6.10.2E: Kernel and Image of a Linear Transformation Exercises
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Exercises
Exercise \(\PageIndex{1}\)
For each matrix \(A\), find a basis for the kernel and image of \(T_{A}\), and find the \(rank \;\) and \(nullity \;\) of \(T_{A}\).
- \(\left[ \begin{array}{rrrr} 1 & 2 & -1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & -3 & 2 & 0 \end{array} \right]\)
- \(\left[ \begin{array}{rrrr} 2 & 1 & -1 & 3 \\ 1 & 0 & 3 & 1 \\ 1 & 1 & -4 & 2 \end{array} \right]\)
- \(\left[ \begin{array}{rrr} 1 & 2 & -1 \\ 3 & 1 & 2 \\ 4 & -1 & 5 \\ 0 & 2 & -2 \end{array} \right]\)
- \(\left[ \begin{array}{rrr} 2 & 1 & 0 \\ 1 & -1 & 3 \\ 1 & 2 & -3 \\ 0 & 3 & -6 \end{array} \right]\)
- Answer
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- \(\left\{\left[\begin{array}{r}-3 \\ 7 \\ 1 \\ 0\end{array}\right],\left[\begin{array}{r}1 \\ 1 \\ 0 \\ -1\end{array}\right]\right\} ;\left\{\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right],\left[\begin{array}{r}0 \\ 1 \\ -1\end{array}\right]\right\} ; 2,2\)
- \(\left\{\left[\begin{array}{r}-1 \\ 2 \\ 1\end{array}\right]\right\} ;\left\{\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 1\end{array}\right],\left[\begin{array}{r}0 \\ 1 \\ -1 \\ -2\end{array}\right]\right\} ; 2,1\)
Exercise \(\PageIndex{2}\)
In each case, (i) find a basis of \(\text{ker }T\), and (ii) find a basis of \(im \;T\). You may assume that \(T\) is linear.
- \(T :{P}_{2} \to \mathbb{R}^2\); \(T(a + bx + cx^{2}) = (a, b)\)
- \(T :{P}_{2} \to \mathbb{R}^2\); \(T(p(x)) = (p(0), p(1))\)
- \(T : \mathbb{R}^3 \to \mathbb{R}^3\); \(T(x, y, z) = (x + y, x + y, 0)\)
- \(T : \mathbb{R}^3 \to \mathbb{R}^4\); \(T(x, y, z) = (x, x, y, y)\)
- \(T :{M}_{22} \to\|{M}_{22}\); \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} a + b & b + c \\ c + d & d + a \end{array} \right]\)
- \(T :{M}_{22} \to \mathbb{R}\); \(T\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = a + d\)
- \(T :{P}_{n} \to \mathbb{R}\); \(T(r_{0} + r_{1}x + \cdots + r_{n}x^{n}) = r_{n}\)
- \(T : \mathbb{R}^n \to \mathbb{R}\); \(T(r_{1}, r_{2}, \dots, r_{n}) = r_{1} + r_{2} + \cdots + r_{n}\)
- \(T :{M}_{22} \to\|{M}_{22}\); \(T(X) = XA - AX\), where
- \(A = \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]\)
- Answer
-
- \(\left\{x^2-x\right\} ;\{(1,0),(0,1)\}\)
- \(\{(0,0,1)\} ;\{(1,1,0,0),(0,0,1,1)\}\)
- \(\left\{\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]\right\} ;\{1\}\)
- \(\{(1,0,0, \ldots, 0,-1),(0,1,0, \ldots, 0,-1)\), \(\ldots,(0,0,0, \ldots, 1,-1)\} ;\{1\}\)
- \(\begin{array}{l}\{ {\left.\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\} } \\ \left\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right]\right\}\end{array}\)
Exercise \(\PageIndex{1}\)
Let \(P : V \to \mathbb{R}\) and \(Q : V \to \mathbb{R}\) be linear transformations, where \(V\) is a vector space. Define \(T : V \to \mathbb{R}^2\) by \(T(\mathbf{v}) = (P(\mathbf{v}), Q(\mathbf{v}))\).
- Show that \(T\) is a linear transformation.
