06. 선형변환과 행렬Linear Transformation and Matrix

Solution. (1) Since $\beta, \gamma$ are the standard ordered bases, the coordinate vector of each image is the vector itself.

$$T(1,0) = (2, 3, 1), \quad T(0,1) = (-1, 4, 0)$$

Listing these as columns gives

$$[T]_\beta^\gamma = \begin{pmatrix} 2 & -1 \\ 3 & 4 \\ 1 & 0 \end{pmatrix}$$

(2) Here $\beta = \{(1,2),(2,3)\}$ and $\gamma = \{(1,1,1),(0,1,1),(2,2,3)\}$. First,

$$T(1,2) = (1-2,\, 1,\, 2+2) = (-1,1,4), \quad T(2,3) = (2-3,\, 2,\, 4+3) = (-1,2,7)$$

Writing $(-1,1,4) = a(1,1,1)+b(0,1,1)+c(2,2,3)$ gives $a+2c=-1$, $a+b+2c=1$, $a+b+3c=4$, and solving yields $a=-7$, $b=2$, $c=3$. Thus $T(1,2) = -7(1,1,1)+2(0,1,1)+3(2,2,3)$.

Writing $(-1,2,7) = a(1,1,1)+b(0,1,1)+c(2,2,3)$ gives $a+2c=-1$, $a+b+2c=2$, $a+b+3c=7$, and solving yields $a=-11$, $b=3$, $c=5$.

Hence

$$[T]_\beta^\gamma = \begin{pmatrix} -7 & -11 \\ 2 & 3 \\ 3 & 5 \end{pmatrix}$$

Solution. Order the coordinates by $\alpha = \{E^{11}, E^{12}, E^{21}, E^{22}\}$.

(1) Since $T(A)=A^t$, we have $T(E^{11})=E^{11}$, $T(E^{12})=E^{21}$, $T(E^{21})=E^{12}$, $T(E^{22})=E^{22}$. Their $\alpha$-coordinate vectors are $(1,0,0,0)$, $(0,0,1,0)$, $(0,1,0,0)$, $(0,0,0,1)$, so

$$[T]_\alpha = \begin{pmatrix} 1&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&1 \end{pmatrix}$$

(2) Since $T(A)=\operatorname{tr}(A)$, we get $T(E^{11})=1$, $T(E^{12})=0$, $T(E^{21})=0$, $T(E^{22})=1$, and the coordinate with respect to $\gamma=\{1\}$ is just the value itself. Hence

$$[T]_\alpha^\gamma = \begin{pmatrix} 1 & 0 & 0 & 1 \end{pmatrix}$$

(3) For $\beta=\{1,x,x^2\}$, compute $f'(0)$, $f(1)$, $f''(3)$.

For $f(x)=1$: $f'=0$, $f(1)=1$, $f''=0$, so $T(1) = \begin{pmatrix} 0&2\\0&0\end{pmatrix}$, with $\alpha$-coordinates $(0,2,0,0)$.

For $f(x)=x$: $f'=1$, $f(1)=1$, $f''=0$, so $T(x) = \begin{pmatrix} 1&2\\0&0\end{pmatrix}$, with $\alpha$-coordinates $(1,2,0,0)$.

For $f(x)=x^2$: $f'(0)=0$, $f(1)=1$, $f''(3)=2$, so $T(x^2) = \begin{pmatrix} 0&2\\0&2\end{pmatrix}$, with $\alpha$-coordinates $(0,2,0,2)$.

