Definition 3.6.1. Basis.
A subset \(B\) of a vector space \(V\) is called a basis if
-
\(B\) is linearly independent.
If the basis \(B\) comes equipped with an ordering (i.e., \(B\) is an ordered set), then we call \(B\) and ordered basis
Section 3.6 Bases
Now that we have the notions of span and linear independence in place, we simply combine them to define a basis of a vector space. In the spirit of Section 3.5, a basis of a vector space \(V\) should be understood as a minimal spanning set.
This section includes many theoretical results. There are two in particular that are worth highlighting, especially in regard to computational techniques for abstract vector spaces:
-
If \(B\) is a basis of \(V\) containing exactly \(n\) elements, then any other basis \(B'\) also contains exactly \(n\) elements. (Theorem 3.7.3)
-
If \(B\) is a basis for \(V\text{,}\) then every element of \(V\) can be written as a linear combination of elements of \(B\) in a unique way. (Theorem 3.6.7)
The first result allows us to define the dimension of a vector space as the number of elements in any given basis. The second result allows us to take any \(n\)-dimensional vector space \(V\) with chosen basis \(B=\{\boldv_1, \boldv_2, \dots, \boldv_n\}\) and effectively identify vectors \(\boldv\in V\) with the sequence \((c_1,c_2,\dots, c_n)\in \R^n\text{,}\) where
\begin{equation*}
\boldv=c_1\boldv_1+c_2\boldv_2+\cdots c_n\boldv_n\text{.}
\end{equation*}
This observation has the following consequence: given any \(n\)-dimensional vector space \(V\text{,}\) no matter how exotic, once we choose a basis \(B\) of \(V\text{,}\) we can reduce any and all linear algebraic questions or computations about \(V\) to a corresponding question in \(\R^n\text{.}\) We will elaborate this idea further in Section 4.1.
Subsection Bases of vector spaces
Remark 3.6.2. Some standard bases.
The examples of standard spanning sets in Remark 3.5.7 are easily seen to be linearly independent, and hence are in fact bases. We list them again here, using the same notation, and refer to these as standard bases for the given spaces.
-
Zero space.Let \(V=\{\boldzero\}\text{.}\) The empty \(B=\emptyset=\{ \}\) is a basis for \(V\text{.}\) Note that \(B=\emptyset\) spans \(V\) by definition (Definition 3.5.1), and it satisfies the defining implication of linear independence (Definition 3.5.9) trivially.
-
Tuples.Let \(V=\R^n\text{.}\) The set \(B=\{\bolde_1, \bolde_2,\dots, \bolde_n\}\) is the standard basis of \(\R^n\text{.}\)
-
Matrices.Let \(V=M_{mn}\text{.}\) The set \(B=\{E_{ij}\colon 1\leq i\leq m, 1\leq j\leq n\}\) is the standard basis of \(M_{mn}\text{.}\)
-
Polynomials of bounded degree.Let \(V=P_n\text{.}\) The set \(B=\{x^n, x^{n-1}, \dots, x, 1\}\) is the standard basis of \(P_n\text{.}\)
-
Polynomials.Let \(V=P\text{,}\) the space of all polynomials. The set\begin{equation*} B=\{1, x, x^2, \dots\}=\{x^i\colon i\geq 0\} \end{equation*}is the standard basis of \(P\text{.}\)
Just as with spanning sets, bases are not in general unique: in fact, for any nonzero vector space there are infinitely many different bases.
Example 3.6.3. Some nonstandard bases.
For each \(V\) and \(B\) below, verify that \(B\) is a basis of \(V\text{.}\)
-
\(V=M_{22}\text{,}\)\begin{equation*} B=\left\{ \begin{bmatrix}3\amp 6\\ 3\amp -6 \end{bmatrix} , \begin{bmatrix}0\amp -1\\ -1\amp 0 \end{bmatrix} , \begin{bmatrix}0\amp -8\\ -12\amp -4 \end{bmatrix} , \begin{bmatrix}1\amp 0\\ -1\amp 2 \end{bmatrix} \right\}\text{.} \end{equation*}
Solution.
Each verification amounts to showing, using the techniques from Section 3.5, that the given \(B\) spans the given \(V\) and is linearly independent. We illustrate with (1) and (2), leaving (3) to the reader.