- Show that \(\text{ker }T = \text{ker }P \cap \text{ker }Q\), the set of vectors in both \(\text{ker }P\) and \(\text{ker }Q\).
- Answer
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b. \(T(\mathbf{v}) = \mathbf{0} = (0, 0)\) if and only if \(P(\mathbf{v}) = 0\) and \(Q(\mathbf{v}) = 0\); that is, if and only if \(\mathbf{v}\) is in \(\text{ker }P \cap \text{ker }Q\).
Exercise \(\PageIndex{4}\)
In each case, find a basis
\(B = \{\mathbf{e}_{1}, \dots, \mathbf{e}_{r}, \mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) of \(V\) such that \(\{\mathbf{e}_{r+1}, \dots, \mathbf{e}_{n}\}\) is a basis of \(\text{ker }T\), and verify Theorem [thm:021572].
- \(T : \mathbb{R}^3 \to \mathbb{R}^4\); \(T(x, y, z) = (x - y + 2z, x + y - z, 2x + z, 2y - 3z)\)
- \(T : \mathbb{R}^3 \to \mathbb{R}^4\); \(T(x, y, z) = (x + y + z, 2x - y + 3z, z - 3y, 3x + 4z)\)
- Answer
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b. \(\text{ker }T = span \;\{(-4, 1, 3)\}\); \(B = \{(1, 0, 0), (0, 1, 0), (-4, 1, 3)\}\), \(im \;T = span \;\{(1, 2, 0, 3), (1, -1, -3, 0)\}\)
Exercise \(\PageIndex{5}\)
Show that every matrix \(X\) in \(\mathbf{M}_{nn}\) has the form \(X = A^{T} - 2A\) for some matrix \(A\) in \(\mathbf{M}_{nn}\). [Hint: The dimension theorem.]
Exercise \(\PageIndex{6}\)
In each case either prove the statement or give an example in which it is false. Throughout, let \(T : V \to W\) be a linear transformation where \(V\) and \(W\) are finite dimensional.
- If \(V = W\), then \(\text{ker }T \subseteq im \;T\).
- If \(dim \;V = 5\), \(dim \;W = 3\), and \(dim \;(\text{ker }T) = 2\), then \(T\) is onto.
- If \(dim \;V = 5\) and \(dim \;W = 4\), then \(\text{ker }T \neq \{\mathbf{0}\}\).
- If \(\text{ker }T = V\), then \(W = \{\mathbf{0}\}\).
- If \(W = \{\mathbf{0}\}\), then \(\text{ker }T = V\).
- If \(W = V\), and \(im \;T \subseteq \text{ker }T\), then \(T = 0\).
- If \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\}\) is a basis of \(V\) and \(T(\mathbf{e}_{1}) = \mathbf{0} = T(\mathbf{e}_{2})\), then \(dim \;(im \;T) \leq 1\).
- If \(dim \;(\text{ker }T) \leq dim \;W\), then \(dim \;W \geq \frac{1}{2} dim \;V\).
- If \(T\) is one-to-one, then \(dim \;V \leq dim \;W\).
- If \(dim \;V \leq dim \;W\), then \(T\) is one-to-one.
- If \(T\) is onto, then \(dim \;V \geq dim \;W\).
- If \(dim \;V \geq dim \;W\), then \(T\) is onto.
- If \(\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{k})\}\) is independent, then \(\{\mathbf{v}_{1}, \dots, \mathbf{v}_{k}\}\) is independent.
- If \(\{\mathbf{v}_{1}, \dots, \mathbf{v}_{k}\}\) spans \(V\), then \(\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{k})\}\) spans \(W\).
- Answer
-
- Yes. \(dim \;(im \;T) = 5 - dim \;(\text{ker }T) = 3\), so \(im \;T = W\) as \(dim \;W = 3\).
- No. \(T = 0: \mathbb{R}^2 \to \mathbb{R}^2\)
- No. \(T : \mathbb{R}^2 \to \mathbb{R}^2\), \(T(x, y) = (y, 0)\). Then \(\text{ker }T = im \;T\)
- Yes. \(dim \;V = dim \;(\text{ker }T) + dim \;(im \;T) \leq dim \;W + dim \;W = 2 dim \;W\)
- No. Consider \(T : \mathbb{R}^2 \to \mathbb{R}^2\) with \(T(x, y) = (y, 0)\).
