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## Solution to Linear Algebra Hoffman & Kunze Chapter 3.2

#### Exercise 3.2.1

(a) Geometrically, in the $x-y$ plane, $T$ is the reflection about the diagonal $x=y$ and $U$ is a projection onto the $x$-axis.

(b) We have

• $(U+T)(x_1,x_2)=(x_2,x_1)+(x_1,0)=(x_1+x_2,x_1)$.
• $(UT)(x_1,x_2)=U(x_2,x_1)=(x_2,0)$.
• $(TU)(x_1,x_2)=T(x_1,0)=(0,x_1)$.
• $T^2(x_1,x_2)=T(x_2,x_1)=(x_1,x_2)$, the identity function.
• $U^2(x_1,x_2)=U(x_1,0)=(x_1,0)$. So $U^2=U$.

#### Exercise 3.2.2

By Theorem 9 part (v), top of page 82, $T$ is invertible if $\{T\epsilon_1,T\epsilon_2,T\epsilon_3\}$ is a basis of $\mathbb C^3$. Since $\mathbb C^3$ has dimension three, it suffices (by Corollary 1 page 46) to show $T\epsilon_1,T\epsilon_2,T\epsilon_3$ are linearly independent. To do this we row reduce the matrix
$$\left[\begin{array}{ccc}1&0&i\\0&1&1\\i&1&0\end{array}\right]$$to row-reduced echelon form. If it reduces to the identity then its rows are independent, otherwise they are dependent. Row reduction follows:
$$\left[\begin{array}{ccc}1&0&i\\0&1&1\\i&1&0\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&0&i\\0&1&1\\0&1&1\end{array}\right] \rightarrow\left[\begin{array}{ccc}1&0&i\\0&1&1\\0&0&0\end{array}\right]$$This is in row-reduced echelon form not equal to the identity. Thus $T$ is not invertible.

#### Exercise 3.2.3

The matrix representation of the transformation is
$$\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] \mapsto \left[\begin{array}{ccc}3&0&0\\1&-1&0\\2&1&1\end{array}\right]\cdot\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]$$where we’ve identified $\mathbb R^3$ with $\mathbb R^{3\times1}$. $T$ is invertible if the matrix of the transformation is invertible. To determine this we row-reduce the matrix – we row-reduce the augmented matrix to determine the inverse for the second part of the Exercise.
$$\left[\begin{array}{ccc|ccc}3&0&0 & 1&0&0 \\ 1&-1&0 & 0&1&0 \\ 2&1&1 & 0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&0 & 0&1&0 \\ 3&0&0 & 1&0&0 \\ 2&1&1 & 0&0&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&0 & 0&1&0 \\ 0&3&0 & 1&-3&0 \\ 0&3&1 & 0&-2&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&-1&0 & 0&1&0 \\ 0&1&0 & 1/3&-1&0 \\ 0&3&1 & 0&-2&1\end{array}\right]$$$$\rightarrow\left[\begin{array}{ccc|ccc}1&0&0 & 1/3&0&0 \\ 0&1&0 & 1/3&-1&0 \\ 0&0&1 & -1&1&1\end{array}\right]$$Since the left side transformed into the identity, $T$ is invertible. The inverse transformation is given by
$$\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] \mapsto \left[\begin{array}{ccc}1/3&0&0\\1/3&-1&0\\-1&1&1\end{array}\right]\cdot\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]$$So$$T^{-1}(x_1,x_2,x_3)=(x_1/3,\ \ x_1/3-x_2,\ \ -x_1+x_2+x_3).$$

#### Exercise 3.2.4

Working with the matrix representation of $T$ we must show
$$(A^2-I)(A-3I)=0$$where
$$A=\left[\begin{array}{ccc}3&0&0\\1&-1&0\\2&1&1\end{array}\right].$$Calculating:
$$A^2=\left[\begin{array}{ccc}3&0&0\\1&-1&0\\2&1&1\end{array}\right]\left[\begin{array}{ccc}3&0&0\\1&-1&0\\2&1&1\end{array}\right]$$$$=\left[\begin{array}{ccc}9&0&0\\2&1&0\\9&0&1\end{array}\right]$$Thus
$$A^2-I=\left[\begin{array}{ccc}8&0&0\\2&0&0\\9&0&0\end{array}\right].$$Also
$$A-3I=\left[\begin{array}{ccc}0&0&0\\1&-4&0\\2&1&-2\end{array}\right]$$Thus
$$(A^2-I)(A-3I)=\left[\begin{array}{ccc}8&0&0\\2&0&0\\9&0&0\end{array}\right]\cdot\left[\begin{array}{ccc}0&0&0\\1&-4&0\\2&1&-2\end{array}\right]$$$$=\left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right].$$