-
Since neither element of \(B=\{(1,1), (1,-1)\}\) is a scalar multiple of the other, the set is linearly independent. To see that \(B\) spans \(\R^2\) we show that for any \((c,d)\in \R^2\) we have\begin{equation*} a(1,1)+b(1,-1)=(c,d) \end{equation*}for some \(a,b\in \R\text{.}\) Indeed we may take \(a=\frac{1}{2}(c+d)\) and \(b=\frac{1}{2}(c-d)\text{.}\) (These formulas were obtained by solving the corresponding system of two equations in the unknowns \(a\) and \(b\text{.}\))
-
To show \(B\) spans \(P_2\) we must show that for any \(dx^2+ex+f\in P_2\) we can find \(a,b,c\in\R\) such that\begin{equation*} a(x^2+x+1)+b(x^2-x)+c(x^2-1)=dx^2+ex+f\text{,} \end{equation*}or\begin{equation*} (a+b+c)x^2+(a-b)x+(a-1)=dx^2+ex+f\text{.} \end{equation*}Equating coefficients yields the system of equations\begin{equation*} \begin{linsys}{3} a\amp +\amp b\amp +\amp c\amp = d\\ a\amp -\amp b\amp \amp \amp =e\\ a\amp \amp \amp -\amp c\amp =f \end{linsys}\text{,} \end{equation*}which corresponds to the matrix equation\begin{equation*} \underset{A}{\begin{amatrix}[rrr]1\amp 1\amp 1\\ 1\amp -1\amp 0\\ 1\amp 0\amp -1\end{amatrix}}\underset{\boldx}{\begin{bmatrix} a\\ b\\ c\end{bmatrix}}=\underset{\boldy}{\begin{bmatrix} d\\ e\\ f\end{bmatrix}}\text{.} \end{equation*}An easy computation shows \(\det A=3\text{,}\) and thus that \(A\) is invertible. We conclude that the system can be solved for \((a,b,c)\) (set \(\boldx=A^{-1}\boldy\)), and thus that \(B\) spans \(P_2\text{.}\)Our work above now can be used to also show that \(B\) is linearly independent. Replacing the arbitrary polynomial \(dx^2+ex+f\) with the zero polynomial \(0x^2+0x+0\text{,}\) we see that a linear combination\begin{equation*} a(x^2+x+1)+b(x^2-x)+c(x^2-1)=\boldzero \end{equation*}corresponds to a solution \(\boldx=(a,b,c)\) to the matrix equation \(A\boldx=\boldzero\text{.}\) Since \(A\) is invertible, we conclude that \(\boldx=\boldzero\) (Theorem 2.5.27), and thus that \(a=b=c=0\text{.}\) This shows \(B\) is linearly independent.
Not every vector space has a finite basis, as we show in the next example.
Example 3.6.4. \(P\) has no finite basis.
Prove that \(P\text{,}\) the space of all real polynomials, does not have a finite basis.
Solution.
We show that no finite set of polynomials can span all of \(P\text{;}\) it follows that \(P\) does not have a finite basis.
Indeed let \(S=\{p_1, p_2, \dots, p_r\}\) be a finite set of polynomials, and let \(n\) be the maximal degree of all the polynomials \(p_i\in S\text{.}\) Then \(p_i\in P_n\) for all \(i\text{,}\) in which case \(\Span S\subseteq P_n\text{:}\) i.e., \(\Span S\) is a subspace of the space of polynomials of degree at most \(n\text{.}\) Since \(P_n\subsetneq P\text{,}\) we conclude that \(\Span S\ne P\text{,}\) as claimed.
Remark 3.6.5.
By Theorem 3.7.16 and its corollaries, we know that if \(V\) has a finite basis, then and subspace of \(V\) also has a finite basis. Let \(X\subseteq \R\) be an interval. Since
\begin{equation*}
P\subseteq C^\infty(X)\subseteq C^n(X)\subseteq C(X)\subseteq F(X,\R)
\end{equation*}
is a chain of subspaces, and since \(P\) does not have a finite basis, we conclude that none of these other function spaces has a finite basis.
Example 3.6.6. Basis for \(\R_{>0}\).
Let \(V=\R_{>0}\text{,}\) and let \(S=\{\boldv\}\text{,}\) where \(\boldv=a\) is any positive real number. Prove: \(S\) is a basis if and only if \(a\ne 1\text{.}\)
Solution.
\((a=1\implies S \text{ not a basis})\).
Suppose \(a=1\text{.}\) Since \(1=\boldzero\in V\text{,}\) we have \(S=\{\boldzero\}\text{.}\) Any set containing the zero vector is linearly dependent (Remark 3.5.10). Thus \(S\) is not a basis.
\((a\ne 1 \implies S \text{ is a basis})\).
Since \(S=\{\boldv\}\) consists of one nonzero element, it is linearly independent (Remark 3.5.10). It remains only to show that \(S\) spans \(\R_{>0}\text{,}\) which amounts to showing that every \(b\in \R_{>0}\) is a scalar multiple of \(\boldv=a\text{.}\) Since by definition scalar multiplication in \(\R_{>0}\) is defined as \(c\boldv=a^c\text{,}\) this is equivalent to showing that every \(b\in \R_{>0}\) can be written in the form \(b=a^c\text{.}\) This fact is a familiar result from calculus, where you learn that the range (or image) of any exponential function \(f(x)=a^x\) is the set of all positive real numbers.
Proceeding directly from the definition, to show a set \(B\) is a basis of \(V\) we have to do two steps: (i) show \(\Span B= V\text{;}\) (ii) show that \(B\) is linearly independent. The following theorem offers gives rise to a one-step technique for proving \(B\) is a basis: show that every element of \(V\) can be written as a linear combination of elements of \(B\) in a unique way.
Theorem 3.6.7. Basis equivalence.
-
Every element \(\boldv\in V\) can be written as a linear combination of elements of \(B\text{,}\) and furthermore this linear combination is unique: i.e., if we have\begin{equation*} \boldv=c_1\boldv_1+c_2\boldv_2+\cdots +c_r\boldv_r=d_1\boldv_1+d_2\boldv_2+\cdots +d_s\boldv_r\text{,} \end{equation*}for some distinct vectors \(\boldv_i\in B\text{,}\) then \(c_i=d_i\) for all \(1\leq i\leq r\text{.}\)
Proof.
Implication: \((1)\implies (2)\).
