ch2

Chapter 2

Linear Algebra in DimensionTwo

We first study linear algebra (mostly) in dimension two. Calculations aresimple here, but yet, we can gain a great deal of intuition.

2.1 Linear Equations in two unknowns

Consider the linear equation:

ax+ by = e (2.1.1)

cx+ dy = f (2.1.2)

We would like to solve the above equation. We may multiply the first equa-tion by d and multiply the second by b and subtract, to find:

(ad− bc)x = de− bf. (2.1.3)

If ad− bc 6= 0, then we may solve the above equation to find x, we may alsofind y:

x =de− bf

ad− bc, y =

−ce+ af

ad− bc. (2.1.4)

Let us use matrix notation for the above. The linear equation (2.1.1) maybe written as follows:

(

a bc d

)(

xy

)

=

(

ef

)

. (2.1.5)

17

We may write this simply as

Ax = b, A =

(

a bc d

)

, x =

(

xy

)

, b =

(

ef

)

. (2.1.6)

If ad− bc 6= 0, (2.1.4) may be expressed in matrix form as:

Bb = x, B =1

ad− bc

(

d −b−c a

)

. (2.1.7)

It can be easily checked that:

AB = BA = I (2.1.8)

where I is the identity matrix. Recall from (1.5.7) that such a matrix B isthe inverse matrix B = A−1. With matrix notation, we see that the solutionto the equation Ax = b is simply given by x = A−1b when ad−bc 6= 0. Thequantity ad− bc is the called the determinant of the matrix A is denoted by:

detA = |A| = ad− bc. (2.1.9)

What happens if detA = ad− bc = 0? Suppose for simplicity that botha and c are non-zero. Then, ad− bc = 0 implies that

b = λa, d = λc (2.1.10)

for some constant λ. Equation (2.1.1), thus looks like:

x+ λy =e

a(2.1.11)

x+ λy =f

c(2.1.12)

The above equations only have a solution when e/a = f/c, in which casethere are infinitely many solutions. If e/a 6= f/c, there are no solutions tothe above equations. The geometric interpretation of this is that lines withthe same slope can have common point(s) if and only if they are the sameline. Otherwise, they are two parallel lines and do not intersect. One wayto write the condition e/a = f/c is to write:

(

ac

)

‖(

ef

)

(2.1.13)

MATH 2574H 18 Yoichiro Mori

where ‖ means that either vector is a constant multiple of the other. Gen-eralizing the above argument a little (the reader should try!), one comes tothe following conclusion. Suppose

detA = 0, A 6= O (2.1.14)

where O is the zero matrix. Let the column vectors of A be:

a1 =

(

ac

)

, a2 =

(

bd

)

. (2.1.15)

Then,

Ax = b

{

has infinitely many solutions if b ‖ a1 or b ‖ a2,

has no solution otherwise.(2.1.16)

When A = O, there is not very much to say:

Ox = b

{

has infinitely many solutions if b = 0,

has no solution otherwise.(2.1.17)

where 0 is the zero vector. One consequence of (2.1.17) and (2.1.16) is thefollowing. If detA = 0 then,

Ax = 0 has infinitely many solutions, (2.1.18)

and

there are some b for which Ax = b has no solution. (2.1.19)

Does A have an inverse if detA = 0? The answer is no, and we givetwo different proofs for this. Since detA = 0, we know from (2.1.18) thatthe equation Ax = 0 has infinitely many solutions. This implies that theremust be a vector c such that

Ac = 0, c 6= 0. (2.1.20)

Now, suppose there is a matrix B satisfying BA = I. Multiplying both sidesof the above from the left with B. We see that

c = 0, (2.1.21)

but this is a contradiction, since c 6= 0.


Another way to show that A−1 cannot exist is the following. Supposethere is a matrix B such that

AB = I. (2.1.22)

Let b be a vector for which Ax = b cannot be solved (such a vector existsfrom (2.1.19)). Given (2.1.22), we have

ABb = b. (2.1.23)

But this implies that Bb is a solution to Ax = b, which is a contradiction.There is thus no matrix B satisfying AB = I.