Listing these coordinate vectors as columns gives

$$[T]_\beta^\alpha = \begin{pmatrix} 0&1&0 \\ 2&2&2 \\ 0&0&0 \\ 0&0&2 \end{pmatrix}$$

Solution. $T(1) = \overline{1} = 1 = 1\cdot 1 + 0\cdot i$ and $T(i) = \overline{i} = -i = 0\cdot 1 + (-1)\cdot i$. So the $\beta$-coordinate vectors are $(1,0)$ and $(0,-1)$, and

$$[T]_\beta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

Solution. Since $v_0=0$, $T(v_1)=v_1+v_0=v_1$, and for $i\geq 2$, $T(v_i)=v_i+v_{i-1}$. Hence the $\beta$-coordinate vector of $T(v_i)$ has a $1$ in position $i$ only when $i=1$, and $1$'s in positions $i-1$ and $i$ when $i \geq 2$, with all other entries $0$. Listing these as columns, $[T]_\beta$ is the $n \times n$ matrix whose diagonal entries and entries immediately above the diagonal are all $1$, and all other entries are $0$.

$$[T]_\beta = \begin{pmatrix} 1&1&0&\cdots&0 \\ 0&1&1& &0 \\ \vdots& &\ddots&\ddots&\vdots \\ 0&0& &1&1 \\ 0&0&\cdots&0&1 \end{pmatrix}$$

That is, the $(i,j)$ entry is $1$ if $i=j$ or $j=i+1$, and $0$ otherwise.

Proof. Suppose $n \geq 1$ and scalars $c_1, \ldots, c_n \in \mathbb{R}$ satisfy

$$c_1 T_1 + c_2 T_2 + \cdots + c_n T_n = T_0$$

where $T_0$ is the zero transformation in $\mathcal{L}(V)$. Applying this equation to $f(x) = x^n \in V$, and using that $T_j(x^n) = f^{(j)}(x) = \dfrac{n!}{(n-j)!}x^{n-j}$ for $j \leq n$, we get

$$0 = \sum_{j=1}^{n} c_j \frac{n!}{(n-j)!} x^{n-j}$$

as an identity of polynomials in $x$. Since $x^{n-1}, x^{n-2}, \ldots, x^0$ are polynomials of distinct degrees, they are linearly independent in $P(\mathbb{R})$, so every coefficient above must vanish: $c_j \cdot \dfrac{n!}{(n-j)!} = 0$ for each $j = 1, \ldots, n$. As $\dfrac{n!}{(n-j)!} \neq 0$, we get $c_j = 0$. This shows $\{T_j : 1 \leq j \leq n\}$ is linearly independent, and since $n$ was arbitrary, every finite subset of $\{T_j : j \geq 1\}$ is linearly independent. Hence $\{T_j : j \geq 1\}$ is a linearly independent subset of $\mathcal{L}(V)$.

Proof. Let $a, b$ be scalars with $aT + bU = T_0$ (the zero transformation). For every $v \in V$,

$$aT(v) = -bU(v)$$

Since $\operatorname{Im}T$ and $\operatorname{Im}U$ are subspaces, the left side lies in $\operatorname{Im}T$ and the right side lies in $\operatorname{Im}U$. As these are equal, this common value lies in $\operatorname{Im}T \cap \operatorname{Im}U = 0$, so $aT(v) = 0$ and $bU(v) = 0$ for every $v \in V$.

Since $T$ is nonzero, there is $v_0 \in V$ with $T(v_0) \neq 0$, and $aT(v_0)=0$ forces $a=0$. Likewise, since $U$ is nonzero, $b=0$. Hence $\{T, U\}$ is a linearly independent subset of $\mathcal{L}(V,W)$.

Solution. (1) First,

$$2B = \begin{pmatrix} 2&0&-6\\8&2&4\end{pmatrix}, \quad 3C = \begin{pmatrix} 3&3&12\\-3&-6&0\end{pmatrix}, \quad 2B+3C = \begin{pmatrix} 5&3&6\\5&-4&4\end{pmatrix}$$

Hence

$$A(2B+3C) = \begin{pmatrix} 1&3\\2&-1\end{pmatrix}\begin{pmatrix} 5&3&6\\5&-4&4\end{pmatrix} = \begin{pmatrix} 20&-9&18\\5&10&8\end{pmatrix}$$

(2) Since $C^t = \begin{pmatrix} 1&-1\\1&-2\\4&0\end{pmatrix}$,

$$C^tA = \begin{pmatrix} 1&-1\\1&-2\\4&0\end{pmatrix}\begin{pmatrix} 1&3\\2&-1\end{pmatrix} = \begin{pmatrix} -1&4\\-3&5\\4&12\end{pmatrix}$$

Solution. $A, B$ are $5\times 6$ and $C$ is $6\times 3$.

