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

    Base change
We know, due to fact, 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 (the coefficient tuple). How do these coordinates behave when we change the bases? This is answered by the following statement.

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote bases of $V$. Suppose that
    $v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$
with coefficients $c_{ij}\in K$, which we collect into the $n\times n$-matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}={\left(c_{ij}\right)}_{ij}\,.$
$u$
${\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}$
${\mathfrak {v}}$
    ${\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}=M_{\mathfrak {w}}^{\mathfrak {v}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}={\begin{pmatrix}c_{11}&c_{12}&\ldots &c_{1n}\\c_{21}&c_{22}&\ldots &c_{2n}\\\vdots &\vdots &\ddots &\vdots \\c_{n1}&c_{n2}&\ldots &c_{nn}\end{pmatrix}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\,$
${\mathfrak {w}}$

### Proof

This follows directly from
    $u=\sum _{j=1}^{n}s_{j}v_{j}=\sum _{j=1}^{n}s_{j}{\left(\sum _{i=1}^{n}c_{ij}w_{i}\right)}=\sum _{i=1}^{n}{\left(\sum _{j=1}^{n}s_{j}c_{ij}\right)}w_{i}\,,$
and the definition of matrix multiplication.
$\Box$
If for a basis ${\mathfrak {v}}$, we consider the corresponding bijective mapping (see remark)
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,$
    ${\begin{matrix}K^{n}&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}}{\longrightarrow }}&K^{n}&\\&\!\!\!\!\!\Psi _{\mathfrak {v}}\searrow &\downarrow \Psi _{\mathfrak {w}}\!\!\!\!\!&\\&&V&\!\!\!\!\!\!\!\!\\\end{matrix}}$

## Definition

Let $K$ denote a field and let $V$ denote a $K$-vector space of dimension $n$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote two bases of $V$. Let
    $v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$
with coefficients $c_{ij}\in K$. Then the $n\times n$-matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}=(c_{ij})_{ij}\,$
${\mathfrak {v}}$
${\mathfrak {w}}$

## Remark

The $j$-th column of a transformation matrix $M_{\mathfrak {w}}^{\mathfrak {v}}$ 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
    $v=cw\,,$
we have $M_{\mathfrak {w}}^{\mathfrak {v}}=c={\frac {v}{w}}$, where the fraction is well-defined. This might help in memorizing the order of the bases in this notation.
Another important relation is
    ${\mathfrak {v}}={{\left(M_{\mathfrak {w}}^{\mathfrak {v}}\right)}^{\text{tr}}}{\mathfrak {w}}\,.$
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 fact as decisive.
In case
    $V=K^{n}\,,$
if ${\mathfrak {e}}$ is the standard basis and ${\mathfrak {v}}$ some further basis, we obtain the transformation matrix $M_{\mathfrak {v}}^{\mathfrak {e}}$ of the base change from ${\mathfrak {e}}$ to ${\mathfrak {v}}$ by expressing $e_{j}$ as a linear combination of the basis vectors $v_{1},\ldots ,v_{n}$, and write down the corresponding tuple as column. The inverse transformation matrix, $M_{\mathfrak {e}}^{\mathfrak {v}}$, consists simply in $v_{1},\ldots ,v_{n}$, written as columns.

## Example

We consider in $\mathbb {R} ^{2}$ the standard basis,
    ${\mathfrak {u}}={\begin{pmatrix}1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\end{pmatrix}}\,,$
and the basis
    ${\mathfrak {v}}={\begin{pmatrix}1\\2\end{pmatrix}},\,{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$
The basis vectors of ${\mathfrak {v}}$ can be expressed directly with the standard basis, namely
    $v_{1}={\begin{pmatrix}1\\2\end{pmatrix}}=1{\begin{pmatrix}1\\0\end{pmatrix}}+2{\begin{pmatrix}0\\1\end{pmatrix}}\,\,{\text{ and }}\,\,v_{2}={\begin{pmatrix}-2\\3\end{pmatrix}}=-2{\begin{pmatrix}1\\0\end{pmatrix}}+3{\begin{pmatrix}0\\1\end{pmatrix}}.$
Therefore, we get immediately
    $M_{\mathfrak {u}}^{\mathfrak {v}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}\,.$
For example, the vector which has with respect to ${\mathfrak {v}}$ the coordinates $(4,-3)$, has the coordinates
    $M_{\mathfrak {u}}^{\mathfrak {v}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}10\\-1\end{pmatrix}}\,$
with respect to the standard basis ${\mathfrak {u}}$. The transformation matrix $M_{\mathfrak {v}}^{\mathfrak {u}}$ is more difficult to compute: We have to write the standard vectors as linear combinations of $v_{1}$ and $v_{2}$. A direct computation (solving two linear systems) yields
    ${\begin{pmatrix}1\\0\end{pmatrix}}={\frac {3}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}-{\frac {2}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,$
and
    ${\begin{pmatrix}0\\1\end{pmatrix}}={\frac {2}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}+{\frac {1}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$
Hence,
    $M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {2}{7}}\\-{\frac {2}{7}}&{\frac {1}{7}}\end{pmatrix}}\,.$

