Linear Algebra/Representing Linear Maps with Matrices

Consider a map h with domain \Re^2 and codomain \Re^3 (fixing as the bases for these spaces) that is determined by this action on the vectors in the domain's basis. To compute the action of this map on any vector at all from the domain, we first express h(\vec{\beta}_1) and h(\vec{\beta}_2) with respect to the codomain's basis: and (these are easy to check). Then, as described in the preamble, for any member \vec{v} of the domain, we can express the image h(\vec{v}) in terms of the h(\vec{\beta})'s. Failed to parse (syntax error): h(\vec{v}) =h(c_1\cdot \begin{bmatrix} 2 \\ 0 \end{bmatrix}+c_2\cdot \begin{bmatrix} 1 \\ 4 \end{bmatrix}) \\ =c_1\cdot h(\begin{bmatrix} 2 \\ 0 \end{bmatrix})+c_2\cdot h(\begin{bmatrix} 1 \\ 4 \end{bmatrix}) \\ =c_1\cdot ( 0\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \!-\frac{1}{2}\begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix} \!+1\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\! ) +c_2\cdot ( 1\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \!-1\begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix} \!+0\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\! ) \\ =(0c_1+1c_2)\cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} +(-\frac{1}{2}c_1-1c_2)\cdot \begin{bmatrix} 0 \\ -2 \\ 0 \end{bmatrix} +(1c_1+0c_2)\cdot \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}

Thus,

Definition

Suppose that ( V ) and ( W ) are vector spaces of dimensions ( n ) and ( m ) with bases ( B ) and ( D ), and that ( \map{h}{V}{W} ) is a linear map. If Failed to parse (syntax error): {h( \vec{\beta}_1 )}{D} = \begin{bmatrix} h_{1,1 \end{bmatrix} \\ h_{2,1} \\ \vdots \\ h_{m,1}}_D \;\ldots\; {h( \vec{\beta}_n )}{D} = \begin{bmatrix} h_{1,n \end{bmatrix} \\ h_{2,n} \\ \vdots \\ h_{m,n}}_D

then Failed to parse (unknown function\generalmatrix): {h}{B,D}=\generalmatrix{h}{n}{m}_{B,D}

is the matrix representation of ( h ) with respect to ( B, D ).% ^{[Index][Index]}%

If ( \map{h}{\Re^3}{\polyspace_1} ) is given by then where the action of ( h ) on ( B ) is given by and a simple calculation gives showing that this is the matrix representing h with respect to the bases.

Theorem

Assume that ( V ) and ( W ) are vector spaces of dimensions ( m ) and ( n ) with bases ( B ) and ( D ), and that ( \map{h}{V}{W} ) is a linear map. If ( h ) is represented by Failed to parse (unknown function\generalmatrix): {h}{B,D}=\generalmatrix{h}{n}{m}_{B,D}

and ( \vec{v}\in V ) is represented by then the representation of the image of \vec{v} is this. Failed to parse (syntax error): {\, h(\vec{v}) \,}{D} = \begin{bmatrix} h_{1,1 \end{bmatrix}c_1+h_{1,2}c_2+\dots+h_{1,n}c_n \\ h_{2,1}c_1+h_{2,2}c_2+\dots+h_{2,n}c_n \\ \vdots \\ h_{m,1}c_1+h_{m,2}c_2+\dots+h_{m,n}c_n}_D

Proof

Definition

The matrix-vector product^[Index]% ^[Index] of a ( \nbym{m}{n} ) matrix and a ( \nbym{n}{1} ) vector is this. Failed to parse (unknown function\generalmatrix): \generalmatrix{a}{n}{m} \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} = \begin{bmatrix} a_{1,1 \end{bmatrix}c_1+a_{1,2}c_2+\dots+a_{1,n}c_n \\ a_{2,1}c_1+a_{2,2}c_2+\dots+a_{2,n}c_n \\ \vdots \\ a_{m,1}c_1+a_{m,2}c_2+\dots+a_{m,n}c_n}