Suppose \(B\) is a basis. By definition, \(B\) spans \(V\text{,}\) and so every element of \(V\) can be written as a linear combination of elements of \(B\text{.}\) It remains to show that this linear combination is unique in the sense described. This follows from the fact that \(B\) is linearly independent. Indeed, if
\begin{equation*}
c_1\boldv_1+c_2\boldv_2+\cdots +c_r\boldv_r=d_1\boldv_1+d_2\boldv_2+\cdots +d_r\boldv_r\text{,}
\end{equation*}
then after some algebra we have
\begin{equation*}
(c_1-d_1)\boldv_1+(c_2-d_2)\boldv_2+\cdots +(c_r-d_r)\boldv_r=\boldzero\text{.}
\end{equation*}
Since \(B\) is linearly independent and since the \(\boldv_i\) are distinct, we must have \(c_i-d_j=0\text{,}\) and hence \(c_i=d_i\) for all \(1\leq i\leq r\text{.}\)
Implication: \((2)\implies (1)\).
If \(B\) satisfies (2), then clearly it spans \(V\text{.}\) The uniqueness of linear combinations of elements of \(B\) now easily implies \(B\) is linearly independent:
\begin{align*}
c_1\boldv_1+c_2\boldv_2+\cdots c_r\boldv_r=\boldzero \amp \implies
c_1\boldv_1+c_2\boldv_2+\cdots c_r\boldv_r=0\boldv_1+0\boldv_2+\cdots 0\boldv_r\\
\amp \implies c_1=0, c_2=0, \dots, c_r=0 \amp \text{(by uniqueness condition)}\text{.}
\end{align*}
Theorem 3.6.7 yields the following one-step technique for proving a set is a basis.
Procedure 3.6.8. One-step technique for bases.
Let \(V\) be a vector space. To prove a subset \(B\subseteq V\) is a basis it suffices to show that every \(\boldv\in V\) can be written as a linear combination of elements of \(B\) in a unique way, as specified in Theorem 3.6.7.
Example 3.6.9. One-step technique for \(\R^3\).
Use the one step technique to decide whether the set
\begin{equation*}
S=\{\boldv_1=(1,1,-3), \boldv_2=(1,0,-1), \boldv_3=(-1,1,-1), \boldv_4=(1,2,1)\}
\end{equation*}
is a basis of \(\R^3\text{.}\)
Solution.
We ask whether for all elements \(\boldy=(a,b,c)\in \R^3\) we can write
\begin{equation}
\boldy=c_1\boldv_1+c_2\boldv_2+c_3\boldv_3+c_4\boldv_4\tag{3.18}
\end{equation}
for a unique choice of \(c_1,c_2,c_3, c_4\text{.}\) This is equivalent to asking whether the matrix equation
\begin{equation*}
\underset{A}{\begin{amatrix}[rrrr]
1\amp 1\amp -1\amp 1\\
1\amp 0\amp 1\amp 2\\
-3\amp -1\amp -1 \amp 1
\end{amatrix}}\, \underset{\boldx}{\colvec{c_1\\ c_2\\ c_3\\ c_4}}=\underset{\boldy}{\colvec{a\\ b\\ c}}\text{.}
\end{equation*}
has a unique solution \(\boldx=(c_1,c_2,c_3,c_4)\) for any choice of \(\boldy=(a,b,c)\text{.}\) Performing Gaussian elimination on the corresponding augmented matrix yields
\begin{equation*}
\begin{amatrix}[rrrr|r]
1\amp 1\amp -1\amp 1\amp a \\
1\amp 0\amp 1\amp 2\amp b\\
-3\amp -1\amp -1 \amp 1 \amp c \end{amatrix}
\xrightarrow{\phantom{row}}U=
\begin{amatrix}[rrrr|r]
\boxed{1}\amp 1\amp -1\amp 1\amp a \\
0\amp \boxed{1}\amp -2\amp -1\amp a-b\\
0\amp 0\amp 0\amp \boxed{1} \amp (a+2b+c)/6
\end{amatrix}\text{.}
\end{equation*}
Since the third column of \(U\) does not have a leading one, we conclude that the corresponding system has a free variable, and hence that for any given \((a,b,c)\in \R^3\) the equation (3.18) has either no solutions (inconsistent) or infinitely many solutions. In particular, it is not true that there is always a unique solution. Thus \(S\) is not a basis according to the one-step technique.
In fact, our Gaussian elimination analysis tells us exactly how \(S\) fails to be a basis. Since the last column of \(U\) does not have a leading one, the corresponding system is always consistent: i.e., there is always at least one solution \(\boldx=(c_1,c_2,c_3,c_4)\) to (3.18) for each \((a,b,c)\in \R^3\text{.}\) This tells us that \(S\) is a spanning set of \(\R^3\text{.}\) On the other hand, the existence of the free variable tells us that for \((a,b,c)=(0,0,0)=\boldzero\text{,}\) we will have infinitely many choices \(c_1,c_2,c_3,c_4\) satisfying
\begin{equation*}
c_1\boldv_1+c_2\boldv_2+c_3\boldv_3+c_4\boldv_4=\boldzero\text{.}
\end{equation*}
This shows that \(S\) is not linearly independent.
Example 3.6.10. One-step technique for \(P_1\).
Use the one-step technique to decide whether the set
\begin{equation*}
S=\{p(x)=2x+1, x+1\}
\end{equation*}
is a basis of \(P_1\text{.}\)
Solution.