## Lemma

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let ${\mathfrak {u}}=u_{1},\ldots ,u_{n},\,{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote bases of $V$. Then the three transformation matrices fulfill the relation
    $M_{\mathfrak {w}}^{\mathfrak {u}}=M_{\mathfrak {w}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {u}}\,.$
    $M_{\mathfrak {v}}^{\mathfrak {u}}\circ M_{\mathfrak {u}}^{\mathfrak {v}}=E_{n}\,.$

### Proof

$\Box$
    Sum of linear subspaces

## Definition

For a $K$-vector space $V$ and a family of linear subspaces $U_{1},\ldots ,U_{n}\subseteq V$, we define the sum of these linear subspaces by
    $U_{1}+\cdots +U_{n}={\left\{u_{1}+\cdots +u_{n}\mid u_{i}\in U_{i}\right\}}\,.$
This sum is again a linear subspace. In case
    $V=U_{1}+\cdots +U_{n}\,,$
we say that $V$ is the Sum of the linear subspaces $U_{1},\ldots ,U_{n}$. The following theorem describes an important relation between the dimension of the sum of two linear subspaces and the dimension of their intersection.

## Theorem

Let $K$ denote a field and let $V$ denote a $K$-vector space of finite dimension. Let $U_{1},U_{2}\subseteq V$ denote linear subspaces. Then
    $\dim _{\left(U_{1}\right)}+\dim _{\left(U_{2}\right)}=\dim _{\left(U_{1}\cap U_{2}\right)}+\dim _{\left(U_{1}+U_{2}\right)}\,.$

### Proof

Let $w_{1},\ldots ,w_{k}$ be a basis of $U_{1}\cap U_{2}$. On one hand, we can complete this basis, according to fact, to a basis $w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n}$ of $U_{1}$, on the other hand, we can complete it to a basis $w_{1},\ldots ,w_{k},v_{1},\ldots ,v_{m}$ of $U_{2}$. Then
    $w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n},v_{1},\ldots ,v_{m}$
is a generating system of $U_{1}+U_{2}$. We claim that it is even a basis. To see this, let
    $a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}+c_{1}v_{1}+\cdots +c_{m}v_{m}=0\,.$
This implies that the element
    $a_{1}w_{1}+\cdots +a_{k}w_{k}+b_{1}u_{1}+\cdots +b_{n}u_{n}=-c_{1}v_{1}-\cdots -c_{m}v_{m}\,$
belongs to $U_{1}\cap U_{2}$. From this we get directly $b_{i}=0$ for $i=1,\ldots ,n$ and $c_{j}=0$ for $j=1,\ldots ,m$. From this we can infer that also $a_{\ell }=0$ for all $\ell$. Hence, we have
linear independence. This gives altogether
    ${\begin{aligned}\dim _{\left(U_{1}\cap U_{2}\right)}+\dim _{\left(U_{1}+U_{2}\right)}&=k+k+n+m\\&=k+n+k+m\\&=\dim _{\left(U_{1}\right)}+\dim _{\left(U_{2}\right)}.\end{aligned}}$
$\Box$
The intersection of two planes in $\mathbb {R} ^{3}$ is "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.

## Corollary

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let $U_{1},U_{2}\subseteq V$ denote linear subspaces of dimension $\dim _{\left(U_{1}\right)}=n-k_{1}$ bzw. $\dim _{\left(U_{2}\right)}=n-k_{2}$. Then
    $\dim _{\left(U_{1}\cap U_{2}\right)}\geq n-k_{1}-k_{2}\,.$

### Proof

Due to fact, we have
    ${\begin{aligned}\dim _{\left(U_{1}\cap U_{2}\right)}&=\dim _{\left(U_{1}\right)}+\dim _{\left(U_{2}\right)}-\dim _{\left(U_{1}+U_{2}\right)}\\&=n-k_{1}+n-k_{2}-\dim _{\left(U_{1}+U_{2}\right)}\\&\geq n-k_{1}+n-k_{2}-n\\&=n-k_{1}-k_{2}.\end{aligned}}$
$\Box$
Recall that to a linear subspace $U\subseteq V$, the difference $\dim _{\left(V\right)}-\dim _{\left(U\right)}$ is called the 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.