With the matrix from ex:PolyOneToRThree we can calculate where that map sends this vector. This vector is represented, with respect to the domain basis B, by and so this is the representation of the value h(\vec{v}) with respect to the codomain basis D. Failed to parse (syntax error): {h(\vec{v})}{D} =\begin{pmatrix} -1/2 &1 &2 \\ -1/2 &-1 &-2 \end{pmatrix}_{B,D} \begin{bmatrix} 0 \\ 1/2 \\ 2 \end{bmatrix}_B \\ =\begin{bmatrix} (-1/2)\cdot 0+1\cdot (1/2) + 2\cdot 2 \\ (-1/2)\cdot 0-1\cdot (1/2) - 2\cdot 2 \end{bmatrix}_D =\begin{bmatrix} 9/2 \\ -9/2 \end{bmatrix}_D

To find h(\vec{v}) itself, not its representation,

Let ( \map{\pi}{\Re^3}{\Re^2} ) be projection onto the ( xy )-plane. To give a matrix representing this map, we first fix bases. For each vector in the domain's basis, we find its image under the map. Then we find the representation of each image with respect to the codomain's basis (these are easily checked). Finally, adjoining these representations gives the matrix representing ( \pi ) with respect to ( B,D ). We can illustrate \nearbytheorem{th:MatMultRepsFuncAppl} by computing the matrix-vector product representing the following statement about the projection map. Representing this vector from the domain with respect to the domain's basis gives this matrix-vector product. Expanding this representation into a linear combination of vectors from ( D ) checks that the map's action is indeed reflected in the operation of the matrix. (We will sometimes compress these three displayed equations into one

To represent a rotation^[Index]

In the definition the width of the matrix equals the height of the vector. Hence, the first product below is defined while the second is not. One reason that this product is not defined is purely formal:~the definition requires that the sizes match, and these sizes don't match. Behind the formality, though, is a reason why we will leave it undefined— the matrix represents a map with a three-dimensional domain while the

Problem

Multiply the matrix by each vector (or state ``not defined). \begin{exparts*}

1. ( \begin{bmatrix}

2 \\ 1 \\ 0 \end{bmatrix} )

1. ( \begin{bmatrix}

-2 \\ -2 \end{bmatrix} )

1. ( \begin{bmatrix}

0 \\ 0 \\ 0 \end{bmatrix} ) \end{exparts*}

Answer

\begin{exparts*}

1. (

\begin{bmatrix} 1\cdot 2+3\cdot 1+1\cdot 0 \\ 0\cdot 2+(-1)\cdot 1+2\cdot 0 \\ 1\cdot 2+1\cdot 1+0\cdot 0 \end{bmatrix} =\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} )

1. Not defined.
2. ( \begin{bmatrix}

0 \\ 0 \\ 0 \end{bmatrix} )

Problem

Perform, if possible, each matrix-vector multiplication. \begin{exparts*}

1. \begin{pmatrix}

2 &1 \\ 3 &-1/2 \end{pmatrix} \begin{bmatrix} 4 \\ 2 \end{bmatrix}

1. \begin{pmatrix}

1 &1 &0 \\ -2 &1 &0 \end{pmatrix} \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}

1. \begin{pmatrix}

1 &1 \\ -2 &1 \end{pmatrix} \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix} \end{exparts*}

Answer

\begin{exparts*}

1. \begin{bmatrix}

2\cdot 4 +1\cdot 2 \\ 3\cdot 4-(1/2)\cdot 2 \end{bmatrix} =\begin{bmatrix} 10 \\ 11 \end{bmatrix}

1. \begin{bmatrix}

4 \\ 1 \end{bmatrix}

1. Not defined.

\end{exparts*}

This exercise is recommended for all readers.

Problem

Solve this matrix equation.

Answer

Matrix-vector multiplication gives rise to a linear system. </math> Gaussian reduction shows that ( z=1 ), ( y=1 ), and ( x=3 ).

This exercise is recommended for all readers.

Problem

For a homomorphism from ( \polyspace_2 ) to ( \polyspace_3 ) that sends where does ( 1-3x+2x^2 ) go?