Some of you may have realized that the two arguments prove two slightlydifferent statements; the first shows that a matrix B satisfying BA = I (aleft inverse) does not exist, and the second shows that a matrix B satisfyingAB = I (a right inverse) does not exist. Since we have taken the definitionof the inverse matrix in (1.5.7) to be a matrix satisfying both AB = Iand BA = I (a matrix that is a left and a right inverse), negating eitherpossibility will show that an inverse cannot exist.

Let us now summarize our results into a theorem.

Theorem 1. For a 2 × 2 matrix A and a vector b ∈ R2, the following

statements are equivalent.

• detA 6= 0.

• Ax = b can be solved for any b.

• Ax = 0 has only one solution, x = 0.

• The matrix A has an inverse.

The same theorem can also be stated in the following way (can you seewhy the two theorems are equivalent?).

Theorem 2. For a 2 × 2 matrix A and a vector b ∈ R2, the following

statements are equivalent.

• detA = 0.

• There are some vectors b for which Ax = b does not have a solution.

• Ax = 0 has infinitely many solutions.

• The matrix A does not have an inverse.


Example 5. Consider the equation:

Ax = b, A =

(

1 22 5

)

, x =

(

xy

)

, b =

(

ef

)

. (2.1.24)

For this equation, the determinant is equal to detA = 1 6= 0, and therefore,

there is always a unique solution for any b. The inverse of A is:

A−1 =

(

5 −2−2 1

)

. (2.1.25)

On the other hand, consider:

Ax = b, A =

(

1 23 6

)

, x =

(

xy

)

, b =

(

ef

)

. (2.1.26)

In this case, detA = 0 and therefore, there is no inverse. There are solutions

if:(

13

)

‖(

ef

)

or f − 3e = 0, (2.1.27)

and in this case, the solutions are all x satisfying:

x+ 2y = e. (2.1.28)

2.2 Linear Independence/Basis Vectors

Given vectors v1, · · · ,vm ∈ Rn, a linear combination of vectors is given by:

c1v1 + c2v2 + · · ·+ cnvm (2.2.1)

where c1, · · · , cn are scalars.

Definition 1 (linear independence). The vectors v1, · · · ,vm ∈ Rn are lin-

early independent if

c1v1 + c2v2 + · · ·+ cmvm = 0 implies ck = 0 for all k = 1, · · · ,m. (2.2.2)

If this is not the case, the vectors v1, · · · ,vm are linearly dependent.

One may also state linear dependence as follows. The vectors v1, · · · ,vm

are linearly dependent if there are scalar constants c1, · · · , cn, not all of whichare equal to 0, satisfying

c1v1 + c2v2 + · · ·+ cmvm = 0. (2.2.3)


Since at least one of the ck is not equal to 0, say, cj , we may divide by cj toobtain:

vj = − 1

cj(c1v1 + · · · + cj−1vj−1 + cj+1vj+1 + · · ·+ cnvn. (2.2.4)

Thus, linear dependence is equivalent to saying that one of the vectors canbe written as a linear combination of the others.

Example 6. Consider the vectors:

v1 =

(

00

)

,v2 =

(

12

)

,v3 =

(

24

)

,v4 =

(

1−1

)

,v5 =

(

01

)

. (2.2.5)

Then, the zero vector v1 with any other combination of vectors is linearly

dependent. For example,

1 · v1 + 0 · v2 = 0 (2.2.6)

and therefore the two vectors v1,v2 are linearly dependent. Vectors v2 and

v3 are linearly dependent because:

2v2 − v3 = 0. (2.2.7)

Let us now examine v2 and v4. We must consider the expression:

c1v2 + c2v4 = c1

(

12

)

+ c2

(

1−1

)

=

(

1 21 −1

)(

c1c2

)

=

(

00

)

. (2.2.8)

We may see the above as an equation for c1 and c2. We may use our result

from Theorem 1 to see that c1 = 0, c2 = 0 is the only solution to the above

equation. Therefore, v2 and v4 are linearly independent. Now, consider the

vectors v2,v4 and v5. In this case, we have:

v2 − v4 − 3v5 = 0 (2.2.9)

and therefore the three vectors are linearly dependent.