- No. Same example as (j).
- No. Define \(T : \mathbb{R}^2 \to \mathbb{R}^2\) by \(T(x, y) = (x, 0)\). If \(\mathbf{v}_{1} = (1, 0)\) and \(\mathbf{v}_{2} = (0, 1)\), then \(\mathbb{R}^2 = span \;\{\mathbf{v}_{1}, \mathbf{v}_{2}\}\) but \(\mathbb{R}^2 \neq span \;\{T(\mathbf{v}_{1}), T(\mathbf{v}_{2})\}\).
Exercise \(\PageIndex{7}\)
Show that linear independence is preserved by one-to-one transformations and that spanning sets are preserved by onto transformations. More precisely, if \(T : V \to W\) is a linear transformation, show that:
- If \(T\) is one-to-one and \(\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}\) is independent in \(V\), then \(\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{n})\}\) is independent in \(W\).
- If \(T\) is onto and \(V = span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}\), then \(W = span \;\{T(\mathbf{v}_{1}), \dots, T(\mathbf{v}_{n})\}\).
- Answer
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b. Given \(\mathbf{w}\) in \(W\), let \(\mathbf{w} = T(\mathbf{v})\), \(\mathbf{v}\) in \(V\), and write \(\mathbf{v} = r_{1}\mathbf{v}_{1} + \cdots + r_{n}\mathbf{v}_{n}\). Then \(\mathbf{w} = T(\mathbf{v}) = r_{1}T(\mathbf{v}_{1}) + \cdots + r_{n}T(\mathbf{v}_{n})\).
Exercise \(\PageIndex{8}\)
Given \(\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}\) in a vector space \(V\), define \(T : \mathbb{R}^n \to V\) by \(T(r_{1}, \dots, r_{n}) = r_{1}\mathbf{v}_{1} + \cdots + r_{n}\mathbf{v}_{n}\). Show that \(T\) is linear, and that:
- \(T\) is one-to-one if and only if \(\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}\) is independent.
- \(T\) is onto if and only if \(V = span \;\{\mathbf{v}_{1}, \dots, \mathbf{v}_{n}\}\).
- Answer
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b. \(im \;T = \left\lbrace \sum_{i} r_i\mathbf{v}_i \mid r_i \mbox{ in } \mathbb{R} \right\rbrace = span \;\{\mathbf{v}_i\}\).
Exercise \(\PageIndex{9}\)
Let \(T : V \to V\) be a linear transformation where \(V\) is finite dimensional. Show that exactly one of (i) and (ii) holds:
(i) \(T(\mathbf{v}) = \mathbf{0}\) for some \(\mathbf{v} \neq \mathbf{0}\) in \(V\);
(ii) \(T(\mathbf{x}) = \mathbf{v}\) has a solution \(\mathbf{x}\) in \(V\) for every \(\mathbf{v}\) in \(V\).
Exercise \(\PageIndex{10}\)
Let \(T :\|{M}_{nn} \to \mathbb{R}\) denote the trace map: \(T(A) = tr \;A\) for all \(A\) in \(\mathbf{M}_{nn}\). Show that
\(dim \;(\text{ker }T) = n^{2} - 1\).
- Answer
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\(T\) is linear and onto. Hence \(1 = dim \;\mathbb{R} = dim \;(im \;T) = dim \;(\mathbf{M}_{nn}) - dim \;(\text{ker }T) = n^{2} - dim \;(\text{ker }T)\)
Exercise \(\PageIndex{11}\)
Show that the following are equivalent for a linear transformation \(T : V \to W\).
\(\text{ker }T = V\)
\(im \;T = \{\mathbf{0}\}\)
\(T = 0\)
Exercise \(\PageIndex{12}\)
Let \(A\) and \(B\) be \(m \times n\) and \(k \times n\) matrices, respectively. Assume that \(A\mathbf{x} = \mathbf{0}\) implies \(B\mathbf{x} = \mathbf{0}\) for every \(n\)-column \(\mathbf{x}\). Show that \(rank \;A \geq rank \;B\).