Answer

Here are two ways to get the answer.

First, obviously 1-3x+2x^2=1\cdot 1-3\cdot x+2\cdot x^2, and so we can apply the general property of preservation of combinations to get h(1-3x+2x^2) =h(1\cdot 1-3\cdot x+2\cdot x^2) =1\cdot h(1)-3\cdot h(x)+2\cdot h(x^2) =1\cdot (1+x)-3\cdot (1+2x)+2\cdot (x-x^3) =-2-3x-2x^3.

The other way uses the computation scheme developed in this subsection. Because we know where these elements of the space go, we consider this basis ( B={1,x,x^2} ) for the domain. Arbitrarily, we can take ( D={1,x,x^2,x^3} ) as a basis for the codomain. With those choices, we have that and, as the matrix-vector multiplication calculation gives this. Thus, ( h(1-3x+2x^2) =-2\cdot 1-3\cdot x+0\cdot x^2-2\cdot x^3 =-2-3x-2x^3 ), as above.

This exercise is recommended for all readers.

Problem

Assume that ( \map{h}{\Re^2}{\Re^3} ) is determined by this action. Using the standard bases, find

1. the matrix representing this map;
2. a general formula for ( h(\vec{v}) ).

Answer

Again, as recalled in the subsection, with respect to \stdbasis_i, a column vector represents itself.

1. To represent ( h ) with respect

to ( \stdbasis_2,\stdbasis_3 ) we take the images of the basis vectors from the domain, and represent them with respect to the basis for the codomain. Failed to parse (unknown function\stdbasis): {\,h(\vec{e}_1)\,}{\stdbasis_3} ={\begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix}}{\stdbasis_3} =\begin{bmatrix} 2 \\ 2 \\ 0 \end{bmatrix} {\,h(\vec{e}_2)\,}{\stdbasis_3} ={\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}}{\stdbasis_3} =\begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}

These are adjoined to make the matrix. Failed to parse (unknown function\stdbasis): {h}{\stdbasis_2,\stdbasis_3}= \begin{pmatrix} 2 &0 \\ 2 &1 \\ 0 &-1 \end{pmatrix}

1. For any ( \vec{v} ) in the domain ( \Re^2 ),

Failed to parse (unknown function\stdbasis): {\vec{v}}{\stdbasis_2} ={\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}}{\stdbasis_2} =\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}

and so Failed to parse (unknown function\stdbasis): {\,h(\vec{v})\,}{\stdbasis_3} =\begin{pmatrix} 2 &0 \\ 2 &1 \\ 0 &-1 \end{pmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} =\begin{bmatrix} 2v_1 \\ 2v_1+v_2 \\ -v_2 \end{bmatrix}

is the desired representation.

This exercise is recommended for all readers.

Problem

Let ( \map{d/dx}{\polyspace_3}{\polyspace_3} ) be the derivative transformation.

1. Represent ( d/dx ) with respect to ( B,B ) where

( B={1,x,x^2,x^3} ).

1. Represent ( d/dx ) with respect to ( B,D ) where

( D={1,2x,3x^2,4x^3} ).

Answer

\begin{exparts*}

We must first find the image of each vector from the domain's basis, and then represent that image with respect to the codomain's basis. Those representations are then adjoined to make the matrix representing the map.

1. Proceeding as in the prior item, we represent the images

of the domain's basis vectors and adjoin to make the matrix. \end{exparts*}

This exercise is recommended for all readers.

Problem

Represent each linear map with respect to each pair of bases.

1. ( \map{d/dx}{\polyspace_n}{\polyspace_n} ) with respect to

( B,B ) where ( B={1,x,\dots,x^n} ), given by

1. ( \map{\int}{\polyspace_n}{\polyspace_{n+1

Problem

Represent the identity map on any nontrivial space with respect to ( B,B ), where ( B ) is any basis.

Answer

Where the space is ( n )-dimensional,

Problem

Represent, with respect to the natural basis, the transpose transformation on the space ( \matspace_{2} ) of 2 matrices.