Take an arbitrary element \(ax+b\in P_1\) and consider the polynomial equation
\begin{equation*}
c_1p(x)+c_2q(x)=ax+b
\end{equation*}
The usual remark about polynomial equality implies that this is equivalent to the matrix equation
\begin{equation*}
\begin{bmatrix}
2\amp 1\\ 1\amp 1
\end{bmatrix}
\begin{bmatrix}
c_1 \\ c_2
\end{bmatrix}
=
\begin{bmatrix}
a\\ b
\end{bmatrix}
=\boldy\text{.}
\end{equation*}
The matrix on the left is invertible, allowing us to solve:
\begin{equation*}
\begin{bmatrix}
c_1 \\ c_2
\end{bmatrix}
=\begin{bmatrix}
1\amp -1\\ -1\amp 2
\end{bmatrix}
\begin{bmatrix}
a\\ b
\end{bmatrix}
=\begin{bmatrix}
a-b\\ -a+2b
\end{bmatrix}\text{.}
\end{equation*}
We conclude that any \(ax+b\in P_1\) can be written as \(c_1p+c_2q\) in a unique way: namely, with \(c_1=a-b\) and \(c_2=-a+2b\text{.}\) Thus \(S\) is a basis.
Video example: deciding if a set is a basis.
Besides deciding whether a given set is a basis, we will often want to come up with a basis of a given space on our own. The following “by inspection” technique often comes in handy in cases where the elements of the vector space can be described in a simple parametric manner.
Procedure 3.6.13. By-inspection basis technique.
To produce a basis \(B\) of a vector space \(V\) “by inspection”, proceed as follows.
-
Paremetric description.Give a simple parametric description of the general element of \(V\text{.}\)
-
Spanning set.If your parametric description is simple enough, you should be able to extract from it a natural spanning set \(B\) of \(V\text{.}\)
-
Linear independence.If your parametric description is free of redundancies, then \(B\) will likely be linearly independent. Verify this using Procedure 3.5.12.
Example 3.6.14. By-inspection basis technique.
Let \(W\subseteq M_{22}\) be the subspace of all symmetric matrices. Use Procedure 3.6.13 to produce a basis of \(W\text{.}\)
Solution.
We follow the three steps of Procedure 3.6.13.
-
A general \(2\times 2\) symmetric matrix can be described parametrically as\begin{equation*} A=\begin{bmatrix} a\amp b\\ b\amp c \end{bmatrix}\text{.} \end{equation*}
-
We have\begin{equation} \begin{bmatrix} a\amp b\\ b\amp c \end{bmatrix}= a \begin{bmatrix} 1\amp 0\\ 0\amp 0 \end{bmatrix} + b \begin{bmatrix} 0\amp 1\\ 1\amp 0 \end{bmatrix} + c \begin{bmatrix} 0\amp 0\\ 0\amp 1 \end{bmatrix}\text{.}\tag{3.19} \end{equation}It follows immediately that the set \(B=\{A_1, A_2, A_3\}\) is a spanning set, where\begin{equation*} A_1= \begin{bmatrix} 1\amp 0\\ 0\amp 0 \end{bmatrix}, A_2=\begin{bmatrix} 0\amp 1\\ 1\amp 0 \end{bmatrix} , A_3=\begin{bmatrix} 0\amp 0\\ 0\amp 1 \end{bmatrix}\text{.} \end{equation*}
-
The expression (3.19) tells us that\begin{align*} aA_1+bA_2+cA_3 = \boldzero\amp \iff \begin{bmatrix} a\amp b\\ b\amp c \end{bmatrix}=\begin{bmatrix} 0\amp 0\\ 0\amp 0 \end{bmatrix} \\ \amp\iff a=b=c=0 \text{.} \end{align*}This proves \(B\) is linearly independent.
Since \(B\) is a linearly independent spanning set of \(W\text{,}\) it is a basis of \(W\text{.}\)
Subsection Bases and linear transformations
Theorem 3.6.15. Bases and linear transformations.
-
Let \(T\) and \(T'\) be linear transformations from \(V\) to \(W\text{.}\) If \(T(\boldu)=T'(\boldu)\) for all \(\boldu\in B\text{,}\) then \(T=T'\text{.}\)
-
Let \(W\) be a vector space. Any mapping\begin{equation*} \boldu\mapsto \boldw_\boldu \end{equation*}assigning each element of \(\boldu\in B\) to a chosen element \(\boldw_\boldu\in W\) extends uniquely to a linear transformation \(T\colon V\rightarrow W\) satisfying\begin{equation*} T(\boldu)=\boldw_\boldu \end{equation*}for all \(\boldu\in B\text{.}\) In more detail, given any \(\boldv\in V\text{,}\) if \(\boldv=c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r\text{,}\) where \(\boldu_i\in B\) and \(c_i\ne 0\) for all \(1\leq i\leq r\text{,}\) then\begin{equation} T(\boldv)=c_1T(\boldu_{1})+c_2T(\boldu_{2})+\cdots +c_rT(\boldu_{r})\text{.}\tag{3.20} \end{equation}
Proof.
Proof of (i).
Assume \(T\) and \(T'\) are linear transformations from \(V\) to \(W\) satisfying \(T(\boldu)=T'(\boldu)\) for all \(\boldu\in B\text{.}\) Given any \(\boldv\in V\) we can write \(\boldv=c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r\text{.}\) It follows that
\begin{align*}
T(\boldv) \amp = T(c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r)\\
\amp = c_1T(\boldu_1)+c_2T(\boldu_2)+\cdots +c_rT(\boldu_r) \amp (T \text{ is linear})\\
\amp =c_1T'(\boldu_1)+c_2T'(\boldu_2)+\cdots +c_rT'(\boldu_r)\\
\amp = T'(c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r) \amp (T' \text{ is linear}) \\
\amp = T'(\boldv)\text{.}
\end{align*}
Since \(T(\boldv)=T'(\boldv)\) for all \(\boldv\in V\text{,}\) we have \(T=T'\text{.}\)
Proof of (ii).
