# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 9/latex

\setcounter{section}{9}
\subtitle {Base change}
We know, due to Theorem 8.4 , that in a finite-dimensional vector space, any two bases have the same length, the same number of vectors. Every vector has, with respect to every basis, unique coordinates \extrabracket {the coefficient tuple} {.} How do these coordinates behave when we change the bases? This is answered by the following statement.
\factsituation {Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let \mathcor {\mathfrak{ v } = v_1 , \ldots , v_n} {and} {\mathfrak{ w } = w_1 , \ldots , w_n} denote bases of $V$.} \factcondition {Suppose that \mathrelationchaindisplay {\relationchain {v_j } { =} { \sum_{ i = 1 }^{ n } c_{ij} w_i } { } { } { } { } { } { } } with coefficients \mathrelationchain {\relationchain { c_{ij} } { \in }{ K } { }{ } { }{ } { }{ } }{,} which we collect into the $n \times n$-matrix \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ v } }_{ \mathfrak{ w } } } { =} { { \left( c_{ij} \right) }_{ij} } { } { } { } { } { } { } }{.}} \factconclusion {Then a vector $u$, which has the coordinates $\begin{pmatrix} s_{1 } \\ \vdots\\ s_{ n } \end{pmatrix}$ with respect to the basis $\mathfrak{ v }$, has the coordinates \mathrelationchaindisplay {\relationchain {\begin{pmatrix} t _{1 } \\ \vdots\\ t _{ n } \end{pmatrix} } { =} { M^{ \mathfrak{ v } }_{ \mathfrak{ w } } \begin{pmatrix} s_{1 } \\ \vdots\\ s_{ n } \end{pmatrix} } { =} { \begin{pmatrix} c_{11 } & c_{1 2} & \ldots & c_{1 n } \\ c_{21 } & c_{2 2} & \ldots & c_{2 n } \\ \vdots & \vdots & \ddots & \vdots \\ c_{ n 1 } & c_{ n 2 } & \ldots & c_{ n n } \end{pmatrix} \begin{pmatrix} s_{1 } \\ \vdots\\ s_{ n } \end{pmatrix} } { } { } { } { } } with respect to the basis $\mathfrak{ w }$.} \factextra } {
This follows directly from \mathrelationchaindisplay {\relationchain { u } { =} { \sum_{ j = 1 }^{ n } s_j v_j } { =} { \sum_{ j = 1 }^{ n } s_j { \left( \sum_{ i = 1 }^{ n } c_{ij} w_i \right) } } { =} { \sum_{ i = 1 }^{ n } { \left( \sum_{ j = 1 }^{ n } s_j c_{ij} \right) } w_i } { } { } }{,} and the definition of matrix multiplication.
If for a basis $\mathfrak{ v }$, we consider the corresponding bijective mapping \extrabracket {see Remark 7.12 } \mathdisp {\Psi_ \mathfrak{ v } \colon K^n \longrightarrow V} { , } then we can express the preceding statement as saying that the triangle \mathdisp {\begin{matrix}K^n & \stackrel{ M^{ \mathfrak{ v } }_{ \mathfrak{ w } } }{\longrightarrow} & K^n & \\ & \!\!\! \!\! \Psi_ \mathfrak{ v } \searrow & \downarrow \Psi_ \mathfrak{ w } \!\!\! \!\! & \\ & & V & \!\!\!\!\! \!\!\! \\ \end{matrix}} { } commutes\extrafootnote {The commutativity of such a diagram of arrows and mappings means that all composed mappings coincide as long as their domain and codomain coincide. In this case, it simply means that \mathrelationchain {\relationchain { \Psi_ \mathfrak{ v } } { = }{ \Psi_ \mathfrak{ w } \circ M^{ \mathfrak{ v } }_{ \mathfrak{ w } } } { }{ } { }{ } { }{ } } holds} {.} {.}