Answer

Taking this as the natural basis the transpose map acts in this way so that representing the images with respect to the codomain's basis and adjoining those column vectors together gives this.

Problem

Assume that ( B={\vec{\beta}_1,\vec{\beta}_2,\vec{\beta}_3,\vec{\beta}_4} ) is a basis for a vector space. Represent with respect to ( B,B ) the transformation that is determined by each.

1. ( \vec{\beta}_1\mapsto\vec{\beta}_2 ),

( \vec{\beta}_2\mapsto\vec{\beta}_3 ), ( \vec{\beta}_3\mapsto\vec{\beta}_4 ), ( \vec{\beta}_4\mapsto\zero )

1. ( \vec{\beta}_1\mapsto\vec{\beta}_2 ),

( \vec{\beta}_2\mapsto\zero ), ( \vec{\beta}_3\mapsto\vec{\beta}_4 ), ( \vec{\beta}_4\mapsto\zero )

1. ( \vec{\beta}_1\mapsto\vec{\beta}_2 ),

( \vec{\beta}_2\mapsto\vec{\beta}_3 ), ( \vec{\beta}_3\mapsto\zero ), ( \vec{\beta}_4\mapsto\zero )

Answer

1. With respect to the basis of the codomain, the images of

the members of the basis of the domain are represented as Failed to parse (unknown function\zero): {\vec{\beta}_2}{B}=\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} {\vec{\beta}_3}{B}=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} {\vec{\beta}_4}{B}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} {\zero}{B}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

and consequently, the matrix representing the transformation is this.

\begin{pmatrix} 0 &0 &0 &0 \\ 1 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &1 &0 \end{pmatrix}

\begin{pmatrix} 0 &0 &0 &0 \\ 1 &0 &0 &0 \\ 0 &1 &0 &0 \\ 0 &0 &0 &0

Problem

exam:RepsOfRigidPlaneMaps shows how to represent the rotation transformation of the plane with respect to the standard basis. Express these other transformations also with respect to the standard basis.

1. the dilation map d_s, which multiplies

all vectors by the same scalar s^[Index]

1. the reflection map f_\ell, which reflects all

all vectors across a line \ell through the origin

Answer

1. The picture of \map{d_s}{\Re^2}{\Re^2} is this.

This map's effect on the vectors in the standard basis for the domain is and those images are represented with respect to the codomain's basis (again, the standard basis) by themselves. Failed to parse (unknown function\stdbasis): {\begin{bmatrix} s \\ 0 \end{bmatrix}}{\stdbasis_2}=\begin{bmatrix} s \\ 0 \end{bmatrix} {\begin{bmatrix} 0 \\ s \end{bmatrix}}{\stdbasis_2}=\begin{bmatrix} 0 \\ s \end{bmatrix}

Thus the representation of the dilation map is this. Failed to parse (unknown function\stdbasis): {d_s}{\stdbasis_2,\stdbasis_2} =\begin{pmatrix} s &0 \\ 0 &s \end{pmatrix}

1. The picture of \map{f_\ell}{\Re^2}{\Re^2} is this.

Some calculation (see Exercise~I.\ref{exer:RigidPlaneMapsAutos}) shows that when the line has slope k (the case of a line with undefined slope is separate but easy) and so the matrix representing reflection is this. Failed to parse (unknown function\stdbasis): {f_\ell}{\stdbasis_2,\stdbasis_2} =\frac{1}{1+k^2}\cdot\begin{pmatrix} 1-k^2 &2k \\ 2k &-(1-k^2) \end{pmatrix}

This exercise is recommended for all readers.

Problem

Consider a linear transformation of ( \Re^2 ) determined by these two.

1. Represent this transformation with respect to the standard

bases.

1. Where does the transformation send this vector?

1. Represent this transformation with respect to these bases.

1. Using ( B ) from the prior item,

represent the transformation with respect to ( B,B ).

Answer

Call the map ( \map{t}{\Re^2}{\Re^2} ).