- Answer
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The condition means \(\text{ker }(T_{A}) \subseteq \text{ker}(T_{B})\), so \(dim \;\left[\text{ker}(T_{A})\right] \leq dim \;\left[\text{ker}(T_{B})\right]\). Then Theorem [thm:021499] gives \(dim \;\left[im \;(T_{A})\right] \geq dim \;\left[im \;(T_{B})\right]\); that is, \(rank \;A \geq rank \;B\).
Exercise \(\PageIndex{13}\)
Let \(A\) be an \(m \times n\) matrix of \(rank \;r\). Thinking of \(\mathbb{R}^n\) as rows, define \(V = \{\mathbf{x}\) in \(\mathbb{R}^m \mid \mathbf{x}A = \mathbf{0}\}\). Show that \(dim \;V = m - r\).
Exercise \(\PageIndex{14}\)
Consider
\[V = \left\lbrace \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] \, \middle| \, a + c = b + d \right\rbrace \nonumber \]
- Consider \(S :\|{M}_{22} \to \mathbb{R}\) with \(S\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = a + c - b - d\). Show that \(S\) is linear and onto and that \(V\) is a subspace of \(\mathbf{M}_{22}\). Compute \(dim \;V\).
- Consider \(T : V \to \mathbb{R}\) with \(T\left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] = a + c\). Show that \(T\) is linear and onto, and use this information to compute \(dim \;(\text{ker }T)\).
Exercise \(\PageIndex{15}\)
Define \(T :\|{P}_{n} \to \mathbb{R}\) by \(T\left[p(x)\right] =\) the sum of all the coefficients of \(p(x)\).
- Use the dimension theorem to show that \(dim \;(\text{ker }T) = n\).
- Conclude that \(\{x - 1, x^{2} - 1, \dots, x^{n} - 1\}\) is a basis of \(\text{ker }T\).
- Answer
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b. \(B = \{x - 1, \dots, x^{n} - 1\}\) is independent (distinct degrees) and contained in \(\text{ker }T\). Hence \(B\) is a basis of \(\text{ker }T\) by (a).
Exercise \(\PageIndex{16}\)
Use the dimension theorem to prove Theorem 1.3.1: If \(A\) is an \(m \times n\) matrix with \(m < n\), the system \(A\mathbf{x} = \mathbf{0}\) of \(m\) homogeneous equations in \(n\) variables always has a nontrivial solution.
Exercise \(\PageIndex{17}\)
Let \(B\) be an \(n \times n\) matrix, and consider the subspaces \(U = \{A \mid A \mbox{ in }\|{M}_{mn}, AB = 0\}\) and \(V = \{AB \mid A \mbox{ in }\|{M}_{mn}\}\). Show that \(dim \;U + dim \;V = mn\).
Exercise \(\PageIndex{18}\)
Let \(U\) and \(V\) denote, respectively, the spaces of even and odd polynomials in \(\mathbf{P}_{n}\). Show that \(dim \;U + dim \;V = n + 1\). [Hint: Consider \(T :\|{P}_{n} \to\|{P}_{n}\) where \(T\left[p(x)\right] = p(x) - p(-x)\).]
Exercise \(\PageIndex{19}\)
Show that every polynomial \(f(x)\) in \(\mathbf{P}_{n-1}\) can be written as \(f(x) = p(x + 1) - p(x)\) for some polynomial \(p(x)\) in \(\mathbf{P}_{n}\). [Hint: Define \(T :\|{P}_{n} \to\|{P}_{n-1}\) by \(T\left[p(x)\right] = p(x + 1) - p(x)\).]
Exercise \(\PageIndex{20}\)
Let \(U\) and \(V\) denote the spaces of symmetric and skew-symmetric \(n \times n\) matrices. Show that \(dim \;U + dim \;V = n^{2}\).