Let us now consider the general case of vectors in R2. First, let us

consider just one vector v1. This vector is linearly independent if and onlyif v1 6= 0.

Next, consider the case of two vectors

v1 =

(

ac

)

,v2 =

(

bd

)

. (2.2.10)


If two vectors are linearly dependent, we have

c1v1 + c2v2 = 0 (2.2.11)

where not both c1 and c2 are zero. If c1 6= 0, this means that:

v1 = −c2c1v2. (2.2.12)

The vector v1 is a constant multiple of v2. If c2 6= 0, then v2 is a constantmultiple of v1. Thus, the geometric meaning of linear dependence of twovectors is that the two vectors are colinear (we shall think of the zero vectoras being colinear to all vectors). In fact, it is easy to convince ourselves thattwo vectors are colinear, or linearly dependent, if and only if:

ad− bc = 0. (2.2.13)

If we introduce the matrix

A =

(

a bc d

)

(2.2.14)

this condition is just detA = 0. An equivalent statement is that two vectorsare linearly independent if and only if detA 6= 0.linear dependence of twovectors means that the two vectors are colinear (we shall think of the zerovector as being colinear to all vectors).

Another way to see this is the following. We are interested in lookingfor scalar constants c1 and c2 that satisfy:

c1v1 + c2v2 =

(

a bc d

)(

c1c2

)

=

(

00

)

. (2.2.15)

The above can be seen as an equation for c1 and c2. From Theorem 1, we seethat the only solution to the above is c1 = 0, c2 = 0 if and only if detA 6= 0.

Suppose v1,v2 are linearly independent. Let us consider the followingproblem. Given any vector v3, can we write v3 as a linear combination ofv1 and v2? We want to find constants c1 and c2 that satisfy

v3 = c1v1 + c2v2. (2.2.16)

Let

v1 =

(

ac

)

, v2 =

(

bd

)

, v3 =

(

ef

)

. (2.2.17)

Finding c1 and c2 satisfying (2.2.16) is equivalent to solving the followinglinear equation:

(

a bc d

)(

c1c2

)

=

(

ef

)

. (2.2.18)


Since v1 and v2 are linearly independent, ad − bc 6= 0. This equation thushas a unique solution. Thus, any vector in R

2 can be written uniquelyas a linear combination of linearly independent vectors v1 and v2. A setof linearly independent vectors in R

n is called a set of basis vectors if anyvector in R

n can be expressed as a linear combination of these vectors. Wethus see that any two linearly independent vectors form a set of basis vectorsin R

2.

We may use this fact to show that a set of 3 or more vectors can neverbe linearly independent. Suppose that the vectors v1, · · · ,vm,m ≥ 3 arelinearly independent. Then, there are two vectors in this set, say, v1 andv2 that are linearly independent. Since v1 and v2 are linearly independent,they form a basis, and therefore, v3 can be written as a linear combinationof v1 and v2. This shows that the three vectors v1,v2 and v3 are linearlydependent. This is a contradiction.

Let us state our observations as a theorem.

Theorem 3. Two vectors

v1 =

(

ac

)

, v2 =

(

bd

)

, (2.2.19)

are linearly independent if and only if

detA 6= 0, A =

(

a bc d

)

(2.2.20)

Any two linearly independent vectors in R2 form a basis of R2. The maxi-

mum number of linearly independent vectors in R2 is 2.

The first part of the above theorem can also be seen as saying thatdetA 6= 0 is equivalent to the column vectors of A being linearly independent.It is in fact easy to see that the row vectors are also linearly independent ifand only if detA 6= 0. We thus have two more conditions we can add to theequivalent conditions in Theorem 1.

Theorem 4. For a 2× 2 matrix A, the following statements are equivalent.

• detA 6= 0.

• The column vectors of A are linearly independent.

• The row vectors of A are linearly independent.