That there can me at most one such \(T\colon V\rightarrow W\) follows from (i). Thus we need only show that such a \(T\) exists.
Since any \(\boldv\in V\) has a unique expression of the form
\begin{equation*}
\boldv=c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r\text{,}
\end{equation*}
where \(c_i\ne 0\) for all \(1\leq i\leq r\text{,}\) the formula in (3.20) defines a function \(T\colon V\rightarrow W\) in a well-defined manner. Note also that the formula still applies even if some of the coefficients are equal to 0: if \(c_i=0\text{,}\) then \(c_iT(\boldv_i)=\boldzero\text{,}\) and the right-hand side of (3.20) is unchanged. We will use this fact below.
We now show that \(T\) is linear. Given \(\boldv, \boldv'\in V\) we can find a common collection of elements \(\boldu_1,\boldu_2,\dots, \boldu_r\in B\) for which
\begin{align*}
\boldv \amp = c_1\boldu_1+c_2\boldu_2+\cdots +c_r\boldu_r\\
\boldv'\amp=d_1\boldu_1+d_2\boldu_2+\cdots +d_r\boldu_r
\end{align*}
for some \(c_i, d_i\in \R\text{.}\) We can no longer assume that \(c_i\ne 0\) and \(d_i\ne 0\) for all \(1\leq i\leq r\text{,}\) but as observed above we still have
\begin{align*}
T(\boldv) \amp = c_1T(\boldu_1)+c_2T(\boldu_2)+\cdots +c_rT(\boldu_r)\\
T(\boldv')\amp=d_1T(\boldu_1)+d_2T(\boldu_2)+\cdots +d_rT(\boldu_r) \text{.}
\end{align*}
Given any \(c, d\in \R\text{,}\) we have
\begin{align*}
T(c\boldv+d\boldv')\amp=T(cc_1\boldu_1+cc_2\boldu_2+\cdots +cc_r\boldu_r+dd_1\boldu_1+dd_2\boldu_2+\cdots +dd_r\boldu_r) \\
\amp= T\left((cc_1+dd_1)\boldu_1+(cc_2+dd_2)\boldu_2+\cdots +(cc_r+dd_r)\boldu_r\right) \\
\amp =(cc_1+dd_1)T(\boldu_1)+(cc_2+dd_2)T(\boldu_2)+\cdots +(cc_r+dd_r)T(\boldu_r) \amp (\knowl{./knowl/xref/eq_bases_transformations.html}{\text{(3.20)}}) \\
\amp= c(c_1T(\boldu_1)+c_2T(\boldu_2)+\cdots +c_rT(\boldu_r))+d(d_1T(\boldu_1)+d_2T(\boldu_2)+\cdots +d_rT(\boldu_r)\\
\amp =cT(\boldv)+dT(\boldv')\text{.}
\end{align*}
Thus \(T\) is a linear transformation.
Remark 3.6.16. Transformations determined by behavior on basis.
Let’s paraphrase the two results of Theorem 3.6.15.
-
A linear transformation \(T\colon V\rightarrow W\) is completely determined by its behavior on a basis \(B\subseteq V\text{.}\) Once we know the images \(T(\boldu)\) for all \(\boldu\in B\text{,}\) the image \(T(\boldv)\) for any other \(\boldv\in V\) is then completely determined. Put another way, if two linear transformations out of \(V\) agree on the elements of a basis \(B\subseteq V\text{,}\) then they agree for all elements of \(V\text{.}\)
-
Once we have a basis \(B\subseteq V\) on hand, it is easy to construct linear transformations \(T\colon V\rightarrow W\text{:}\) simply choose images \(T(\boldu)\in W\) for all \(\boldu\in B\) in any manner you like, and then define \(T(\boldv)\) for any element \(\boldv\in V\) using (3.20).
Example 3.6.17. Composition of reflections.
Let \(r_0\colon \R^2\rightarrow\R^2\) be reflection across the \(x\)-axis, and let \(r_{\pi/2}\colon \R^2\rightarrow \R^2\) be reflection across the \(y\)-axis. (See Exercise 20.) Use an argument in the spirit of statement (i) from Remark 3.6.16 to show that
\begin{equation*}
r_{\pi/2}\circ r_{0}=\rho_{\pi}\text{.}
\end{equation*}
(Note: this equality can also be shown using our matrix formulas for rotations and reflections. See Exercise 21. )
Solution.
Since \(r_0\) and \(r_{\pi/2}\) are both linear transformations (Exercise 20), so is the composition \(T=r_{\pi/2}\circ r_{0}\text{.}\) We wish to show \(T=\rho_{\pi}\text{.}\) Since \(\rho_{\pi}\) is also a linear transformation, it suffices by Theorem 3.6.15 to show that \(T\) and \(\rho_\pi\) agree on a basis of \(\R^2\text{.}\) Take the standard basis \(B=\{(1,0), (0,1)\}\text{.}\) Compute:
\begin{align*}
T(1,0) \amp=r_{\pi/2}(r_{0}(1,0)) \\
\amp =r_{\pi/2}(1,0) \\
\amp =(-1,0)\\
\amp =\rho_{\pi}(1,0)\\
T(0,1) \amp \\
T(0,1) \amp=r_{\pi/2}(r_{0}(0,1)) \\
\amp =r_{\pi/2}(0,-1) \\
\amp =(0,-1)\\
\amp =\rho_{\pi}(0,1)\text{.}
\end{align*}
Since \(T\) and \(\rho_\pi\) agree on the basis \(B\text{,}\) we have \(T=\rho_\pi\text{.}\)
As a corollary to Theorem 3.6.15 we can at last complete the partial description of linear transformations of the form \(T\colon \R^n\rightarrow \R^m\) given in Theorem 3.2.9.