\inputdefinition { } {
Let $K$ denote a field and let $V$ denote a $K$-vector space of dimension $n$. Let \mathcor {\mathfrak{ v } = v_1 , \ldots , v_n} {and} {\mathfrak{ w } = w_1 , \ldots , w_n} denote two bases of $V$. Let \mathrelationchaindisplay {\relationchain { v_j } { =} { \sum_{ i = 1 }^{ n } c_{ij} w_i } { } { } { } { } { } { } } with coefficients \mathrelationchain {\relationchain { c_{ij}
} { \in }{ K } { }{ } { }{ } { }{ } }{.} Then the $n \times n$-matrix \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ v } }_{ \mathfrak{ w } } } { =} {(c_{ij})_{ij} } { } { } { } { } { } { } }
\inputremark {
The $j$-th column of a transformation matrix \mathl{M^{ \mathfrak{ v } }_{ \mathfrak{ w } }} consists of the coordinates of $v_j$ with respect to the basis $\mathfrak{ w }$. The vector $v_j$ has the coordinates $e_j$ with respect to the basis $\mathfrak{ v }$, and when we apply the matrix to $e_j$ we get the $j$-th column of the Matrix, and this is just the coordinate tuple of $v_j$ with respect to the basis $\mathfrak{ w }$.
For a one-dimensional space and \mathrelationchaindisplay {\relationchain {v } { =} {cw } { } { } { } { } { } { } }{,} we have \mathrelationchain {\relationchain { M^{ \mathfrak{ v } }_{ \mathfrak{ w } } } { = }{ c } { = }{ { \frac{ v }{ w } } } { }{ } { }{ } }{,} where the fraction is well-defined. This might help in memorizing the order of the bases in this notation.
Another important relation is \mathrelationchaindisplay {\relationchain { \mathfrak{ v } } { =} { { { \left( M^{ \mathfrak{ v } }_{ \mathfrak{ w } } \right) } ^{ \text{tr} } } \mathfrak{ w } } { } { } { } { } { } { } }{.} Note that here, the matrix is not applied to an $n$-tuple of $K$, but to an $n$-tuple of $V$, yielding a new $n$-tuple of $V$. This equation be an argument to define the transformation matrix the other way round, however, we consider the behavior in Lemma 9.1
as decisive.
In case \mathrelationchaindisplay {\relationchain {V } { =} {K^n } { } { } { } { } { } { } }{,} if $\mathfrak{ e }$ is the standard basis and $\mathfrak{ v }$ some further basis, we obtain the transformation matrix \mathl{M^{ \mathfrak{ e } }_{ \mathfrak{ v } }} of the base change from $\mathfrak{ e }$ to $\mathfrak{ v }$ by expressing $e_j$ as a linear combination of the basis vectors \mathl{v_1 , \ldots , v_n}{,} and write down the corresponding tuple as column. The inverse transformation matrix, \mathl{M^{ \mathfrak{ v } }_{ \mathfrak{ e } }}{,} consists simply in \mathl{v_1 , \ldots , v_n}{,} written as columns.