- Answer
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Define \(T :{M}_{nn} \to {M}_{nn}\) by \(T(A) = A - A^{T}\) for all \(A\) in \(\mathbf{M}_{nn}\). Then \(\text{ker }T = U\) and \(im \;T = V\) by Example [exa:021376], so the dimension theorem gives \(n^{2} = dim \;\mathbf{M}_{nn} = dim \;(U) + dim \;(V)\).
Exercise \(\PageIndex{21}\)
Assume that \(B\) in \(\mathbf{M}_{nn}\) satisfies \(B^{k} = 0\) for some \(k \geq 1\). Show that every matrix in \(\mathbf{M}_{nn}\) has the form \(BA - A\) for some \(A\) in \(\mathbf{M}_{nn}\). [Hint: Show that \(T :{M}_{nn} \to {M}_{nn}\) is linear and one-to-one where
\(T(A) = BA - A\) for each \(A\).]
Exercise \(\PageIndex{22}\)
Fix a column \(\mathbf{y} \neq \mathbf{0}\) in \(\mathbb{R}^n\) and let
\(U = \{A\) in \(\mathbf{M}_{nn} \mid A\mathbf{y} = \mathbf{0}\}\). Show that \(dim \;U = n(n - 1)\).
- Answer
-
Define \(T :\|{M}_{nn} \to \mathbb{R}^n\) by \(T(A) = A\mathbf{y}\) for all \(A\) in \(\mathbf{M}_{nn}\). Then \(T\) is linear with \(\text{ker }T = U\), so it is enough to show that \(T\) is onto (then \(dim \;U = n^{2} - dim \;(im \;T) = n^{2} - n\)). We have \(T(0) = \mathbf{0}\). Let \(\mathbf{y} = \left[ \begin{array}{cccc} y_{1} & y_{2} & \cdots & y_{n} \end{array} \right]^{T} \neq \mathbf{0}\) in \(\mathbb{R}^n\). If \(y_{k} \neq \mathbf{0}\) let \(\mathbf{c}_{k} = y_{k}^{-1}\mathbf{y}\), and let \(\mathbf{c}_{j} = \mathbf{0}\) if \(j \neq k\). If \(A = \left[ \begin{array}{cccc} \mathbf{c}_{1} & \mathbf{c}_{2} & \cdots & \mathbf{c}_{n} \end{array} \right]\), then \(T(A) = A\mathbf{y} = y_{1}\mathbf{c}_{1} + \cdots + y_{k}\mathbf{c}_{k} + \cdots + y_{n}\mathbf{c}_{n} = \mathbf{y}\). This shows that \(T\) is onto, as required.
Exercise \(\PageIndex{23}\)
If \(B\) in \(\mathbf{M}_{mn}\) has \(rank \;r\), let \(U = \{A\) in \(\mathbf{M}_{nn} \mid BA = 0\}\) and \(W = \{BA \mid A\) in \(\mathbf{M}_{nn}\}\). Show that \(dim \;U = n(n - r)\) and \(dim \;W = nr\). [Hint: Show that \(U\) consists of all matrices \(A\) whose columns are in the \(null \) space of \(B\). Use Example \(\PageIndex{7}\).]
Exercise \(\PageIndex{24}\)
Let \(T : V \to V\) be a linear transformation where \(dim \;V = n\). If \(\text{ker }T \cap im \;T = \{\mathbf{0}\}\), show that every vector \(\mathbf{v}\) in \(V\) can be written \(\mathbf{v} = \mathbf{u} + \mathbf{w}\) for some \(\mathbf{u}\) in \(\text{ker }T\) and \(\mathbf{w}\) in \(im \;T\). [Hint: Choose bases \(B \subseteq \text{ker }T\) and \(D \subseteq im \;T\), and use Exercise 6.3.33.]