#### Exercise 3.2.5

An (ordered) basis for $\mathbb C^{2\times2}$ is given by
$$A_{11}=\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\quad A_{21}=\left[\begin{array}{cc}0&0\\1&0\end{array}\right]$$$$A_{12}=\left[\begin{array}{cc}0&1\\0&0\end{array}\right],\quad A_{22}=\left[\begin{array}{cc}0&0\\0&1\end{array}\right].$$If we identify $\mathbb C^{2\times 2}$ with $\mathbb C^4$ by
$$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\mapsto (a,b,c,d)$$then since
$$A_{11}\mapsto A_{11}-4A_{21}$$$$A_{21}\mapsto -A_{11}+4A_{21}$$$$A_{12}\mapsto A_{12}-4A_{22}$$$$A_{22}\mapsto -A_{12}+4A_{22}$$the matrix of the transformation is given by
$$\left[\begin{array}{cccc} 1&-4&0&0\\ -1&4&0&0\\ 0&0&1&-4\\ 0&0&-1&4\end{array}\right].$$To find the rank of $T$ we row-reduce this matrix:
$$\rightarrow\left[\begin{array}{cccc} 1&-4&0&0\\ 0&0&1&-4\\ 0&0&0&0\\ 0&0&0&0\end{array}\right].$$It has rank two so the rank, so the rank of $T$ is $2$.

Note that $T^2(A)=T(T(A))=T(BA)=B(BA)=B^2A$. Thus $T^2$ is given by multiplication by a matrix just as $T$ is, but multiplication with $B^2$ instead of $B$. Explicitly
$$B^2=\left[\begin{array}{cc}1&-1\\-4&4\end{array}\right]\left[\begin{array}{cc}1&-1\\-4&4\end{array}\right]$$$$=\left[\begin{array}{cc}5&-5\\-20&20\end{array}\right].$$

#### Exercise 3.2.6

Let $\{\alpha_1,\alpha_2,\alpha_3\}$ be a basis for $\Bbb R^3$. Then $T(\alpha_1),T(\alpha_2),T(\alpha_3)$ must be linearly dependent in $\Bbb R^2$, because $\Bbb R^2$ has dimension $2$. So suppose $$b_1T(\alpha_1)+b_2T(\alpha_2)+b_3T(\alpha_3)=0$$ and not all $b_1,b_2,b_3$ are zero. Then
$$b_1\alpha_1+b_2\alpha_2+b_3\alpha_3\not=0$$ and
$$UT(b_1\alpha_1+b_2\alpha_2+b_3\alpha_3)$$$$=U(T(b_1\alpha_1+b_2\alpha_2+b_3\alpha_3))$$$$=U(b_1T(\alpha_1)+b_2T(\alpha_2)+b_3T(\alpha_3)$$$$=U(0)=0.$$Thus (by the definition at the bottom of page 79) $UT$ is not non-singular and thus by Theorem 9, page 81, $UT$ is not invertible.

The obvious generalization is that if $n>m$ and $T:\Bbb R^n\rightarrow\Bbb R^m$ and $U:\Bbb R^m\rightarrow\Bbb R^n$ are linear transformations, then $UT$ is not invertible. The proof is an immediate generalization the proof of the special case above, just replace $\alpha_3$ with $\dots,\alpha_n$.

#### Exercise 3.2.7

Identify $\mathbb R^2$ with $\mathbb R^{2\times1}$ and let $T$ and $U$ be given by the matrices
$$A=\left[\begin{array}{cc}1&0\\0&0\end{array}\right],\quad B=\left[\begin{array}{cc}0&1\\0&0\end{array}\right].$$More precisely, for
$$X=\left[\begin{array}{c}x\\y\end{array}\right].$$Let $T$ be given by $X\mapsto AX$ and let $U$ be given by $X\mapsto BX$. Thus $TU$ is given by $X\mapsto ABX$ and $UT$ is given by $X\mapsto BAX$. But $BA=0$ and $AB\not=0$ so we have the desired example.

#### Exercise 3.2.8

If $T^2=0$ then the range of $T$ must be contained in the null space of $T$ since if $y$ is in the range of $T$ then $y=Tx$ for some $x$ so $Ty=T(Tx)=T^2x=0$. Thus $y$ is in the null space of $T$.

