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$, which in fact is all of $\mathbb C^3$.

#### Exercise 3.3.3

Let $W$ be the set of all $2\times2$ complex Hermitian matrices, that is, the set of $2\times2$ complex matrices $A$ such that $A_{ij}=\overline{A_{ji}}$ (the bar denoting complex conjugation). As we pointed out in Example 6 of Chapter 2, $W$ is a vector space over the field of real numbers, under the usual operations. Verify that $$(x,y,z,t)\rightarrow\left[\begin{array}{cc}t+x&y+iz\\y-iz&t-x\end{array}\right]$$ is an isomorphism of $\mathbb R^4$ onto $W$.

Solution: The function is linear since the four components are all linear combinations of the components of the domain $(x,y,z,t)$. Identify $\mathbb C^{2\times2}$ with $\mathbb C^4$ by $A\mapsto(A_{11},A_{12},A_{21},A_{22})$. Then the matrix of the transformation is given by
$$\left[\begin{array}{cccc}1&0&0&1\\0&1&i&0\\0&1&-i&0\\-1&0&0&1\end{array}\right].$$ As usual, the transformation is an isomorphism if the matrix is invertible. We row-reduce to veryify the matrix is invertible. We will row-reduce the augmented matrix in order to find the inverse explicitly:
$$\left[\begin{array}{cccc|cccc}1&0&0&1&1&0&0&0\\0&1&i&0&0&1&0&0\\0&1&-i&0&0&0&1&0\\-1&0&0&1&0&0&0&1\end{array}\right].$$This reduces to
$$\rightarrow\left[\begin{array}{cccc|cccc}1&0&0&0&1/2&0&0&-1/2\\0&1&0&0&0&1/2&1/2&0\\0&0&1&0&0&-i/2&i/2&0\\0&0&0&1&1/2&0&0&1/2\end{array}\right].$$Thus the inverse transformation is
$$\left[\begin{array}{cc}x&y\\z&w\end{array}\right]\mapsto\left(\frac{x-w}{2},\ \ \frac{y+z}2,\ \ \frac{i(z-y)}2,\ \ \frac{x+w}2\right).$$

#### Exercise 3.3.4

Show that $F^{m\times n}$ is isomorphic to $F^{mn}$.

Solution: Define the bijection $\sigma$ from $\{(a,b)\mid a,b\in\mathbb N, 1\leq a\leq m, 1\leq b\leq n\}$ to $\{1,2,\dots,mn\}$ by $(a,b)\mapsto (a-1)n+b$. Define the function $G$ from $F^{m\times n}$ to $F^{mn}$ as follows. Let $A\in F^{m\times n}$. Then map $A$ to the $mn$-tuple that has $A_{ij}$ in the $\sigma(i,j)$ position. In other words $$A\mapsto(A_{11},A_{12},A_{13},\dots,A_{1n},A_{21},A_{22},A_{23},\dots,A_{2n},\dots \dots, A_{nn}).$$ Since addition in $F^{m\times n}$ and in $F^{mn}$ is performed compenent-wise, $G(A+B)=G(A)+G(B)$. Similarly since scalar multiplication factors out of vectors component-wise in the same way in $F^{m\times n}$ as in $F^{mn}$, we also have $G(cA)=cG(A)$. Thus $G$ is a linear function. $G$ is clearly one-to-one (as well as clearly onto), and both $F^{m\times n}$ and $F^{mn}$ have dimension $mn$ (by Example 17, page 45 and Exercise 2.3.12, page 49), thus (by Theorem 9, page 81) it follows that $G$ has an inverse and therefore is an isomorphism.

#### Exercise 3.3.5

Let $V$ be the set of complex numbers regarded as a vector space over the field of real numbers (Exercise 1). We define a function $T$ from $V$ into the space of $2\times2$ real matrices, as follows. If $z=x+iy$ with $x$ and $y$ real numbers, then
$$T(z)=\left[\begin{array}{cc}x+7y&5y\\-10y&x-7y\end{array}\right].$$(a) Verify that $T$ is a one-one (real) linear transformation of $V$ into the space of $2\times2$ matrices.
(b) Verify that $T(z_1z_2)=T(z_1)T(z_2)$.
(c) How would you describe the range of $T$?