(1) For $AB$ to be defined, the number of columns of $A$ ($6$) must equal the number of rows of $B$ ($5$), but $6 \neq 5$, so $AB$ is undefined, and hence $B+AB$ is not possible.

(2) $A^t$ is $6 \times 5$ and $B$ is $5 \times 6$, so $A^tB$ is possible, of size $6 \times 6$.

(3) $B^t$ is $6 \times 5$ and $A$ is $5 \times 6$, so $B^tA$ is possible, of size $6 \times 6$. Then, since $C$ is $6 \times 3$, $(B^tA)C$ is also possible, of size $6 \times 3$.

Solution. (1) Let $A = \begin{pmatrix} 1&0\\0&0\end{pmatrix}$ and $B = \begin{pmatrix} 0&0\\1&0\end{pmatrix}$. Then $A \neq O$, $B \neq O$, and

$$AB = \begin{pmatrix} 1&0\\0&0\end{pmatrix}\begin{pmatrix} 0&0\\1&0\end{pmatrix} = \begin{pmatrix} 0&0\\0&0\end{pmatrix} = O$$

(2) Let $A = \begin{pmatrix} 0&1\\0&0\end{pmatrix}$. Then $A \neq O$ and

$$A^2 = \begin{pmatrix} 0&1\\0&0\end{pmatrix}\begin{pmatrix} 0&1\\0&0\end{pmatrix} = \begin{pmatrix} 0&0\\0&0\end{pmatrix} = O$$

Proof. Let $A = (a_{ij})$, $B = (b_{ij})$ be $n \times n$ matrices.

(1) Since $(AB)_{ii} = \sum_{k=1}^n a_{ik}b_{ki}$,

$$\operatorname{tr}(AB) = \sum_{i=1}^n \sum_{k=1}^n a_{ik}b_{ki} = \sum_{k=1}^n \sum_{i=1}^n b_{ki}a_{ik} = \sum_{k=1}^n (BA)_{kk} = \operatorname{tr}(BA)$$

(2) The $(i,i)$ entry of $A^t$ equals the $(i,i)$ entry of $A$, i.e., $(A^t)_{ii} = a_{ii}$, so

$$\operatorname{tr}(A^t) = \sum_{i=1}^n (A^t)_{ii} = \sum_{i=1}^n a_{ii} = \operatorname{tr}(A)$$

(3) Suppose there exist $n \times n$ matrices $A, B$ (with scalars in $\mathbb{R}$ or $\mathbb{C}$) such that $AB - BA = I$. Taking the trace of both sides and using (1),

$$\operatorname{tr}(AB - BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = 0$$

while $\operatorname{tr}(I) = n$. Hence $n = 0$, which is a contradiction since $\mathbb{R}$ and $\mathbb{C}$ have characteristic $0$ and $n \geq 1$. Therefore no such $A, B$ exist.

Solution. (1) Let $\beta = \{1, x, x^2\}$ and let $\gamma$ be the standard ordered basis of $\mathbb{R}^3$.

Since $U(1) = U(1+0x+0x^2) = (1,0,1)$, $U(x) = U(0+1x+0x^2) = (1,0,-1)$, $U(x^2) = U(0+0x+1x^2) = (0,1,0)$,

$$[U]_\beta^\gamma = \begin{pmatrix} 1&1&0\\0&0&1\\1&-1&0 \end{pmatrix}$$

Since $T(1) = 0\cdot(3+x) + 2\cdot 1 = 2$, $T(x) = 1\cdot(3+x) + 2x = 3+3x$, $T(x^2) = 2x(3+x)+2x^2 = 6x+4x^2$, the $\beta$-coordinates are $(2,0,0)$, $(3,3,0)$, $(0,6,4)$, respectively, so

$$[T]_\beta = \begin{pmatrix} 2&3&0\\0&3&6\\0&0&4 \end{pmatrix}$$

Computing the product $[U]_\beta^\gamma [T]_\beta$,

$$[U]_\beta^\gamma[T]_\beta = \begin{pmatrix} 1&1&0\\0&0&1\\1&-1&0 \end{pmatrix}\begin{pmatrix} 2&3&0\\0&3&6\\0&0&4 \end{pmatrix} = \begin{pmatrix} 2&6&6\\0&0&4\\2&0&-6 \end{pmatrix}$$