1. To represent this map with respect to the standard bases, we

must find, and then represent, the images of the vectors \vec{e}_1 and \vec{e}_2 from the domain's basis. The image of \vec{e}_1 is given.

One way to find the image of \vec{e}_2 is by eye— we can see this.

A more systemmatic way to find the image of \vec{e}_2 is to use the given information to represent the transformation, and then use that representation to determine the image. Taking this for a basis, the given information says this. Failed to parse (unknown function\stdbasis): {t}{C,\stdbasis_2} \begin{pmatrix} 2 &-1 \\ 0 &0 \end{pmatrix}

As we have that Failed to parse (unknown function\stdbasis): {t(\vec{e}_2)}{\stdbasis_2} =\begin{pmatrix} 2 &-1 \\ 0 &0 \end{pmatrix}_{C,\stdbasis_2} \begin{bmatrix} 1 \\ -1 \end{bmatrix}_C =\begin{bmatrix} 3 \\ 0 \end{bmatrix}_{\stdbasis_2}

and consequently we know that t(\vec{e}_2)=3\cdot\vec{e}_1 (since, with respect to the standard basis, this vector is represented by itself). Therefore, this is the representation of t with respect to \stdbasis_2,\stdbasis_2. Failed to parse (unknown function\stdbasis): {t}{\stdbasis_2,\stdbasis_2} =\begin{pmatrix} -1 &3 \\ 0 &0 \end{pmatrix}_{\stdbasis_2,\stdbasis_2}

1. To use the matrix developed in the prior item, note that

Failed to parse (unknown function\stdbasis): {\begin{bmatrix} 0 \\ 5 \end{bmatrix}}{\stdbasis_2}=\begin{bmatrix} 0 \\ 5 \end{bmatrix}_{\stdbasis_2}

and so we have this is the representation, with respect to the codomain's basis, of the image of the given vector. Failed to parse (unknown function\stdbasis): {t(\begin{bmatrix} 0 \\ 5 \end{bmatrix})}{\stdbasis_2} =\begin{pmatrix} -1 &3 \\ 0 &0 \end{pmatrix}_{\stdbasis_2,\stdbasis_2} \begin{bmatrix} 0 \\ 5 \end{bmatrix}_{\stdbasis_2} =\begin{bmatrix} 15 \\ 0 \end{bmatrix}_{\stdbasis_2}

Because the codomain's basis is the standard one, and so vectors in the codomain are represented by themselves, we have this.

1. We first find the image of each member of ( B ), and then

represent those images with respect to ( D ). For the first step, we can use the matrix developed earlier. Failed to parse (unknown function\stdbasis): {\begin{bmatrix} 1 \\-1 \end{bmatrix}}{\stdbasis_2} =\begin{pmatrix} -1 &3 \\ 0 &0 \end{pmatrix}_{\stdbasis_2,\stdbasis_2} \begin{bmatrix} 1 \\ -1 \end{bmatrix}_{\stdbasis_2} =\begin{bmatrix} -4 \\ 0 \end{bmatrix}_{\stdbasis_2} \text{so} t(\begin{bmatrix} 1 \\ -1 \end{bmatrix})=\begin{bmatrix} -4 \\ 0 \end{bmatrix}

Actually, for the second member of B there is no need to apply the matrix because the problem statement gives its image. Now representing those images with respect to D is routine. Thus, the matrix is this.

1. We know the images of the members of the domain's basis

from the prior item. We can compute the representation of those images with respect to the codomain's basis. Thus this is the matrix.

Problem

Suppose that ( \map{h}{V}{W} ) is nonsingular so that by \nearbytheorem{th:OOHomoEquivalence}, for any basis ( B={\vec{\beta}_1,\dots,\vec{\beta}_n}\subset V ) the image ( h(B)={h(\vec{\beta}_1),\dots,h(\vec{\beta}_n)} ) is a basis for ( W ).

1. Represent the map h with respect to B,h(B).
2. For a member \vec{v} of the domain, where

the representation of \vec{v} has components c_1, \ldots, c_n, represent the image vector ( h(\vec{v}) ) with respect to the image basis h(B).