Corollary 3.6.18. Matrix transformations.
Given any linear transformation \(T\colon \R^n\rightarrow \R^m\) there is a unique \(m\times n\) matrix \(A\) such that \(T=T_A\text{.}\) In fact we have
\begin{equation*}
A=\begin{bmatrix}\vert\amp \vert\amp \amp \vert \\ T(\bolde_1)\amp T(\bolde_2)\amp \cdots \amp T(\bolde_n)\\ \vert\amp \vert\amp \amp \vert \end{bmatrix}\text{,}
\end{equation*}
where \(B=\{\bolde_1, \bolde_2, \dots, \bolde_n\}\) is the standard basis of \(\R^n\text{.}\) As a result, in the special case where the domain and codomain are both spaces of tuples, all linear transformations are matrix transformations.
Proof.
Let \(B=\{\bolde_1, \bolde_2, \dots, \bolde_n\}\) be the standard basis of \(\R^n\text{,}\) and let \(A\) be the \(m\times n\) matrix defined as
\begin{equation*}
A=\begin{bmatrix}\vert\amp \vert\amp \amp \vert \\ T(\bolde_1)\amp T(\bolde_2)\amp \cdots \amp T(\bolde_n)\\ \vert\amp \vert\amp \amp \vert \end{bmatrix}\text{.}
\end{equation*}
In other words, the \(j\)-th column of \(A\) is \(T(\bolde_j)\text{,}\) considered as an \(m\times 1\) column vector. The corresponding matrix transformation \(T_A\colon \R^n\rightarrow \R^m\) is linear by Theorem 3.2.9. Since \(T\) is linear by assumption, Theorem 3.6.15 applies: to show \(T=T_A\) we need only show that \(T(\bolde_j)=T_A(\bolde_j)\) for all \(1\leq j\leq n\text{.}\) We have
\begin{align*}
T_A(\bolde_j) \amp =A\bolde_j \amp (\knowl{./knowl/xref/d_matrix_transform.html}{\text{Definition 3.2.8}})\\
\amp=(j\text{-th column of } A) \amp (\knowl{./knowl/xref/th_column_method.html}{\text{Theorem 2.1.24}}) \\
\amp = T(\bolde_j) \amp (\text{def. of } A)\text{.}
\end{align*}
Thus \(T=T_A\text{,}\) as claimed.
Besides rounding out our theoretical discussion of linear transformations from \(\R^n\) to \(\R^m\text{,}\) computationally Corollary 3.6.18 provides a recipe for computing a “matrix formula” for a linear transformation \(T\colon \R^n\colon \rightarrow \R^m\text{.}\) In other words, it tells us how to build the \(A\text{,}\) column by column, such that \(T\boldx=A\boldx\) for all \(\boldx\in R^n\text{.}\) For reasons that will be made more clear in Section 4.2, we will call \(A\) the standard matrix of \(T\text{.}\)
Definition 3.6.19. Standard matrix of linear \(T\colon \R^n\rightarrow \R^m\).
Let \(T\colon \R^n\rightarrow \R^m\) be a linear transformation. The standard matrix of \(T\) is the unique \(m\times n\) matrix \(A\) satisfying \(T=T_A\text{.}\) Equivalently, \(A\) is the unique matrix satisfying
\begin{equation*}
T(\boldx)=A\boldx
\end{equation*}
for all \(\boldx\in \R^n\text{.}\)
Example 3.6.20. Standard matrix computation.
The function \(T\colon \R^3\rightarrow \R^2\) defined as \(T(x,y,z)=T(x+y+z, 2x+3y-4z)\) is linear.
Solution.
We have
\begin{align*}
A \amp = \begin{amatrix}[ccc]\vert\amp \vert\amp \vert\\
T((1,0,0))\amp T((0,1,0))\amp T((0,0,1)) \\
\vert\amp \vert\amp vert \end{amatrix} \\
\amp =
\begin{amatrix}[rrr]
1 \amp 1 \amp 1 \\ 2\amp 3 \amp -4 \end{amatrix} \text{.}
\end{align*}
Let \(\boldx=(-2,3,4)\text{.}\) Since \(A\) provides a “matrix formula” for \(T\) we have
\begin{align*}
T(\boldx) \amp = A\boldx \\
\amp =
\begin{amatrix}[rrr]
1 \amp 1 \amp 1 \\ 2\amp 3 \amp -4 \end{amatrix}
\colvec{-2\\ 3\\ 4}\\
\amp = \colvec{5\\ -9} \text{.}
\end{align*}
Thus \(T((-2,3,4))=(5,-9)\text{,}\) as you can confirm.
Example 3.6.21. Rotation matrices revisited.
Fix an angle \(\alpha\text{.}\) Taking for granted that the rotation operation \(\rho_\alpha\colon\R^2\rightarrow\R^2\) is a linear transformation, re-derive the matrix formula for \(\rho_\alpha\colon \R^2\rightarrow\R^2\text{:}\) i.e., compute \(A\text{,}\) the standard matrix of \(\rho_\alpha\text{.}\)
Solution.