2.3 Linear Transformation

An important way to view square matrices is as linear transformations. Takea 2 × 2 matrix A and a vector in x ∈ R

2. We may view A as a map thattakes the vector x to the vector Ax. Letting

A =

(

a bc d

)

, x =

(

xy

)

(2.3.1)

we see that the vector x is mapped to:

Ax = xv1 + yv2,v1 =

(

ac

)

, v2 =

(

bd

)

. (2.3.2)

The unit vectors e1 = (1, 0)T and e2 = (0, 1)T are thus mapped to thevectors v1 and v2. The unit coordinate square spanned by e1 and e2 getsmapped to the parallelogram spanned by the vectors v1 and v2. As a whole,the orthogonal coordinate plane R

2 gets mapped to a slanted coordinateplane with coordinate axes aligned with v1 and v2.

Let us interpret the results in the previous sections in terms of this lineartransformation. First, suppose detA 6= 0. From a geometric point of view,this means that v1 and v2 are not parallel (or linearly independent). Fromthis, it is graphically clear that every point in R

2 gets mapped one-to-oneonto R

2. This implies that A should have an inverse, and that Ax = b

should be uniquely solvable for every b. Note also that the only point thatgets mapped to the origin 0 is 0.

Now, suppose detA = 0. This means that (a, c)T and (b, d)T are colinear,or linearly dependent. There are two cases to consider.

First, suppose either one of v1 and v2 is a nonzero vector (say, v1).Then, we see that A maps R

2 to a line ℓ that goes through the origin andis parallel to v1. Since

v2 = λv1 for some λ (2.3.3)

we see that points satisfying x+λy = 0 get sent to the origin. The equationAx = b will have a solution only if b lies on the line ℓ.

The second case is when a = b = c = d = 0. In this case, all points inR2 get sent to the origin.We thus see that depending on the matrix A, the linear transformation

will map the plane either to the whole plane R2, a line or the origin. The

set which the plane is mapped to is called the image of A and is written asImA. We thus see that matrices can be classified by the dimension of theimage (we shall later define dimension, but for now, think of it in the usual


intuitive way; a plane has dimension 2, a line has dimension 1, etc), whichwe call the rank of a matrix. If the matrix A maps the plane to the plane,the rank of A is 2. If A maps the plane to a line, the rank of A is 1. If Amaps the plane to the origin, the rank of A is 0.

A notion that is intimately related to the image is the nullspace (orkernel) of a matrix A and is written as KerA. The nullspace is the set ofpoints that are sent to the origin. If the matrix A has rank 2, then the onlypoint sent to the origin is the origin. The kernel is the origin alone, and thedimension of the kernel is 0. If the matrix A has rank 1, then a whole planegets sent to a line. The dimension of the kernel is 1. When the rank of A isO, then the whole plane gets mapped to the origin, and therefore, the kernelis the whole plane and the dimension of the kernel is 2.

We describe this as a theorem:

Theorem 5. Consider a linear transformation given by a 2× 2 matrix A.

1. If detA 6= 0, then the ImA is the whole plane and the kernel is just the

origin. The rank of A is 2 and the the dimension of the kernel is 0.

2. If detA = 0 and A 6= O, then ImA is a line and the kernel is a line.

The rank of A is 1 and the dimension of the kernel is 1.

3. If A = O, then ImA is the origin, and the kernel is the whole plane.

The rank of A is 0 and the dimension of the kernel is 2.

An interesting consequence of this is that:

rankA+ dimkerA = 2, (2.3.4)

where rankA is the rank of A and dimkerA is the dimension of the kernel ofA.

There is an interesting geometric meaning to the determinant. It is thesigned area of the parallelogram spanned by the vectors v1 and v2. It issigned in the sense that:

det

(

1 00 1

)

= 1, det

(

0 11 0

)

= −1. (2.3.5)

Even though the parallelogram (square) spanned by (1, 0)T and (0, 1)T isthe same as that spanned by (0, 1)T and (1, 0)T, the latter determinant hasa negative sign. The sign is carrying information on orientation. Supposeyou turn the first vector anti-clockwise until it overlaps with the secondvector. If you can do this by turning the first vector less than 180 degrees


(or π radians), then the determinant is positive. If more than 180 degreesis required, then the determinant is negative.

The “parallelogram area” interpretation of the determinant makes itclear why detA detects whether the vectors v1 and v2 are linearly indepen-dent. If the two vectors are non-parallel (linearly independent), then theparallelogram formed by the two vectors should have a non-zero area. If thetwo vectors are parallel (linearly dependent), the two vectors collapse ontoa line, and the resulting parallelogram does not have any area.