Answer

1. The images of the members of the domain's basis are

and those images are represented with respect to the codomain's basis in this way. Hence, the matrix is the identity.

1. Using the matrix in the prior item,

the representation is this.

Problem

Give a formula for the product of a matrix and ( \vec{e}_i ), the column vector that is all zeroes except for a single one in the ( i )-th position.

Answer

The product Failed to parse (syntax error): \begin{pmatrix} h_{1,1} &\ldots &h_{1,i} &\ldots &h_{1,n} \\ h_{2,1} &\ldots &h_{2,i} &\ldots &h_{2,n} \\ &\vdots \\ h_{m,1} &\ldots &h_{m,i} &\ldots &h_{1,n} \end{pmatrix} \begin{bmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} h_{1,i \end{bmatrix} \\ h_{2,i} \\ \vdots \\ h_{m,i}}

gives the ( i )-th column of the matrix.

This exercise is recommended for all readers.

Problem

For each vector space of functions of one real variable, represent the derivative transformation with respect to ( B,B ).

1. ( {a\cos x+b\sin x o a,b\in\Re} ),

( B={\cos x,\sin x} )

1. ( {ae^x+be^{2x} o a,b\in\Re} ),

Problem

Find the range of the linear transformation of ( \Re^2 ) represented with respect to the standard bases by each matrix. \begin{exparts*}

1. \begin{pmatrix}

1 &0 \\ 0 &0 \end{pmatrix}

1. \begin{pmatrix}

0 &0 \\ 3 &2 \end{pmatrix}

1. a matrix of the form

\begin{pmatrix} a &b \\ 2a &2b \end{pmatrix} \end{exparts*}

Answer

1. It is the set of vectors of the codomain represented with

respect to the codomain's basis in this way. As the codomain's basis is \stdbasis_2, and so each vector is represented by itself, the range of this transformation is the x-axis.

1. It is the set of vectors of the codomain represented

in this way. With respect to \stdbasis_2 vectors represent themselves so this range is the y~axis.

1. The set of vectors represented with

respect to \stdbasis_2 as is the line y=2x, provided either a or b is not zero, and is the set consisting of just the origin if both are zero.

This exercise is recommended for all readers.

Problem

Can one matrix represent two different linear maps?

Problem

Prove \nearbytheorem{th:MatMultRepsFuncAppl}.

Answer

We mimic ex:TypLinMapRepByMat, just replacing the numbers with letters.

Write ( B ) as ( {\vec{\beta}_1,\ldots,\vec{\beta}_n} ) and ( D ) as ( {\vec{\delta}_1,\dots,\vec{\delta}_m} ). By definition of representation of a map with respect to bases, the assumption that means that h(\vec{\beta}_i)=h_{i,1}\vec{\delta}_1+\dots+h_{i,n}\vec{\delta}_n. And, by the definition of the representation of a vector with respect to a basis, the assumption that means that ( \vec{v}=c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n ). Substituting gives Failed to parse (syntax error): h(\vec{v}) =h(c_1\cdot\vec{\beta}_1+\dots+c_n\cdot\vec{\beta}_n) \\ =c_1\cdot h(\vec{\beta}_1)+\dots+c_n\cdot \vec{\beta}_n \\ =c_1\cdot (h_{1,1}\vec{\delta}_1+\dots+h_{m,1}\vec{\delta}_m) +\dots +c_n\cdot (h_{1,n}\vec{\delta}_1+\dots+h_{m,n}\vec{\delta}_m) \\ =(h_{1,1}c_1+\dots+h_{1,n}c_n)\cdot\vec{\delta}_1 +\cdots +(h_{m,1}c_1+\dots+h_{m,n}c_n)\cdot\vec{\delta}_m

and so h(\vec{v}) is represented as required.

This exercise is recommended for all readers.

Problem

exam:RepsOfRigidPlaneMaps shows how to represent rotation of all vectors in the plane through an angle ( \theta ) about the origin, with respect to the standard bases.

