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

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 (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

${}M_{\mathfrak {w}}^{\mathfrak {v}}={\left(c_{ij}\right)}_{ij}\,.$

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

${}{\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}=M_{\mathfrak {w}}^{\mathfrak {v}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}={\begin{pmatrix}c_{11}&c_{12}&\ldots &c_{1n}\\c_{21}&c_{22}&\ldots &c_{2n}\\\vdots &\vdots &\ddots &\vdots \\c_{n1}&c_{n2}&\ldots &c_{nn}\end{pmatrix}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\,$

with respect to the basis ${}{\mathfrak {w}}$.

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 7.12 )

$\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,$

then we can express the preceding statement as saying that the triangle

${\begin{matrix}K^{n}&{\stackrel {M_{\mathfrak {w}}^{\mathfrak {v}}}{\longrightarrow }}&K^{n}&\\&\!\!\!\!\!\Psi _{\mathfrak {v}}\searrow &\downarrow \Psi _{\mathfrak {w}}\!\!\!\!\!&\\&&V&\!\!\!\!\!\!\!\!\\\end{matrix}}$

commutes.

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

${}M_{\mathfrak {w}}^{\mathfrak {v}}=(c_{ij})_{ij}\,$

is called the transformation matrix of the base change from ${}{\mathfrak {v}}$ to ${}{\mathfrak {w}}$.

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 coordinate tuple ${}e_{j}$ 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 might be an argument to define the transformation matrix the other way around; however, we consider the behavior in Lemma 9.1 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 each ${}e_{j}$ as a linear combination of the basis vectors ${}v_{1},\ldots ,v_{n}$, and writing down the corresponding tuples as columns. 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 that has the coordinates ${}(4,-3)$ with respect to ${}{\mathfrak {v}}$, 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

${}M_{\mathfrak {w}}^{\mathfrak {u}}=M_{\mathfrak {w}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {u}}\,.$

In particular, we have

${}M_{\mathfrak {v}}^{\mathfrak {u}}\circ M_{\mathfrak {u}}^{\mathfrak {v}}=E_{n}\,.$

Proof

See Exercise 9.9 .

$\Box$

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

${}\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}=\dim _{K}{\left(U_{1}\cap U_{2}\right)}+\dim _{K}{\left(U_{1}+U_{2}\right)}\,.$

Proof

Let ${}w_{1},\ldots ,w_{k}$ be a basis of ${}U_{1}\cap U_{2}$. On one hand, we can extend this basis, according to Theorem 8.10 , to a basis ${}w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n}$ of ${}U_{1}$, on the other hand, we can extend 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 the equation before, we can then infer that also ${}a_{\ell }=0$ holds for all ${}\ell$. Hence, we have linear independence. This gives altogether

${}{\begin{aligned}\dim _{K}{\left(U_{1}\cap U_{2}\right)}+\dim _{K}{\left(U_{1}+U_{2}\right)}&=k+k+n+m\\&=k+n+k+m\\&=\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}.\end{aligned}}$

$\Box$

The intersection of two planes (through the origin) 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 dimensions ${}\dim _{K}{\left(U_{1}\right)}=n-k_{1}$ and ${}\dim _{K}{\left(U_{2}\right)}=n-k_{2}$. Then

${}\dim _{K}{\left(U_{1}\cap U_{2}\right)}\geq n-k_{1}-k_{2}\,.$

Proof

Due to Theorem 9.7 , we have

${}{\begin{aligned}\dim _{K}{\left(U_{1}\cap U_{2}\right)}&=\dim _{K}{\left(U_{1}\right)}+\dim _{K}{\left(U_{2}\right)}-\dim _{K}{\left(U_{1}+U_{2}\right)}\\&=n-k_{1}+n-k_{2}-\dim _{K}{\left(U_{1}+U_{2}\right)}\\&\geq n-k_{1}+n-k_{2}-n\\&=n-k_{1}-k_{2}.\end{aligned}}$

$\Box$

Recall that, for a linear subspace ${}U\subseteq V$, the difference ${}\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}$ 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

of the solution space of the system is at least ${}n-k$.

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 Corollary 9.8 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.

1. Every vector ${}v\in V$ has a representation

   ${}v=u_{1}+u_{2}+\cdots +u_{m}\,,$

   where ${}u_{i}\in U_{i}$.

2. ${}U_{i}\cap {\left(\sum _{j\neq i}U_{j}\right)}=0$ for all ${}i$.

If the sum of the ${}U_{i}$ is direct, then we also write ${}U_{1}\oplus \cdots \oplus U_{m}$ 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

${}V=U_{1}\oplus \cdots \oplus U_{k}\,.$

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 Theorem 8.10 , 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 also write ${}V^{I}$. This is just the mapping space ${}\operatorname {Map} \,{\left(I,V\right)}$.

Each vector space ${}V_{j}$ is a linear subspace inside the direct product, namely as the set of all tuples

$(x_{i})_{i\in I}{\text{ with }}x_{i}=0{\text{ for all }}i\neq j.$

The set of all these tuples that are only (at most) at one place different from ${}0$ generates a linear subspace of the direct product. For ${}I$ infinite, it is 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

${}\bigoplus _{i\in I}V_{i}={\left\{(v_{i})_{i\in I}\mid v_{i}\in V_{i},\,v_{i}\neq 0{\text{ for only finitely many }}i\right\}}\,$

is called the direct sum of the ${}V_{i}$.

We have the linear subspace relation

${}\bigoplus _{i\in I}V_{i}\subseteq \prod _{i\in I}V_{i}\,.$

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 satisfying the property that only finitely many members are different from ${}0$. 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 "isomorphic“ to the direct sum ${}\bigoplus _{i\in I}Kv_{i}$.

Footnotes

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