On the other hand, since $(UT)(f) = U(T(f))$, we have $UT(1) = U(2) = (2,0,2)$, $UT(x) = U(3+3x) = (6,0,0)$, $UT(x^2) = U(6x+4x^2) = (6,4,-6)$, and listing these as columns,

$$[UT]_\beta^\gamma = \begin{pmatrix} 2&6&6\\0&0&4\\2&0&-6 \end{pmatrix}$$

Hence $[UT]_\beta^\gamma = [U]_\beta^\gamma[T]_\beta$.

(2) Since $h(x) = 3-2x+x^2$, $[h(x)]_\beta = (3,-2,1)$. Here $U(h(x)) = U(3-2x+x^2)$ corresponds to $a=3, b=-2, c=1$, so $U(h(x)) = (3-2,\, 1,\, 3+2) = (1,1,5)$ and $[U(h(x))]_\gamma = (1,1,5)$. Indeed,

$$[U]_\beta^\gamma[h(x)]_\beta = \begin{pmatrix} 1&1&0\\0&0&1\\1&-1&0 \end{pmatrix}\begin{pmatrix} 3\\-2\\1 \end{pmatrix} = \begin{pmatrix} 1\\1\\5 \end{pmatrix} = [U(h(x))]_\gamma$$

so $[U(h(x))]_\gamma = [U]_\beta^\gamma[h(x)]_\beta$.

Solution. (1) The standard matrix of clockwise rotation by $\theta$ around the origin is

$$A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$$

Then

$$A^2 = \begin{pmatrix} \cos^2\theta-\sin^2\theta & 2\sin\theta\cos\theta \\ -2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta \end{pmatrix} = \begin{pmatrix} \cos2\theta & \sin2\theta \\ -\sin2\theta & \cos2\theta \end{pmatrix}$$

so $A^2$ corresponds to clockwise rotation by $2\theta$ around the origin.

(2) Reflection about the $y$-axis is $T(x,y) = (-x,y)$, so

$$A = \begin{pmatrix} -1&0\\0&1 \end{pmatrix}, \quad A^2 = \begin{pmatrix} -1&0\\0&1 \end{pmatrix}\begin{pmatrix} -1&0\\0&1 \end{pmatrix} = \begin{pmatrix} 1&0\\0&1 \end{pmatrix} = I$$

so $A^2$ corresponds to the identity transformation.

(3) Projection on the $y$-axis is $T(x,y) = (0,y)$, so

$$A = \begin{pmatrix} 0&0\\0&1 \end{pmatrix}, \quad A^2 = \begin{pmatrix} 0&0\\0&1 \end{pmatrix}\begin{pmatrix} 0&0\\0&1 \end{pmatrix} = \begin{pmatrix} 0&0\\0&1 \end{pmatrix} = A$$

so $A^2$ corresponds to the same transformation as $A$, namely the projection on the $y$-axis itself.

Solution. The columns of $[T]_\beta^\gamma$ are the $\gamma$-coordinate vectors of the images of the vectors in $\beta$. The first column $(1,0)$ means $T(1,0) = 1\cdot(0,1)+0\cdot(2,1) = (0,1)$, and the second column $(1,1)$ means $T(1,1) = 1\cdot(0,1)+1\cdot(2,1) = (2,2)$.