\inputexample {
We consider in $\R^2$ the standard basis, \mathrelationchaindisplay {\relationchain { \mathfrak{ u } } { =} { \begin{pmatrix} 1 \\0 \end{pmatrix} , \, \begin{pmatrix} 0 \\1 \end{pmatrix} } { } { } { } { } { } { } }{,} and the basis \mathrelationchaindisplay {\relationchain { \mathfrak{ v } } { =} { \begin{pmatrix} 1 \\2 \end{pmatrix} , \, \begin{pmatrix} -2 \\3 \end{pmatrix} } { } { } { } { } { } { } }{.} The basis vectors of $\mathfrak{ v }$ can be expressed directly with the standard basis, namely \mathdisp {v_1= \begin{pmatrix} 1 \\2 \end{pmatrix} = 1 \begin{pmatrix} 1 \\0 \end{pmatrix} + 2 \begin{pmatrix} 0 \\1 \end{pmatrix} \text{ and } v_2= \begin{pmatrix} -2 \\3 \end{pmatrix} = -2 \begin{pmatrix} 1 \\0 \end{pmatrix} + 3 \begin{pmatrix} 0 \\1 \end{pmatrix}} { . } Therefore, we get immediately \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ v } }_{ \mathfrak{ u } } } { =} { \begin{pmatrix} 1 & -2 \\ 2 & 3 \end{pmatrix} } { } { } { } { } { } { } }{.} For example, the vector which has with respect to $\mathfrak{ v }$ the coordinates \mathl{(4,-3)}{,} has the coordinates \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ v } }_{ \mathfrak{ u } } \begin{pmatrix} 4 \\-3 \end{pmatrix} } { =} { \begin{pmatrix} 1 & -2 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 4 \\-3 \end{pmatrix} } { =} { \begin{pmatrix} 10 \\-1 \end{pmatrix} } { } { } { } { } } with respect to the standard basis $\mathfrak{ u }$. The transformation matrix \mathl{M^{ \mathfrak{ u } }_{ \mathfrak{ v } }} is more difficult to compute: We have to write the standard vectors as linear combinations of \mathcor {v_1} {and} {v_2} {.} A direct computation \extrabracket {solving two linear systems} yields \mathrelationchaindisplay {\relationchain { \begin{pmatrix} 1 \\0 \end{pmatrix} } { =} { { \frac{ 3 }{ 7 } } \begin{pmatrix} 1 \\2 \end{pmatrix} - { \frac{ 2 }{ 7 } } \begin{pmatrix} -2 \\3 \end{pmatrix} } { } { } { } { } { } { } } and \mathrelationchaindisplay {\relationchain { \begin{pmatrix} 0 \\1 \end{pmatrix} } { =} { { \frac{ 2 }{ 7 } } \begin{pmatrix} 1 \\2 \end{pmatrix} + { \frac{ 1 }{ 7 } } \begin{pmatrix} -2 \\3 \end{pmatrix} } { } { } { } { } { } { } }{.} Hence, \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ u } }_{ \mathfrak{ v } } } { =} { \begin{pmatrix} { \frac{ 3 }{ 7 } } & { \frac{ 2 }{ 7 } } \\ - { \frac{ 2 }{ 7 } } & { \frac{ 1 }{ 7 } } \end{pmatrix} } { } { } { } { } { } { } }{.}
\factsituation {Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let \mathcor {\mathfrak{ u } = u_1 , \ldots , u_n ,\, \mathfrak{ v } = v_1 , \ldots , v_n} {and} {\mathfrak{ w } = w_1 , \ldots , w_n} denote bases of $V$.} \factconclusion {Then the three transformation matrices fulfill the relation \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ u } }_{ \mathfrak{ w } } } { =} { M^{ \mathfrak{ v } }_{ \mathfrak{ w } } \circ M^{ \mathfrak{ u } }_{ \mathfrak{ v } } } { } { } { } { } { } { } }{.}} \factextra {In particular, we have \mathrelationchaindisplay {\relationchain { M^{ \mathfrak{ u } }_{ \mathfrak{ v } } \circ M^{ \mathfrak{ v } }_{ \mathfrak{ u } } } { =} { E_n } { } { } { } { } { } { } }{.}}
\subtitle {Sum of linear subspaces}
\inputdefinition { } {
For a $K$-vector space $V$ and a family of linear subspaces \mathrelationchain {\relationchain { U_1 , \ldots , U_n } { \subseteq }{ V } { }{ } { }{ } { }{ } }{,} we define the \definitionword {sum of these linear subspaces} by \mathrelationchaindisplay {\relationchain { U_1 + \cdots + U_n } { =} { { \left\{ u_1 + \cdots + u_n \mid u _i \in U_i \right\} } } { } { } { } { } { } { } }
This sum is again a linear subspace. In case \mathrelationchaindisplay {\relationchain {V } { =} { U_1 + \cdots + U_n } { } { } { } { } { } { } }{,} we say that $V$ is the Sum of the linear subspaces \mathl{U_1 , \ldots , U_n}{.} The following theorem describes an important relation between the dimension of the sum of two linear subspaces and the dimension of their intersection.