The determinant can also be thought of as a scaling factor for area.As we saw, the unit coordinate square gets mapped to the parallelogramspanned by v1 and v2. Therefore, the area of the unit square is scaled by afactor of detA. It is not difficult to convince oneself that this should actuallybe true for any shape. For example, any circle will be mapped to an ellipse(it turns out) that has an area that is detA times the original circle.

An important property of the determinant is its compatibility with ma-trix multiplication. Suppose A and B are both 2× 2 matrices. Then,

det(AB) = (detA)(detB). (2.3.6)

This can be checked by a somewhat tedious algebraic computation. Thegeometric reason why (2.3.6) holds is the following. Suppose we apply thematrix AB to R

2. For a vector v ∈ R2, we have:

(AB)(v) = A(Bv) (2.3.7)

by the associative law. This implies that the linear transformation by ABis the same as applying the linear transformation by B and A in succession.The magnifying factor for the area, when applying the linear transformationAB should be det(AB). Applying the linear transformation AB to R

2 isthe same as applying the linear transformation B and then A to R

2. In thiscase, the magnifying factor for the area should be detA × detB. The twomagnifying factors should be the same, and we thus have (2.3.6).

Example 7. Consider the matrices:

A =

(

1 22 5

)

, B =

(

1 23 6

)

. (2.3.8)

We see that detA = 1 6= 0 and therefore, the matrix A maps the plane to

the plane. The kernel is just the origin.

The matrix B has determinant equal to 0. Since the matrix is non-zero,

it must have rank 1. To find the image and kernel of B, we may apply B to


the vector (x, y)T:

B

(

xy

)

= (x+ 2y)

(

13

)

. (2.3.9)

Therefore, the kernel is the line:

x+ 2y = 0 (2.3.10)

and the image is the line parallel to (1, 3)T passing through the origin. In

other words, the image is the line:

y = 3x. (2.3.11)

2.4 Examples of Linear transformations

Here, we take a look at some examples of linear transformations. Before weproceed, we make the following useful observation. Suppose

Au1 = v1 and Au2 = v2, (2.4.1)

where u1 and u2 are linearly independent. Let U be the matrix formed bytaking u1 and u2 to be the column vectors, and V be the matrix formed bytaking v1 and v2 to be the column vectors. Then, (2.4.1) can be written as:

AU = V. (2.4.2)

Since the column vectors of U are linearly independent, U has an inverse,and therefore,

A = V U−1. (2.4.3)

A linear transformation A can be found if we know where two linearly inde-pendent vectors are sent.

2.4.1 Scaling Transformation

The matrix:

A =

(

a 00 b

)

(2.4.4)

scales the x axis by a factor a and the y axis by a factor of b. There are manymatrices that have a similar character. For example, consider the matrix:

A =

(

2 11 2

)

. (2.4.5)


At first glance, this does not look like (2.4.4), but,

A

(

11

)

= 3

(

11

)

, A

(

1−1

)

=

(

1−1

)

(2.4.6)

So this matrix scales the (1, 1)T direction by a factor of 3 and the (1,−1)T

direction by a factor of 1.

2.4.2 Rotation

Rotation of the plane about the origin is a linear transformation. To seewhat the matrix may be, we examine where the vectors (1, 0)T and (0, 1)T

may be mapped to by rotation about the origin by an angle θ in the counter-clockwise direction. We see that

(

10

)

7→(

cos(θ)sin(θ)

)

,

(

01

)

7→(

− sin(θ)cos(θ)

)

. (2.4.7)

Therefore, the rotation matrix R(θ) about the origin is given by:

R(θ) =

(

cos(θ) − sin(θ)sin(θ) cos(θ)

)

. (2.4.8)

For example, rotation by π/2, or 90 degrees is given by:

R(π/2) =

(

0 −11 0

)

. (2.4.9)

Note that the determinant of the matrix R(θ) is equal to 1. Rotation doesnot change the area of anything. The inverse matrix of a rotation by degreeθ should be a rotation matrix by degree −θ. Indeed, it can be checked that

R(θ)R(−θ) = R(−θ)R(θ) = I. (2.4.10)

One interesting thing we can do with rotation matrices is the following. Arotation by θ followed by a rotation by ψ should give us a rotation by θ+ψaround the origin. Therefore, we should have:

R(θ + ψ) = R(ψ)R(θ) (2.4.11)

From this, it is easy to deduce the addition rule for the sines and cosines(see Exercise 10 of Section 2.5).