Solution:

(a) The four coordinates of $T(z)$ are written as linear combinations of the coordinates of $z$ (as a vector over $\mathbb R$). Thus $T$ is clearly a linear transformation. To see that $T$ is one-to-one, let $z=x+yi$ and $w=a+bi$ and suppose $T(z)=T(w)$. Then considering the top right entry of the matrix we see that $5y=5b$ which implies $b=y$. It now follows from the top left entry of the matrix that $x=a$. Thus $T(z)=T(w)$ $\Rightarrow$ $z=w$, thus $T$ is one-to-one.

(b) Let $z_1=x+yi$ and $z_2=a+bi$. Then
$$T(z_1z_2)=T((ax-by)+(ay+bx)i)=\left[\begin{array}{cc}(ax-by)+7(ay+bx)&5(ay+bx)\\-10(ay+bx)&(ax-by)-7(ay+bx)\end{array}\right].$$On the other hand,
$$T(z_1)T(z_2)=\left[\begin{array}{cc}x+7y&5y\\-10y&x-7y\end{array}\right]\left[\begin{array}{cc}a+7b&5b\\-10b&a-7b\end{array}\right]$$$$=\left[\begin{array}{cc}(ax-by)+7(ay+bx)&5(ay+bx)\\-10(ay+bx)&(ax-by)-7(ay+bx)\end{array}\right].$$Thus $T(z_1z_2)=T(z_1)T(z_2)$.

(c) The range of $T$ has (real) dimension equal to two by part (a), and so the range of $T$ is isomorphic to $\mathbb C$ as real vector spaces. But both spaces also have a natural multiplication and in part (b) we showed that $T$ respects the multiplication. Thus the range of $T$ is isomorphic to $\mathbb C$ as fields and we have essentially found an isomorphic copy of the field $\mathbb C$ in the algebra of $2\times 2$ real matrices.

#### Exercise 3.3.6

Let $V$ and $W$ be finite-dimensional vector spaces over the field $F$. Prove that $V$ and $W$ are isomorphic if and only if $\dim(V)=\dim(W)$.

Solution: Suppose $\dim(V)=\dim(W)=n$. By Theorem 10, page 84, both $V$ and $W$ are isomorphic to $F^n$, and consequently, since isomorphism is an equivalence relation, $V$ and $W$ are isomorphic to each other. Conversely, suppose $T$ is an isomorphism from $V$ to $W$. Suppose $\dim(W)=n$. Then by Theorem 10 again, there is an isomorphism $S:W\rightarrow F^n$. Thus $ST$ is an isomorphism from $V$ to $F^n$ implying also $\dim(V)=n$.

#### Exercise 3.3.7

Let $V$ and $W$ be vector spaces over the field $F$ and let $U$ be an isomorphism of $V$ onto $W$. Prove that $T\rightarrow UTU^{-1}$ is an isomorphism of $L(V,V)$ onto $L(W,W)$.

Solution: $L(V,V)$ is defined on page 75 as the vector space of linear transformations from $V$ to $V$, and likewise $L(W,W)$ is the vector space of linear transformations from $W$ to $W$.

Call the function $f$. We know $f(T)$ is linear since it is a composition of three linear tranformations $UTU^{-1}$. Thus indeed $f$ is a function from $L(V,V)$ to $L(W,W)$. Now $$f(aT+T')=U(aT+T')U^{-1}=(aUT+UT')U^{-1}=aUTU^{-1}+UT'U^{-1}=af(T)+f(T').$$ Thus $f$ is linear. We just must show $f$ has an inverse. Let $g$ be the function from $L(W,W)$ to $L(V,V)$ given by $g(T)=U^{-1}TU$. Then $gf(T)=U^{-1}(UTU^{-1})U=T$. Similarly $fg=I$. Thus $f$ and $g$ are inverses. Thus $f$ is an isomorphism.