Let \(B=\{\bolde_1, \bolde_2\}=\{(1,0), (0,1)\}\text{.}\) According to Corollary 3.6.18
\begin{align*}
A \amp=\begin{bmatrix}
\vert\amp \vert \\
\rho_\alpha(1,0)\amp \rho_\alpha(0,1) \\
\vert \amp \vert
\end{bmatrix} \\
\amp=\begin{amatrix}[rr]
\cos\alpha \amp -\sin\alpha \\
\sin\alpha \amp \cos\alpha
\end{amatrix} \text{,}
\end{align*}
since \((1,0)\) gets rotated by \(\rho_\alpha\) to \((\cos\alpha, \sin\alpha)\text{,}\) and \((0,1)\) gets rotated to \((-\sin\alpha, \cos\alpha)\text{.}\)
Exercises Exercises
WeBWork Exercises
1.
2.
Find a basis \(\lbrace p(x), q(x) \rbrace\) for the kernel of the linear transformation \(L:{\mathbb P}_3[x]\to {\mathbb R}\) defined by \(L(f(x)) = f'(-5)-f(1)\) where \({\mathbb P}_3[x]\) is the vector space of polynomials in \(x\) with degree less than 3.
3.
4.
One-step basis technique.
For each vector space \(V\) and subset \(S\text{,}\) use the one-step technique (Procedure 3.6.8) to decide whether \(S\) is a basis for \(V\text{.}\)
5.
\(V=\R^3\)
-
\(\displaystyle S=\{(1,1,2),(1,-1,-4), (3,-1,0) \}\)
-
\(\displaystyle S=\{(1,1,2),(1,-1,-4), (1,-3,1) \}\)
6.
\(V=P_2\)
-
\(\displaystyle S=\{2x^2+x+1, x^2-2x-1, x^2+1 \}\)
-
\(\displaystyle S=\{2x^2-3x+1, 4x^2+x+1, 7x-1\}\)
7.
\(V=M_{22}\)
-
\(S=\{A_1, A_2, A_3, A_4 \}\text{,}\) where\begin{equation*} A_1=\begin{bmatrix}1\amp 1\\ 1\amp 1 \end{bmatrix}, \ A_2=\begin{bmatrix}1\amp -1\\ 0\amp 0 \end{bmatrix}, \ A_3=\begin{bmatrix}0\amp -1\\ 1\amp 0 \end{bmatrix}, \ A_4=\begin{bmatrix}1\amp 0\\ 0\amp 0 \end{bmatrix} \end{equation*}
-
\(S=\{A_1, A_2, A_3, A_4 \}\text{,}\) where\begin{equation*} A_1=\begin{bmatrix}1\amp 1\\ 1\amp 1 \end{bmatrix}, \ A_2=\begin{bmatrix}1\amp -1\\ -1\amp 0 \end{bmatrix}, \ A_3=\begin{bmatrix}0\amp 1\\ 1\amp 0 \end{bmatrix}, \ A_4=\begin{bmatrix}1\amp 0\\ 0\amp 0 \end{bmatrix} \end{equation*}
By-inspection basis technique.
For each given \(V\) and subspace \(W\subseteq V\text{,}\) provide a basis for \(W\) using the “by inspection” technique Procedure 3.6.13. In more detail:
-
Give a simple parametric description of the elements of \(W\text{.}\)
-
If your parametric description is simple enough, you should be able to find an obvious spanning set \(S\) of \(W\text{.}\)
-
Argue that your spanning set is linearly independent.
8.
9.
10.
11.
12.
Suppose \(B=\{\boldv_1,\boldv_2,\boldv_3\}\) be a basis for the vector space\(V\text{.}\) Let \(B'=\{\boldu_1,\boldu_2,\boldu_3\}\text{,}\) where
\begin{equation*}
\boldu_1 = \boldv_1, \boldu_2 = \boldv_1 + \boldv_2, \boldu_3 = \boldv_1 +\boldv_2 + \boldv_3\text{.}
\end{equation*}
Prove that \(B'\) is a basis.
13.
Let \(S=\{\boldv_1, \boldv_2, \dots, \boldv_k\}\) be a set of \(k\) distinct elements of \(\R^n\text{,}\) let \(A\) be an invertible \(n\times n\) matrix, and let \(S'=\{A\boldv_1, A\boldv_2,\dots, A\boldv_k\}\text{.}\) Prove that \(S\) is a basis of \(\R^n\) if and only if \(S'\) is a basis of \(\R^n\) as follows.
-
Prove that \(A\boldv_i\ne A\boldv_j\) for all \(i\ne j\text{:}\) i.e., \(S'\) contains \(k\) distinct elements.
-
Prove that if \(\{\boldv_1, \boldv_2, \dots, \boldv_k\}\) is a basis of \(\R^n\text{,}\) then \(\{A\boldv_1, A\boldv_2,\dots, A\boldv_k\}\) is also a basis for any invertible \(n\times n\) matrix \(A\text{.}\)
-
Now prove that if \(\{A\boldv_1, A\boldv_2,\dots, A\boldv_k\}\) is a basis of \(\R^n\) for the invertible matrix \(A\text{,}\) then \(\{\boldv_1, \boldv_2, \dots, \boldv_k\}\) is a basis of \(\R^n\text{.}\)Hint: in (b) you showed that if you start with any basis and apply any invertible matrix to the elements of this basis, then you end up with another basis; think of a useful choice of matrix for the present situation.