2.4.3 Reflection

Take the line y = mx and let us consider reflection across this line. Thisis also a linear transformation. Let us call this matrix A(m). We need toknow the images of two linearly independent vectors. The vector (1,m)T

must be mapped to itself, since (1,m)T lies on the line itself. The vector(m,−1)T, which is perpendicular to the line y = mx, must be mapped to(−m, 1)T. We thus have:

A(m)

(

1m

)

=

(

1m

)

, A(m)

(

m−1

)

=

(

−m1

)

. (2.4.12)

We thus have:

A(m)

(

1 mm −1

)

=

(

1 −mm 1

)

(2.4.13)

We thus see that

A(m) =

(

1 −mm 1

) −1

1 +m2

(

−1 −m−m 1

)

=1

1 +m2

(

1−m2 2m2m m2 − 1

)

.

(2.4.14)We can produce another derivation of this same result using rotations. Thevector y = mx subtends an angle θ (m = tan θ) with respect to the x axis.Thus, reflection across the line y = mx can be seen as a succesion of thefollowing operations.

1. Rotate around the origin by an angle −θ.

2. Reflect across the x-axis. Since the point (x, y)T gets mapped to(x,−y)T, reflection across the x-axis is given by the matrix:

A(0) =

(

1 00 −1

)

. (2.4.15)

3. Rotate around the origin by an angle θ.

Therefore, reflection across the line y = mx is given by:

A(m) = R(θ)A(0)R(−θ) =(

cos2(θ)− sin2(θ) 2 cos(θ) sin(θ)2 cos(θ) sin(θ) −(cos2(θ)− sin2(θ))

)

=

(

cos(2θ) sin(2θ)sin(2θ) − cos(2θ)

)

.

(2.4.16)


It is not difficult to show that the above and (2.4.14) are the same matrix,using m = tan(θ) and some trigonometric identities.

An interesting fact about reflection matrices is that their determinant is−1. A reflection does not change area but reverses the orientation.

2.4.4 Orthogonal Projection

Consider the line y = mx, and let P (m) be the linear transformation thattakes a point on the plane to the nearest point on the line y = mx. Thisis called the orthogonal projection. The vector (1,m)T is mapped to itselfand the vector (m,−1)T is mapped to the origin. Therefore,

P (m)

(

1 mm −1

)

=

(

1 0m 0

)

. (2.4.17)

We see that

P (m) =1

1 +m2

(

1 mm m2

)

. (2.4.18)

Note that the determinant of matrix is 0, as it should be. The whole planegets mapped to the line y = mx. We also see that:

P (m)2 = P (m). (2.4.19)

Applying a projection twice produces the same result.

2.4.5 Magnification and Rotation

The following matrix occurs pretty often:

A =

(

a −bb a

)

(2.4.20)

where (a, b) 6= (0, 0). To understand what this matrix does, we write it inthe following form:

A =√

a2 + b2

(

a√

a2+b2− b

√

a2+b2b

√

a2+b2a

√

a2+b2

)

. (2.4.21)

Setting θ so that

cos(θ) =a√

a2 + b2, sin(θ) =

b√a2 + b2

, (2.4.22)


we may write A as:

A =√

a2 + b2R(θ) (2.4.23)

where R(θ) is a rotation matrix. We thus see that A may be seen as rotationby an angle θ around the origin followed by magnification by a factor of√a2 + b2 (or vice-versa). For example, the matrix:

A =

(

1 −11 1

)

(2.4.24)

has the effect of rotating by π/4 (45 degrees) and magnifying the wholeplane by a factor of

√2.