1. Rotation of all vectors in three-space through an angle

( \theta ) about the ( x )-axis is a transformation of \Re^3. Represent it with respect to the standard bases. Arrange the rotation so that to someone whose feet are at the origin and whose head is at ( (1,0,0) ), the movement appears clockwise.

1. Repeat the prior item, only rotate about the ( y )-axis

instead. (Put the person's head at \vec{e}_2.)

1. Repeat, about the ( z )-axis.
2. Extend the prior item to ( \Re^4 ).

(Hint: `rotate about the ( z )-axis' can be restated as `rotate parallel to the ( xy )-plane'.)

Answer

1. The picture is this.

The images of the vectors from the domain's basis are represented with respect to the codomain's basis (again, \stdbasis_3) by themselves, so adjoining the representations to make the matrix gives this. Failed to parse (unknown function\stdbasis): {r_\theta}{\stdbasis_3,\stdbasis_3}= \begin{pmatrix} 1 &0 &0 \\ 0 &\cos\theta &\sin\theta \\ 0 &-\sin\theta &\cos\theta \end{pmatrix}

1. The picture is similar to the one in the prior answer.

The images of the vectors from the domain's basis are represented with respect to the codomain's basis \stdbasis_3 by themselves, so this is the matrix.

1. To a person standing up, with the vertical z-axis,

a rotation of the xy-plane that is clockwise proceeds from the positive y-axis to the positive x-axis. That is, it rotates opposite to the direction in exam:RepsOfRigidPlaneMaps. The images of the vectors from the domain's basis are represented with respect to \stdbasis_3 by themselves, so the matrix is this.

\begin{pmatrix} \cos\theta &\sin\theta &0 &0 \\ -\sin\theta &\cos\theta &0 &0 \\ 0 &0 &1 &0 \\ 0 &0 &0 &1

Problem

(Schur's Triangularization Lemma)^[Index]%

^[Index]

1. Let ( U ) be a subspace of ( V ) and fix bases

( B_U\subseteq B_V ). What is the relationship between the representation of a vector from ( U ) with respect to ( B_U ) and the representation of that vector (viewed as a member of ( V )) with respect to ( B_V )?

1. What about maps?
2. Fix a basis

( B={\vec{\beta}_1,\dots,\vec{\beta}_n} ) for ( V ) and observe that the spans Failed to parse (unknown function\spanof): \spanof{{\zero}}={\zero}\subset\spanof{{\vec{\beta}_1}} \subset\spanof{{\vec{\beta}_1,\vec{\beta}_2}} \subset \cdots \subset\spanof{B}=V

form a strictly increasing chain of subspaces. Show that for any linear map ( \map{h}{V}{W} ) there is a chain ( W_0={\zero}\subseteq W_1\subseteq \dots \subseteq W_m =W ) of subspaces of ( W ) such that Failed to parse (unknown function\spanof): h(\spanof{{\vec{\beta}_1,\dots,\vec{\beta}_i}})\subset W_i

for each ( i ).

1. Conclude that for every linear map

( \map{h}{V}{W} ) there are bases ( B,D ) so the matrix representing ( h ) with respect to ( B,D ) is upper-triangular

Problem

Is an upper-triangular representation unique?

Answer

Write ( B_U ) as ( {\vec{\beta}_1,\dots,\vec{\beta}_k} ) and then B_V as ( {\vec{\beta}_1,\dots,\vec{\beta}_k, \vec{\beta}_{k+1},\dots,\vec{\beta}_n} ). If Failed to parse (lexing error): {\vec{v}}{B_U}=\begin{bmatrix} c_1 \\ \vdots \\ c_k \end{bmatrix} \text{so that\ } \vec{v}=c_1\cdot\vec{\beta}_1+\cdots+c_k\cdot\vec{\beta}_k

then, because \vec{v}=c_1\cdot\vec{\beta}_1+\dots+c_k\cdot\vec{\beta}_k +0\cdot\vec{\beta}_{k+1}+\dots+0\cdot\vec{\beta}_n.

We must first decide what the question means. Compare ( \map{h}{V}{W} ) with its restriction to the subspace