\factsituation {Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let \mathrelationchain {\relationchain { U_1,U_2 } { \subseteq }{ V } { }{ } { }{ } { }{ } } denote linear subspaces.} \factconclusion {Then \mathrelationchaindisplay {\relationchain { \dim_{ K } { \left( U_1 \right) } + \dim_{ K } { \left( U_2 \right) } } { =} { \dim_{ K } { \left( U_1 \cap U_2 \right) } + \dim_{ K } { \left( U_1 + U_2 \right) } } { } { } { } { } { } { } }{.}} \factextra } {
Let \mathl{w_1 , \ldots , w_k} be a basis of \mathl{U_1 \cap U_2}{.} On one hand, we can complete this basis, according to Theorem 8.10 , to a basis \mathl{w_1 , \ldots , w_k, u_1 , \ldots , u_n} of $U_1$, on the other hand, we can complete it to a basis \mathl{w_1 , \ldots , w_k, v_1 , \ldots , v_m} of $U_2$. Then \mathdisp {w_1 , \ldots , w_k, u_1 , \ldots , u_n , v_1 , \ldots , v_m} { } is a generating system of \mathl{U_1+U_2}{.} We claim that it is even a basis. To see this, let \mathrelationchaindisplay {\relationchain { a_1w_1 + \cdots + a_k w_k + b_1 u_1 + \cdots + b_n u_n + c_1 v_1 + \cdots + c_mv_m } { =} { 0 } { } { } { } { } { } { } }{.} This implies that the element \mathrelationchaindisplay {\relationchain { a_1w_1 + \cdots + a_k w_k + b_1 u_1 + \cdots + b_n u_n } { =} {- c_1 v_1 - \cdots - c_mv_m } { } { } { } { } { } { } } belongs to \mathl{U_1 \cap U_2}{.} From this we get directly \mathrelationchain {\relationchain {b_i } { = }{ 0 } { }{ } { }{ } { }{ } } for \mathrelationchain {\relationchain { i } { = }{ 1 , \ldots , n } { }{ } { }{ } { }{ } } and \mathrelationchain {\relationchain {c_j } { = }{ 0 } { }{ } { }{ } { }{ } } for \mathrelationchain {\relationchain { j } { = }{ 1 , \ldots , m } { }{ } { }{ } { }{ } }{.} From this we can infer that also \mathrelationchain {\relationchain { a_\ell } { = }{ 0 } { }{ } { }{ } { }{ } } for all $\ell$. Hence, we have
linear independence. This gives altogether \mathrelationchainalign {\relationchainalign { \dim_{ K } { \left( U_1 \cap U_2 \right) } + \dim_{ K } { \left( U_1 + U_2 \right) } } { =} { k + k +n +m } { =} { k+n +k+m } { =} { \dim_{ K } { \left( U_1 \right) } + \dim_{ K } { \left( U_2 \right) } } { } { } }{.}
The intersection of two planes in $\R^3$ is \quotationshort{usually} a line; it is the plane itself if the same plane is taken twice, but it is never just a point. This observation is generalized in the following statement.