2.5 Exercises

1. Consider the linear equations:

Ax = b, x =

(

xy

)

, b =

(

ef

)

. (2.5.1)

Discuss the solvability of this equation for the following choices of the2× 2 matrix A. Does the equation have a solution for every b? If not,for what right hand side b does the equation have a solution? If thereis a solution, find what the solution is.

(

1 −12 1

)

,

(

1 −1−1 1

)

,

(

2 4−1 −2

)

,

(

5 51 3

)

,

(

9 33 1

)

(

0 00 0

)

,

(

3 52 3

)

,

(

2 5−4 −10

)

,

(

4 71 2

)

,

(

6 21 1/2

) (2.5.2)

2. Find the inverse matrices of Exercise 1 if they exist.

3. Describe the image and kernel of the matrices of Exercise 1.

4. Examine if the following vectors are linearly independent. If they arelinearly dependent, exhibit a nontrivial linear combination that givesthe zero vector.

(a) (1, 1)T, (1, 2)T.

(b) (2, 3)T, (4, 6)T.

(c) (0, 0)T, (1, 1)T.


(d) (1, 0)T, (0, 3)T.

(e) (2, 1)T, (1, 2)T, (0, 3)T.

(f) (3, 1)T, (1,−1)T, (1, 1)T.

(g) (1, 1)T, (2, 2)T, (3, 4)T.

(h) (−1, 1)T, (2, 1)T, (3, 3)T.

5. Prove the product formula (2.3.6) for the determinant of 2×2 matrices.

6. A matrix A that satisfies the property:

A2 = A (2.5.3)

is called a projection (note that the orthogonal projection is a spe-cial kind of projection, see (2.4.19)). What can the determinant ofa projection be? Hint: Use the product rule for the determinant of

matrices.

7. Consider matrices A and B that are inverses of each other, that is tosay, AB = BA = I.

(a) Show that (detA)(detB) = 1.

(b) Suppose components of both A and B are integers. What canyou say about detA and detB? Hint: Since A and B are inte-

ger matrices, their determinant must also be integers. Therefore,

detA and detB are two integers that, if multiplied, give you the

value of 1.

8. 2× 2 matrices A that satisfy the relation:

AT

(

1 00 −1

)

A =

(

1 00 −1

)

(2.5.4)

are important in Einstein’s theory of relativity. What are the possiblevalues of the determinant of A? Hint: Use the product rule. Note that

the determinant of A and AT are equal.

9. Find the linear transformation with the following properties.

(a) Maps (1, 0)T to (1, 1)T and (0, 1)T to (0, 2)T.

(b) Maps (2, 1)T to (1, 1)T and (3, 1)T to (0, 1)T.

(c) Maps (1,−1)T to (0, 1)T and (−2, 1)T to (1, 2)T.


(d) Rotation around the origin by 3π/4 in the counter-clockwise di-rection.

(e) Rotation around the origin by π/3 in the counter-clockwise di-rection.

(f) Rotation around the origin by π/6 in the counter-clockwise di-rection, followed by a magnification by a factor of 2.

(g) Reflection about the line y = 2x.

(h) Reflection about the line y = x followed by a reflection about theline y = 2x.

(i) Orthogonal projection to the line y = −2x.

(j) Orthogonal projection to the line y = x followed by an orthogonalprojection to the line y = −x.

10. Compute both sides of (2.4.11) to deduce the addition law for sinesand cosines.

11. Suppose A(m) is the matrix for reflection about y = mx and A(n)is the matrix for reflection about y = nx. Show that A(m)A(n) is arotation matrix (that is to say, successive reflection across two linesthrough the origin is in fact a rotation about the origin). What is therotation angle? Hint: Use expression (2.4.16) and use some trigono-

metric identities.

12. Take any vector v ∈ R2. Show that the length of the vector does

not change after application of either a rotation matrix or a reflectionmatrix.

13. Consider the matrix (2.4.20) and write this as:

A = aI + bJ, I =

(

1 00 1

)

, J =

(

0 −11 0

)

. (2.5.5)

(a) Show that IJ = JI.

(b) Using this, compute (aI + bJ)(cI + dJ).

(c) Compare this with the multiplication of two complex numbersa+ bi and c+ di where c and d are real numbers.


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