Exercise \(\PageIndex{25}\)
Let \(T : \mathbb{R}^n \to \mathbb{R}^n\) be a linear operator of \(rank \;1\), where \(\mathbb{R}^n\) is written as rows. Show that there exist numbers \(a_{1}, a_{2}, \dots, a_{n}\) and \(b_{1}, b_{2}, \dots, b_{n}\) such that \(T(X) = XA\) for all rows \(X\) in \(\mathbb{R}^n\), where
\[A = \left[ \begin{array}{cccc} a_{1}b_{1} & a_{1}b_{2} & \cdots & a_{1}b_{n} \\ a_{2}b_{1} & a_{2}b_{2} & \cdots & a_{2}b_{n} \\ \vdots & \vdots & & \vdots \\ a_{n}b_{1} & a_{n}b_{2} & \cdots & a_{n}b_{n} \end{array} \right] \nonumber \]
[Hint: \(im \;T = \mathbb{R}\mathbf{w}\) for \(\mathbf{w} = (b_{1}, \dots, b_{n})\) in \(\mathbb{R}^n\).]
Exercise \(\PageIndex{26}\)
Prove Theorem \(\PageIndex{5}\).
Exercise \(\PageIndex{27}\)
Let \(T : V \to \mathbb{R}\) be a nonzero linear transformation, where \(dim \;V = n\). Show that there is a basis \(\{\mathbf{e}_{1}, \dots, \mathbf{e}_{n}\}\) of \(V\) so that \(T(r_{1}\mathbf{e}_{1} + r_{2}\mathbf{e}_{2} + \cdots + r_{n}\mathbf{e}_{n}) = r_{1}\).
Exercise \(\PageIndex{28}\)
Let \(f \neq 0\) be a fixed polynomial of degree \(m \geq 1\). If \(p\) is any polynomial, recall that
\((p \circ f)(x) = p\left[f(x)\right]\). Define \(T_{f} : P_{n} \to P_{n+m}\) by
\(T_{f}(p) = p \circ f\).
- Show that \(T_{f}\) is linear.
- Show that \(T_{f}\) is one-to-one.
Exercise \(\PageIndex{29}\)
Let \(U\) be a subspace of a finite dimensional vector space \(V\).
- Show that \(U = \text{ker }T\) for some linear operator \(T : V \to V\).
-
Show that \(U = im \;S\) for some linear operator
\(S : V \to V\). [Hint: Theorem [thm:019430] and Theorem [thm:020916].]
- Answer
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b. By Lemma [lem:019415], let \(\{\mathbf{u}_{1}, \dots, \mathbf{u}_{m}, \dots, \mathbf{u}_{n}\}\) be a basis of \(V\) where \(\{\mathbf{u}_{1}, \dots, \mathbf{u}_{m}\}\) is a basis of \(U\). By Theorem [thm:020916] there is a linear transformation \(S : V \to V\) such that \(S(\mathbf{u}_{i}) = \mathbf{u}_{i}\) for \(1 \leq i \leq m\), and \(S(\mathbf{u}_{i}) = \mathbf{0}\) if \(i > m\). Because each \(\mathbf{u}_{i}\) is in \(im \;S\), \(U \subseteq im \;S\). But if \(S(\mathbf{v})\) is in \(im \;S\), write \(\mathbf{v} = r_{1}\mathbf{u}_{1} + \cdots + r_{m}\mathbf{u}_{m} + \cdots + r_{n}\mathbf{u}_{n}\). Then \(S(\mathbf{v}) = r_{1}S(\mathbf{u}_{1}) + \cdots + r_{m}S(\mathbf{u}_{m}) = r_{1}\mathbf{u}_{1} + \cdots + r_{m}\mathbf{u}_{m}\) is in \(U\). So \(im \;S \subseteq U\).
Exercise \(\PageIndex{30}\)
Let \(V\) and \(W\) be finite dimensional vector spaces.
- Show that \(dim \;W \leq dim \;V\) if and only if there exists an onto linear transformation \(T : V \to W\). [Hint: Theorem [thm:019430] and Theorem [thm:020916].]
- Show that \(dim \;W \geq dim \;V\) if and only if there exists a one-to-one linear transformation \(T : V \to W\). [Hint: Theorem [thm:019430] and Theorem [thm:020916].]
- Answer
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Let \(A\) and \(B\) be \(n \times n\) matrices, and assume that \(AXB=0\), \(X \in\|{M}_{nn}\), implies \(X=0\). Show that \(A\) and \(B\) are both invertible. [Hint: Dimension Theorem.]