\factsituation {Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let \mathrelationchain {\relationchain { U_1,U_2 } { \subseteq }{ V } { }{ } { }{ } { }{ } } denote linear subspaces of dimension \mathcor {\dim_{ K } { \left( U_1 \right) } = n-k_1} {and} {\dim_{ K } { \left( U_2 \right) } = n-k_2} {.}} \factconclusion {Then \mathrelationchaindisplay {\relationchain { \dim_{ K } { \left( U_1 \cap U_2 \right) } } { \geq} { n-k_1 -k_2 } { } { } { } { } { } { } }{.}} \factextra } {
Due to Theorem 9.7 , we have \mathrelationchainalign {\relationchainalign { \dim_{ K } { \left( U_1 \cap U_2 \right) } } { =} { \dim_{ K } { \left( U_1 \right) } + \dim_{ K } { \left( U_2 \right) } - \dim_{ K } { \left( U_1 +U_2 \right) } } { =} { n-k_1 + n-k_2 - \dim_{ K } { \left( U_1 +U_2 \right) } } { \geq} { n-k_1 + n-k_2 -n } { =} { n-k_1-k_2 } }{.}
Recall that to a linear subspace \mathrelationchain {\relationchain {U } { \subseteq }{V } { }{ } { }{ } { }{ } }{,} the difference \mathl{\dim_{ K } { \left( V \right) } - \dim_{ K } { \left( U \right) }} is called the \keyword {codimension} of $U$ in $V$. With this concept, we can paraphrase the statement above by saying that the codimension of an intersection of linear subspaces equals at most the sum of their codimensions.
\factsituation {Let a homogeneous system of linear equations with $k$ equations in $n$ variables be given.} \factconclusion {Then the dimension of the solution space of the system is at least \mathl{n-k}{.}} \factextra } {
The solution space of one linear equation in $n$ variables has dimension $n-1$ or $n$. The solution space of the system is the intersection of the solution spaces of the individual equations. Therefore, the statement follows by applying Corollary 9.8
to the individual solution spaces.
\subtitle {Direct sum}
\inputdefinition { } {
Let $K$ denote a field and let $V$ denote a $K$-vector space. Let \mathl{U_1 , \ldots , U_m} be a family of linear subspaces of $V$. We say that $V$ is the \definitionword {direct sum} of the \mathl{U_i}{,} if the following conditions are fulfilled. \enumerationtwo {Every vector \mathrelationchain {\relationchain { v } { \in }{V } { }{ } { }{ } { }{ } } has a representation \mathrelationchaindisplay {\relationchain {v } { =} {u_1+u_2 + \cdots + u_m } { } { } { } { } { } { } }{,} where \mathrelationchain {\relationchain { u_i } { \in }{ U_i } { }{ } { }{ } { }{ } }{.} } { \mathrelationchain {\relationchain {U_i \cap { \left( \sum_{j \neq i} U_j \right) } } { = }{ 0 } { }{ } { }{ } { }{ } } for all $i$.
If the sum of the $U_i$ is direct, then we write also \mathl{U_1 \oplus \cdots \oplus U_m} instead of \mathl{U_1 + \cdots + U_m}{.} For two linear subspaces \mathrelationchaindisplay {\relationchain { U_1,U_2 } { \subseteq} { V } { } { } { } { } { } { } }{,} the second condition just means \mathrelationchain {\relationchain {U_1 \cap U_2 } { = }{ 0 } { }{ } { }{ } { }{ } }{.}
\inputexample {
Let $V$ denote a finite-dimensional $K$-vector space together with a basis \mathl{v_1 , \ldots , v_n}{.} Let \mathrelationchaindisplay {\relationchain { \{ 1 , \ldots , n\} } { =} { I_1 \uplus \ldots \uplus I_k } { } { } { } { } { } { } } be a partition of the index set. Let \mathrelationchaindisplay {\relationchain { U_j } { =} { \langle v_i ,\, i \in I_j \rangle } { } { } { } { } { } { } } be the linear subspaces generated by the subfamilies. Then \mathrelationchaindisplay {\relationchain { V } { =} { U_1 \oplus \cdots \oplus U_k } { } { } { } { } { } { } }{.} The extreme case \mathrelationchain {\relationchain { I_j } { = }{ \{j\} } { }{ } { }{ } { }{ } } yields the direct sum \mathrelationchaindisplay {\relationchain {V } { =} { K v_1 \oplus \cdots \oplus K v_n } { } { } { } { } { } { } } with one-dimensional linear subspaces.