14. Bases for important matrix subspaces.
Let \(V=M_{nn}\text{.}\) For each of the following subspaces \(W\subset M_{nn}\text{,}\) give a basis \(B\) of \(W\text{.}\) You must explicitly describe the elements of your basis as linear combinations of the elements \(E_{ij}\) of the standard basis for \(M_{nn}\text{.}\) No justification needed, as long as your proposed basis is simple enough.
-
Upper triangular matrices.\(\displaystyle W=\{A\in M_{nn}\colon A \text{ is upper triangular}\}\)
-
Symmetric matrices.\(\displaystyle W=\{A\in M_{nn}\colon A^T=A\}\)
-
Skew-symmetric matrices.\(\displaystyle W=\{A\in M_{nn}\colon A^T=-A\}\)
15.
The set \(B=\{\boldv_1=(1,-1), \boldv_2=(1,1)\}\) is a basis of \(\R^2\text{.}\) Suppose the linear transformation \(T\colon \R^2\rightarrow \R^3\) satisfies
\begin{equation*}
T(\boldv_1)=(4,1,2), T(\boldv_2)=(1,0,2)\text{.}
\end{equation*}
Find a formula for \(T(\boldx)\text{,}\) where \(\boldx=(x,y)\) is a general element of \(\R^2\text{.}\)
16.
The set \(B=\{x^2,x, 1\}\) is a basis of \(P_2\text{.}\) Suppose the linear transformation \(T\colon P_2\rightarrow M_{22}\) satisfies
\begin{equation*}
T(x^2)=\begin{amatrix}[rr] 1 \amp 0\\ 0\amp 0 \end{amatrix},
T(x)=\begin{amatrix}[rr]0\amp 1\\ 1\amp 0 \end{amatrix},
T(1)=\begin{amatrix}[rr] 0\amp 0\\ 0\amp 1 \end{amatrix}\text{.}
\end{equation*}
Find a formula for \(T(p(x))\text{,}\) where \(p(x)=ax^2+bx+c\) is a general element of \(P_2\text{.}\)
17.
The set \(B=\{x^2+1,x+1, 1\}\) is a basis of \(P_2\text{.}\) Suppose the linear transformation \(T\colon P_2\rightarrow M_{22}\) satisfies
\begin{equation*}
T(x^2+1)=\begin{amatrix}[rr] 2 \amp 1\\ 2\amp 5 \end{amatrix},
T(x+1)=\begin{amatrix}[rr]0\amp 1\\ 2\amp 3 \end{amatrix},
T(1)=\begin{amatrix}[rr] 1\amp 1\\ 1\amp 1 \end{amatrix}\text{.}
\end{equation*}
Show that the general formula for \(T\) is given by
\begin{equation*}
T(p(x))=\begin{amatrix}[cc] p(-1)\amp p(0)\\ p(1)\amp p(2) \end{amatrix}\text{.}
\end{equation*}
Hint.
Use (1) of Theorem 3.6.15.
18.
Suppose \(T\colon V\rightarrow V\) is a linear transformation, and \(B\) is a basis of \(V\) for which \(T(\boldv)=\boldzero\) for all \(\boldv\in B\text{.}\) Show that \(T=0_{V,W}\text{:}\) i.e., \(T\) is the zero transformation from \(V\) to \(W\text{.}\)
Hint.
Use (1) of Theorem 3.6.15.
19.
Suppose \(T\colon V\rightarrow V\) is a linear transformation, and \(B\) is a basis of \(V\) for which \(T(\boldv)=\boldv\) for all \(\boldv\in B\text{.}\) Show that \(T=\id_V\text{:}\) i.e., \(T\) is the identity transformation of \(V\text{.}\)
Hint.
Use (1) of Theorem 3.6.15.
20.
Let \(T\colon \R^n\rightarrow \R^n\) be a linear transformation. Assume there is a basis \(B\) of \(\R^n\) and a constant \(c\in \R\) such that \(T(\boldv)=c\boldv\) for all \(\boldv\in B\text{.}\) Prove: \(T=T_{A}\text{,}\) where
\begin{equation*}
A=cI_n=\begin{amatrix}[cccc]
c \amp 0\amp \dots\amp 0\\
0\amp c\amp \dots \amp 0 \\
\vdots \amp \amp \amp \vdots\\
0\amp 0\amp \dots \amp c
\end{amatrix}\text{.}
\end{equation*}
Hint.
Use (1) of Theorem 3.6.15.
Matrix transformations.
For each linear transformation \(T\colon \R^n\rightarrow \R^m\) and \(\boldx\in \R^n\) : (a) compute the standard matrix \(A\) of \(T\) using Corollary 3.6.18; (b) compute \(T(\boldx)\) using \(A\text{.}\) You may take for granted that the given \(T\) is linear.
21.
\begin{align*}
T\colon \R^2\amp \rightarrow\R^4 \amp \boldx\amp=(1,3) \\
(x,y) \amp\mapsto (2x-y, 2y, x+y, x)
\end{align*}
22.
\begin{align*}
T\colon \R^4\amp \rightarrow \R^3 \amp
\boldx=(0,2,4,-1)\\
(x_1,x_2,x_3,x_4)\amp\mapsto (2x_1-x_2+x_4, x_2-x_3, x_1+3x_2-x_3-x_4)
\end{align*}