Write an arbitrary $(x,y) \in \mathbb{R}^2$ as a linear combination of $\beta$: $(x,y) = a(1,0)+b(1,1)$ gives $a+b=x$, $b=y$, so $a=x-y$, $b=y$. Since $T$ is linear,

$$T(x,y) = aT(1,0) + bT(1,1) = (x-y)(0,1) + y(2,2) = (2y,\ x-y+2y) = (2y,\ x+y)$$

That is, $T(x,y) = (2y, x+y)$. Check: the $\gamma$-coordinate of $T(1,0)=(0,1)$ is $(1,0)$ since $(0,1)=1\cdot(0,1)+0\cdot(2,1)$, and the $\gamma$-coordinate of $T(1,1)=(2,2)$ is $(1,1)$ since $(2,2)=1\cdot(0,1)+1\cdot(2,1)$. These match the two columns of the given $[T]_\beta^\gamma$.

Solution. (1) For $T : \mathbb{R}^2 \to \mathbb{R}^3$, since $\dim \mathbb{R}^2 = 2 \lt 3 = \dim \mathbb{R}^3$, $T$ cannot be onto (its rank is at most $2$, so its image cannot be all of $\mathbb{R}^3$). Hence $T$ is not bijective, so $T$ is not invertible.

(2) The standard matrix of $T : \mathbb{R}^3 \to \mathbb{R}^3$ is

$$A = \begin{pmatrix} 3&0&-2\\0&1&0\\3&4&0 \end{pmatrix}$$

To find $\operatorname{Ker}T$: if $T(a_1,a_2,a_3)=(3a_1-2a_3,\,a_2,\,3a_1+4a_2)=(0,0,0)$, then $a_2=0$; from $3a_1+4a_2=3a_1=0$ we get $a_1=0$; and from $3a_1-2a_3=-2a_3=0$ we get $a_3=0$. So $\operatorname{Ker}T=0$, hence $T$ is one-to-one, and since $\dim\mathbb{R}^3=\dim\mathbb{R}^3$, $T$ is also onto. Therefore $T$ is invertible.

Solution. Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

(1) $\dim F^3 = 3$ and $\dim P_3(F) = 4$ (a basis for $P_3(F)$ is $\{1,x,x^2,x^3\}$). Since $3 \neq 4$, $F^3$ and $P_3(F)$ are not isomorphic.

(2) $V = \{A \in M_{2\times 2}(\mathbb{R}) : \operatorname{tr}(A)=0\}$ is a subspace defined by a single linear condition, so $\dim V = \dim M_{2\times 2}(\mathbb{R}) - 1 = 4-1=3$ (indeed, $\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}\right\}$ is a basis for $V$). Since $\dim\mathbb{R}^3=3$, $\dim V = \dim \mathbb{R}^3$, so $V$ and $\mathbb{R}^3$ are isomorphic.

Solution. Define $\Phi : V \to F^3$ by

$$\Phi\left( \begin{pmatrix} a & a+b \\ 0 & c \end{pmatrix} \right) = (a,b,c)$$

Since the $(1,1)$ entry of $\begin{pmatrix}a&a+b\\0&c\end{pmatrix}$ is $a$, the $(2,2)$ entry is $c$, and $b$ is the $(1,2)$ entry minus $a$, the triple $a,b,c$ is uniquely determined by the matrix, so $\Phi$ is well defined.

Linearity. Let $M_1 = \begin{pmatrix}a_1&a_1+b_1\\0&c_1\end{pmatrix}$, $M_2 = \begin{pmatrix}a_2&a_2+b_2\\0&c_2\end{pmatrix} \in V$, $t \in F$. Then

$$M_1 + tM_2 = \begin{pmatrix} a_1+ta_2 & (a_1+ta_2)+(b_1+tb_2) \\ 0 & c_1+tc_2 \end{pmatrix} \in V$$

so

$$\Phi(M_1+tM_2) = (a_1+ta_2,\, b_1+tb_2,\, c_1+tc_2) = (a_1,b_1,c_1)+t(a_2,b_2,c_2) = \Phi(M_1)+t\Phi(M_2)$$

Hence $\Phi$ is linear.

Injectivity. If $\Phi(M)=0$, then $a=b=c=0$, so $M=O$. Hence $\operatorname{Ker}\Phi=0$ and $\Phi$ is one-to-one.

