Solution. Suppose \(\sum _{i}c_iTv_i=0\). By linearity, \[T\left (\sum _i c_iv_i\right )=\sum _i c_iTv_i=0,\] so \(\sum _i c_iv_i\in \text {null}\,T\). Since \(T\) is injective, \(\text {null}\,T=\{0\}\) by Proposition 6.2.4, so \(\sum _i c_iv_i=0\). The independence of \((v_1,\dots ,v_n)\) now forces every \(c_i=0\). Hence \((Tv_1,\dots ,Tv_n)\) is independent.

Without injectivity this fails badly. Take \(T:\mathbb {R}^2\to \mathbb {R}^2\) with \(T(x,y)=(x,0)\) and the independent pair \(v_1=(1,0)\), \(v_2=(0,1)\). Then \(Tv_1=(1,0)\) and \(Tv_2=(0,0)\), and any set containing \(0\) is dependent.

Solution. Let \(w\in W\). Since \(T\) is surjective there is \(v\in V\) with \(Tv=w\). Since \((v_1,\dots ,v_n)\) spans \(V\), we may write \(v=\sum _i c_iv_i\). Applying \(T\) and using linearity, \[w=Tv=T\left (\sum _i c_iv_i\right )=\sum _i c_iTv_i.\] So every \(w\in W\) is a linear combination of \(Tv_1,\dots ,Tv_n\), which is what it means for them to span \(W\).

Taken with the previous problem: injective maps preserve independence, surjective maps preserve spanning, and therefore an invertible map carries a basis to a basis.

Solution.

(a)

Linear. Each output coordinate is a homogeneous linear expression in \(x,y,z\), so \(T\) is given by matrix multiplication by \(\begin {pmatrix}2&-1&3\\7&5&-6\end {pmatrix}\), and matrix multiplication is linear.

(b)

Linear. The zero map satisfies \(T(u+v)=0=0+0=Tu+Tv\) and \(T(\lambda u)=0=\lambda \cdot 0\). It is the matrix of zeros.

(c)

Not linear. A linear map must send \(0\) to \(0\), since \(T(0)=T(0+0)=T(0)+T(0)\). Here \(T(0,0,0)=(1,1)\neq (0,0)\). (Such a map is affine, not linear: it is a linear map followed by a translation.)

Solution. Take \[f(x,y)=\sqrt [3]{x^3+y^3}.\] Homogeneity holds, since \[f(\alpha x,\alpha y)=\sqrt [3]{\alpha ^3x^3+\alpha ^3y^3} =\alpha \sqrt [3]{x^3+y^3}=\alpha f(x,y)\] for every real \(\alpha \), including negative ones (the real cube root is defined for all reals and is odd).

But \(f\) is not additive: with \(u=(1,0)\) and \(v=(0,1)\), \[f(u)+f(v)=1+1=2,\qquad f(u+v)=f(1,1)=\sqrt [3]{2}\approx 1.26.\] So the two conditions in the definition of a linear map are genuinely independent — neither implies the other, and scaling correctly is no guarantee of adding correctly.

Solution.

(a)

The columns are the coordinates of the images: \[M=\begin {pmatrix}1&1\\2&-1\\-1&0\end {pmatrix}.\]

(b)

Compute the images of the new domain basis directly: \begin {align*} T(-x_1+x_2)&=-(y_1+2y_2-y_3)+(y_1-y_2)=-3y_2+y_3,\\ T(2x_1-x_2)&=2(y_1+2y_2-y_3)-(y_1-y_2)=y_1+5y_2-2y_3. \end {align*}

Now re-express in the new codomain basis. Writing \(y'_1=y_1\), \(y'_2=y_1+y_2\), \(y'_3=y_1+y_2+y_3\), we invert: \[y_1=y'_1,\qquad y_2=y'_2-y'_1,\qquad y_3=y'_3-y'_2.\] Substituting, \begin {align*} -3y_2+y_3&=-3(y'_2-y'_1)+(y'_3-y'_2)=3y'_1-4y'_2+y'_3,\\ y_1+5y_2-2y_3&=y'_1+5(y'_2-y'_1)-2(y'_3-y'_2)=-4y'_1+7y'_2-2y'_3. \end {align*}

Hence \[M'=\begin {pmatrix}3&-4\\-4&7\\1&-2\end {pmatrix}.\] The same answer follows from Theorem 6.3.5 as \(M'=Q^{-1}MP\) with \(P=\begin {pmatrix}-1&2\\1&-1\end {pmatrix}\) and \(Q=\begin {pmatrix}1&1&1\\0&1&1\\0&0&1\end {pmatrix}\).

Solution. First find \(\dim \text {null}\,T\). The conditions express \(x_1\) in terms of \(x_2\), and \(x_3\) in terms of \(x_4\), leaving \(x_2\) and \(x_4\) free. A general element is \[(5x_2,\ x_2,\ 7x_4,\ x_4)=x_2(5,1,0,0)+x_4(0,0,7,1),\] and those two vectors are independent, so \(\dim \text {null}\,T=2\).

By Theorem 6.2.8, \[\dim \text {range}\,T=\dim \mathbb {F}^4-\dim \text {null}\,T=4-2=2.\] But range\(T\) is a subspace of \(\mathbb {F}^2\), which has dimension \(2\), and a subspace of the same finite dimension as the whole space must be the whole space. Hence range\(T=\mathbb {F}^2\) and \(T\) is surjective.

Solution. Suppose such a \(T\) existed. The stated set is described by three independent constraints — \(x_1=3x_2\), \(x_3=x_4\) and \(x_4=x_5\) — leaving \(x_2\) and \(x_3\) free. Explicitly its elements are \[x_2(3,1,0,0,0)+x_3(0,0,1,1,1),\] so it has dimension \(2\).

Then Theorem 6.2.8 would give \[\dim \text {range}\,T=5-2=3.\] But range\(T\) is a subspace of \(\mathbb {F}^2\), so \(\dim \text {range}\,T\leq 2\). This is a contradiction, so no such \(T\) exists.

Compare the previous problem: the arithmetic is the same, and the whole question of whether the map can exist is settled by whether \(\dim V-\dim \text {null}\,T\) fits inside the codomain.

Solution.

If \(S\) and \(T\) are invertible

Then \(T^{-1}S^{-1}\) is a two-sided inverse of \(ST\): \[(ST)(T^{-1}S^{-1})=S(TT^{-1})S^{-1}=SS^{-1}=I,\] and similarly \((T^{-1}S^{-1})(ST)=I\).

If \(ST\) is invertible

Let \(R=(ST)^{-1}\), so \(STR=I\) and \(RST=I\).

\(S\) is surjective, because for any \(v\in V\) we have \(S(TRv)=(STR)v=v\). By Theorem 6.2.8 applied to \(S:V\to V\), surjectivity forces \(\dim \text {null}\,S=0\), so \(S\) is also injective, hence invertible by Proposition 6.4.2.

\(T\) is injective, because if \(Tv=0\) then \(v=(RST)v=RS(Tv)=RS0=0\). By rank–nullity again, injectivity of a map from \(V\) to itself forces surjectivity, so \(T\) is invertible.

Note where finite dimensionality is used: it is what allows injectivity and surjectivity to imply each other for a map of a space to itself. In infinite dimensions the statement is false — the shift operators on sequence spaces give \(ST=I\) with neither factor invertible.