To give an example of an operator where $T^2=0$ but $T\not=0$, let $V=\mathbb R^{2\times1}$ and let $T$ be given by the matrix
$$A=\left[\begin{array}{cc}0&1\\0&0\end{array}\right].$$Specifically, for
$$X=\left[\begin{array}{c}x\\y\end{array}\right].$$Let $T$ be given by $X\mapsto AX$. Since $A\not=0$, $T\not=0$. Now $T^2$ is given by $X\mapsto A^2X$, but $A^2=0$. Thus $T^2=0$.

#### Exercise 3.2.9

By the comments in the Appendix on functions, at the bottom of page 389, we see that simply because $TU=I$ as functions, then necessarily $T$ is onto and $U$ is one-to-one. It then follows immediately from Theorem 9, page 81, that $T$ is invertible. Now $TT^{-1}=I=TU$ and multiplying on the left by $T^{-1}$ we get $T^{-1}TT^{-1}=T^{-1}TU$ which implies $(I)T^{-1}=(I)U$ and thus $U=T^{-1}$.

Let $V$ be the space of polynomial functions in one variable over $\Bbb R$. Let $D$ be the differentiation operator and let $T$ be the operator “multiplication by $x$” (exactly as in Example 11, page 80). As shown in Example 11, $UT=I$ while $TU\not=I$. Thus this example fulfills the requirement.

#### Exercise 3.2.10

Let $\mathcal B=\{\alpha_1,\dots,\alpha_n\}$ be a basis for $F^{n\times1}$ and let $\mathcal B’=\{\beta_1,\dots,\beta_m\}$ be a basis for $F^{m\times1}$. We can define a linear transformation from $F^{n\times1}$ to $F^{m\times1}$ uniquely by specifying where each member of $\mathcal B$ goes in $F^{m\times1}$. If $m<n$ then we can define a linear transformation that maps at least one member of $\mathcal B$ to each member of $\mathcal B’$ and maps at least two members of $\mathcal B$ to the same member of $\mathcal B’$. Any linear transformation so defined must necessarily be onto without being one-to-one. Similarly, if $m>n$ then we can map each member of $\mathcal B$ to a unique member of $\mathcal B’$ with at least one member of $\mathcal B’$ not mapped to by any member of $\mathcal B$. Any such transformation so defined will necessarily be one-to-one but not onto.

#### Exercise 3.2.11

Let $\{\alpha_1,\dots,\alpha_n\}$ be a basis for $V$. Then the rank of $T$ is the number of linearly independent vectors in the set $\{T\alpha_1,\dots,T\alpha_n\}$. Suppose the rank of $T$ equals $k$ and suppose WLOG that $\{T\alpha_1,\dots,T\alpha_k\}$ is a linearly independent set (it might be that $k=1$, pardon the notation). Then $\{T\alpha_1,\dots,T\alpha_k\}$ give a basis for the range of $T$. It follows that $\{T^2\alpha_1,\dots,T^2\alpha_k\}$ span the range of $T^2$ and since the dimension of the range of $T^2$ is also equal to $k$, $\{T^2\alpha_1,\dots,T^2\alpha_k\}$ must be a basis for the range of $T^2$. Now suppose $v$ is in the range of $T$. Then $v=c_1T\alpha_1+\cdots+c_kT\alpha_k$. Suppose $v$ is also in the null space of $T$. Then $$0=T(v)=T(c_1T\alpha_1+\cdots+c_kT\alpha_k)=c_1T^2\alpha_1+\cdots+c_kT^2\alpha_k.$$ But $\{T^2\alpha_1,\dots,T^2\alpha_k\}$ is a basis, so $T^2\alpha_1,\dots,T^2\alpha_k$ are linearly independent, thus it must be that $c_1=\cdots=c_k=0$, which implies $v=0$. Thus we have shown that if $v$ is in both the range of $T$ and the null space of $T$ then $v=0$, as required.

#### Exercise 3.2.12

We showed in Exercise 2.3.12, page 49, that the dimension of $V$ is $mn$ and the dimension of $W$ is $pn$. By Theorem 9 page (iv) we know that an invertible linear transformation must take a basis to a basis. Thus if there’s an invertible linear transformation between $V$ and $W$ it must be that both spaces have the same dimension. Thus if $T$ is inverible then $pn=mn$ which implies $p=m$. The matrix $B$ is then invertible because the assignment $B\mapsto BX$ is one-to-one (Theorem 9 (ii), page 81) and non-invertible matrices have non-trivial solutions to $BX=0$ (Theorem 13, page 23). Conversely, if $p=n$ and $B$ is invertible, then we can define the inverse transformation $T^{-1}$ by $T^{-1}(A)=B^{-1}A$ and it follows that $T$ is invertible.