Surjectivity. For any $(a,b,c) \in F^3$, $M = \begin{pmatrix}a&a+b\\0&c\end{pmatrix} \in V$ satisfies $\Phi(M)=(a,b,c)$, so $\Phi$ is onto.

Hence $\Phi$ is linear and bijective, so it is an isomorphism from $V$ to $F^3$.

Proof. (1) By definition of the inverse, $AA^{-1}=A^{-1}A=I$. This equation shows $A$ is a two-sided inverse of $A^{-1}$, so by uniqueness of inverses, $(A^{-1})^{-1}=A$.

(2)

$$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$$

$$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$$

So $B^{-1}A^{-1}$ is a two-sided inverse of $AB$, and by uniqueness of inverses, $(AB)^{-1}=B^{-1}A^{-1}$.

(3) Taking the transpose of $AA^{-1}=I$ gives $(A^{-1})^tA^t = I^t = I$, and taking the transpose of $A^{-1}A=I$ gives $A^t(A^{-1})^t = I$. Hence $(A^{-1})^t$ is a two-sided inverse of $A^t$, so $(A^t)^{-1}=(A^{-1})^t$.

(4) We prove by induction on $n$ that $A^n(A^{-1})^n = I = (A^{-1})^nA^n$. For $n=0$, $A^0(A^{-1})^0 = I \cdot I = I$, so the claim holds. Suppose it holds for $n=k$, i.e., $A^k(A^{-1})^k=I=(A^{-1})^kA^k$. Then

$$A^{k+1}(A^{-1})^{k+1} = A\big(A^k(A^{-1})^k\big)A^{-1} = A\cdot I \cdot A^{-1} = AA^{-1} = I$$

$$(A^{-1})^{k+1}A^{k+1} = A^{-1}\big((A^{-1})^kA^k\big)A = A^{-1}\cdot I \cdot A = A^{-1}A = I$$

so the claim also holds for $n=k+1$. Hence $A^n(A^{-1})^n=I=(A^{-1})^nA^n$ for all $n \in \mathbb{Z}^{\geq 0}$, and by uniqueness of inverses, $(A^n)^{-1}=(A^{-1})^n$.

(5) For $c \in F-\{0\}$, $c^{-1} \in F$ exists, and

$$(cA)(c^{-1}A^{-1}) = (cc^{-1})(AA^{-1}) = 1\cdot I = I, \quad (c^{-1}A^{-1})(cA) = (c^{-1}c)(A^{-1}A) = 1\cdot I = I$$

so by uniqueness of inverses, $(cA)^{-1}=c^{-1}A^{-1}$.

Proof. (1) Suppose $A$ is invertible and $AB=O$. Multiplying both sides on the left by $A^{-1}$,

$$A^{-1}(AB) = A^{-1}O \ \Longrightarrow\ (A^{-1}A)B = O \ \Longrightarrow\ IB = O \ \Longrightarrow\ B = O$$

(2) Suppose $AB=O$ for some nonzero matrix $B$. If $A$ were invertible, then by (1) $B=O$, contradicting $B \neq O$. Hence $A$ is not invertible.

Proof. Let $B$ be an $n \times n$ invertible matrix and $\Phi(A) = B^{-1}AB$.

Linearity. For $A_1, A_2 \in M_{n\times n}(F)$ and $t \in F$,

$$\Phi(A_1+tA_2) = B^{-1}(A_1+tA_2)B = B^{-1}A_1B + tB^{-1}A_2B = \Phi(A_1)+t\Phi(A_2)$$

so $\Phi$ is linear.

Bijectivity. Define $\Psi(A) = BAB^{-1}$, a map $\Psi : M_{n\times n}(F) \to M_{n\times n}(F)$. Then

$$\Phi(\Psi(A)) = B^{-1}(BAB^{-1})B = (B^{-1}B)A(B^{-1}B) = A, \quad \Psi(\Phi(A)) = B(B^{-1}AB)B^{-1} = (BB^{-1})A(BB^{-1}) = A$$

so $\Psi = \Phi^{-1}$ and $\Phi$ is bijective. Hence $\Phi$ is linear and bijective, so it is an isomorphism.