\factsituation {Let $V$ be a finite-dimensional $K$-vector space and let \mathrelationchain {\relationchain { U } { \subseteq }{ V } { }{ } { }{ } { }{ } } be a linear subspace.} \factconclusion {Then there exists a linear subspace \mathrelationchain {\relationchain { W } { \subseteq }{ V } { }{ } { }{ } { }{ } } such that we have the direct sum decomposition \mathrelationchaindisplay {\relationchain { V } { =} { U \oplus W } { } { } { } { } { } { } }{.}} \factextra } {
Let \mathl{v_1 , \ldots , v_k} denote a basis of $U$. We can extend this basis, according to Theorem 8.10 , to a basis \mathl{v_1 , \ldots , v_k, v_{k+1} , \ldots , v_n} of $V$. Then \mathrelationchaindisplay {\relationchain { W } { =} { \langle v_{k+1} , \ldots , v_n \rangle } { } { } { } { } { } { } } fulfills all the properties of a direct sum.
In the preceding statement, the linear subspace $W$ is called a \keyword {direct complement} for $U$ \extrabracket {in $V$} {.} In general, there are many different direct complements.
\subtitle {Direct sum and product}
Recall that for a family \mathcond {M_i} {i \in I} {,} of sets $M_i$, the product set \mathl{\prod_{i \in I} M_i} is defined. If all \mathrelationchain {\relationchain { M_i } { = }{ V_i } { }{ } { }{ } { }{ } } are $K$-vector spaces over a field $K$, then this is, using componentwise addition and scalar multiplication, again a $K$-vector space. This is called the \keyword {direct product of vector spaces} {.} If it is always the same space, say \mathrelationchain {\relationchain {M_i } { = }{V } { }{ } { }{ } { }{ } }{,} then we write also \mathl{V^{I}}{.} This is just the mapping space \mathl{\operatorname{Map} \, { \left( I , V \right) }}{.}
The vector space $V_j$ is a linear subspace inside the direct product, namely as the set of all tuples \mathdisp {(x_i)_{i \in I} \text{ with } x_i = 0 \text{ for alle } i \neq j} { . } The set of all these tuples, which are only \extrabracket {at most} at one place different from $0$, generate a linear subspace, which is for $I$ infinite not the direct product.
\inputdefinition { } {
Let $I$ denote a set and let $K$ denote a field. Suppose that for every \mathrelationchain {\relationchain { i } { \in }{ I } { }{ } { }{ } { }{ } }{,} a $K$-vector space $V_i$ is given. Then the set \mathrelationchaindisplay {\relationchain { \bigoplus_{i \in I} V_i } { =} { { \left\{ (v_i)_{ i \in I} \mid v_i \in V_i , \, v_i \neq 0 \text{ for only finitely many } i \right\} } } { } { } { } { } { } { } }
If we always have the same vector space, then we write \mathl{V^{(I)}} for this direct sum. In particular, \mathrelationchaindisplay {\relationchain { V^{(I)} } { \subseteq} {V^{I} } { } { } { } { } { } { } } is a linear subspace. For $I$ finite, there is no difference, but for an infinite index set, this inclusion is strict. For example, \mathl{\R^\N} is the space of all real sequences, but \mathl{\R^{(\N)}} consists only of those sequences with the property that only finitely many members are different from $0$. The polynomial ring \mathl{K[X]} is the direct sum of the vector spaces \mathl{KX^n,\, n \in \N}{.} Every $K$-vector space with a basis \mathcond {v_i} {i \in I} {,} is \quotationshort{isomorphic} to the direct sum \mathl{\bigoplus_{i \in I} Kv_i}{.}