## Corollary

Let a homogeneous system of linear equations with $k$ equations in $n$ variables be given. Then the dimension
$n-k$

### Proof

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 fact to the individual solution spaces.
$\Box$
    Direct sum

## Definition

Let $K$ denote a field and let $V$ denote a $K$-vector space. Let $U_{1},\ldots ,U_{m}$ be a family of linear subspaces of $V$. We say that $V$ is the direct sum of the $U_{i}$, if the following conditions are fulfilled.
* Every vector $v\in V$ has a representation where $u_{i}\in U_{i}$.
    $v=u_{1}+u_{2}+\cdots +u_{m}\,,$
* $U_{i}\cap {\left(\sum _{j\neq i}U_{j}\right)}=0$ for all $i$.
If the sum of the $U_{i}$ is direct, then we write also $U_{1}\oplus \cdots \oplus U_{m}$ instead of $U_{1}+\cdots +U_{m}$. For two linear subspaces
    $U_{1},U_{2}\subseteq V\,,$
the second condition just means $U_{1}\cap U_{2}=0$.

## Example

Let $V$ denote a finite dimensional $K$-vector space together with a basis $v_{1},\ldots ,v_{n}$. Let
    $\{1,\ldots ,n\}=I_{1}\uplus \ldots \uplus I_{k}\,$
be a partition of the index set. Let
    $U_{j}=\langle v_{i},\,i\in I_{j}\rangle \,$
be the linear subspaces generated by the subfamilies. Then Dann ist
    $V=U_{1}\oplus \cdots \oplus U_{k}\,.$
The extreme case $I_{j}=\{j\}$ yields the direct sum
    $V=Kv_{1}\oplus \cdots \oplus Kv_{n}\,$
with one-dimensional linear subspaces.

## Lemma

Let $V$ be a finite dimensional $K$-vector space and let $U\subseteq V$ be a linear subspace. Then there exists a linear subspace $W\subseteq V$ such that we have the direct sum decomposition
    $V=U\oplus W\,$

### Proof

Let $v_{1},\ldots ,v_{k}$ denote a basis of $U$. We can extend this basis, according to fact, to a basis $v_{1},\ldots ,v_{k},v_{k+1},\ldots ,v_{n}$ of $V$. Then
    $W=\langle v_{k+1},\ldots ,v_{n}\rangle \,$
fulfills all the properties of a direct sum.
$\Box$
In the preceding statement, the linear subspace $W$ is called a direct complement for $U$ (in $V$). In general, there are many different direct complements.
    Direct sum and product
Recall that for a family $M_{i}$, $i\in I$, of sets $M_{i}$, the product set $\prod _{i\in I}M_{i}$ is defined. If all $M_{i}=V_{i}$ 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 direct product of vector spaces. If it is always the same space, say $M_{i}=V$, then we write also $V^{I}$. This is just the mapping space $\operatorname {Map} \,{\left(I,V\right)}$.
The vector space $V_{j}$ is a linear subspace inside the direct product, namely as the set of all tuples
    $(x_{i})_{i\in I}{\text{ with }}x_{i}=0{\text{ for alle }}i\neq j.$
The set of all these tuples, which are only (at most) at one place different from $0$, generate a linear subspace, which is for $I$ infinite not the direct product.

## Definition

Let $I$ denote a set and let $K$ denote a field. Suppose that for every $i\in I$, a $K$-vector space $V_{i}$ is given. Then the set
    $\bigoplus _{i\in I}V_{i}={\left\{(v_{i})_{i\in I}\mid v_{i}\in V_{i},\,v_{i}\neq 0{\text{ for only finitely many }}i\right\}}\,$
$V_{i}$
If we always have the same vector space, then we write $V^{(I)}$ for this direct sum. In particular,
    $V^{(I)}\subseteq V^{I}\,$
is a linear subspace. For $I$ finite, there is no difference, but for an infinite index set, this inclusion is strict. For example, $\mathbb {R} ^{\mathbb {N} }$ is the space of all real sequences, but $\mathbb {R} ^{(\mathbb {N} )}$ consists only of those sequences with the property that only finitely many members are different fro $0$. The polynomial ring $K[X]$ is the direct sum of the vector spaces $KX^{n},\,n\in \mathbb {N}$. Every $K$-vector space with a basis $v_{i}$, $i\in I$, is "isomorph“ to the direct summe $\bigoplus _{i\in I}Kv_{i}$.
    Footnotes
* ↑ The commutativity of such a diagram of arrows and mappings means that all composed mappings coincide as long astheir domain and codomain coincide. In this case, it simply means that $\Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}$ holds.