Linear Algebra. Session 11 - math.tamu.eduroquesol/Math_304...Dr. Marco A Roque Sol Linear Algebra....

Abstract Linear Algebra

Linear Algebra. Session 11

Dr. Marco A Roque Sol

04 / 21 / 2020

Dr. Marco A Roque Sol Linear Algebra. Session 11

Abstract Linear Algebra Orthogonal Polynomials

Examples

Inner Product.

Find the distance from the point (2, 1,−2) to the plane Π, givenby:.

6(x − 1) + 2(y − 3) + 3(z + 4) = 0.

Solution

The Solution of the above equation is : xyz

= α

10−2

+ β

03−2

+

13−4

so the vectors x1 = (1, 0,−2)T and x2 = (0, 3,−2)T span theplane Π.



Examples

The orthogonal projection,p, of v = (2, 1,−2)T − (1, 3,−4)T =(1,−2, 2)T onto the plane Π is given by:

p =< v, x1 >

< x1, x1 >x1 +

< v, x2 >

< x2, x2 >x2

where< v, x1 >=< 1,−2, 2 > · < 1, 0,−2 >= −3

< v, x2 >=< 1,−2, 2 > · < 0, 3,−2 >= −10

< x1, x1 >=< 1, 0,−2 > · < 1, 0,−2 >= 5

< x2, x2 >=< 0, 3,−2 > · < 0, 3,−2 >= 13



Examples

It follows that

p = −35x1 − 10

13x2 = −35(1, 0,−2)T − 10

13(0, 3,−2)T ⇒

p = (−35 ,−

1013 ,

20865 )T

Finally, the distance from (2, 1,−2) to the palne Π is :

||o = v − p|| = ||(1,−2, 2)T − (−3

5,−10

13,

208

65)T || =

||o|| = ||(8/5,−16/13,−78/65)T || = 65||(104,−80,−78)T ||

||o|| = 65√

1042 + 802 + 782 = 65√

23300 �



Examples

Orthogonal Subspaces.

(a) Let S be the subspace of R3 spanned by the vectorsx = (x1, x2, x3)T and y = (y1, y2, y3)T , and let A be the matrixdefined by:

A =

(x1 x2 x3y1 y2 y3

)(a) Show that S⊥ = N(A).

(b) Find the orthogonal complement of the subspace of R3

spanned by x = (1, 2, 1)T and y = (1,−1, 2)T .

Solution

a) A vector v = (v1, v2, v3) belongs to S⊥ if and only if



Examples

< v, x >= 0

< v, y >= 0

but that is true if and only if

v1x1 + v2x2 + v3x3 = 0

v1y1 + v2y2 + v3y3 = 0

if and only if, in matrix notation

(x1 x2 x3y1 y2 y3

) v1v2v3

=

(00

)



Examples

if and only if

Av = 0

if and only if v ∈ S⊥ �

b) Using a), we are looking for the solutions of

Av =

(x1 x2 x3y1 y2 y3

) v1v2v3

=

(1 2 11 −1 2

) v1v2v3

=

(00

)

using elementary row operations we find the row-echelon form ofthe matrix A



Examples

(1 2 10 −3 1

) v1v2v3

⇒ (1 2 10 1 −1/3

) v1v2v3

=

(00

)

and we obtain the system

v1 + 2v2 + v3 = 0v2 − 1

3v3 = 0

whose solution is the one-dimensional subspace:

N(A) =

v ∈ R3/ v = α

−513

, α ∈ R



Examples

Least Squares Solutions.

Find the least squares solution of each of the following system:

−x1 + x2 = 102x1 + x2 = 5x1 − 2x2 = 20

Solution

In matrix notation, the above system can be written as

Ax = b⇒

−1 12 11 −2

( x1x2

)=

105

20



Examples

and we know that a vector y is a lest square approximation for thesystem if, it minimizes ||r(y)|| = ||Ay − b||, and this is happeningif y satisfies the normal equations. namely,

ATAy = ATb

(−1 2 11 1 −2

) −1 12 11 −2

( y1y2

)=

(−1 2 11 1 −2

) 105

20



Examples

(6 −1−1 6

)(y1y2

)=

(20−25

)⇒

y =

(2.71−3.71

)�



Examples

Inner Product.

In C [−π, π] with inner product defined by

< f , g >=1

π

∫ π

−πf (x)g(x)dx ,

show that cos(mx) and sin(nx) are orthogonal and that both areunit vectors. Determine the distance between the two vectors.

Solution

< cos(mx), sin(nx) >=1

π

∫ π

−πcos(mx) sin(nx)dx

and using the Trig. Identitiy

cos a sin b =1

2[sin(a + b)x − sin(a− b)x ]

we obtain:Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

< cos(mx), sin(nx) >=1

2π

∫ π

−π[sin(m + n)x − sin(m − n)x ]dx =

1

2π

∫ π

−πsin(m + n)x − 1

2π

∫ π

−πsin(m − n)x = 0− 0 = 0

On the other hand

< cos(mx), cos(mx) >=1

π

∫ π

−πcos2(mx)dx =

1

2π

∫ π

−π[1+cos(2mx)]dx

=1

2π

∫ π

−π1dx +

1

2π

∫ π

−πcos(2mx)dx = 1 + 0 = 1.



Examples

< sin(nx), sin(nx) >1

π

∫ π

−πsin2(mx)dx =

1

2π

∫ π

−π[1− cos(2mx)]dx

=1

2π

∫ π

−π1dx − 1

2π

∫ π

−πcos(2mx)dx = 1− 0 = 1

Now the distance between x = cos(mx) and y = sin(nx) is givenbiy

||x− y||2 =< x− y, x− y >=1

π

∫ π

−π[cos(mx)− sin(nx)]2dx =

1

π

∫ π

−π[cos2(mx)−2 cos(mx) sin(nx) + sin2(mx)]dx = 1−0 + 1 = 2

Therefore ||x− y|| =√

2 �



Examples

Norms

Determine whether the following define norms on C [a, b]:

(a) ||f || = |f (a)|+ |f (b)|

(b) ||f || =∫ ba |f (x)|dx

(c) ||f || = maxa≤x≤b |f (x)|

Solution

Remmember that a function || · || : V→ R is called a norm definedin the vector space V, if satisfies:

||x|| ≥ 0, ||x|| = 0 only for x = 0

||rx|| = |r |||x|| for all r ∈ R.

||x + y|| ≤ ||x||+ ||y||.



Examples

a) ||f || = |f (a)|+ |f (b)| .

||f || = |f (a)|+ |f (b)| = 0⇒ |f (a)| = 0, |f (b)| = 0, but notnecessarily that the function f (x) = 0 in the whole interval,Therefore, this function, does not define a norm ... !!!!

b) ||f || =∫ ba |f (x)|dx .

||f || =∫ ba |f (x)|dx > 0 provided f (x) 6= 0. Now, if

||f || =∫ ba |f (x)|dx = 0 this implies, from Calculus, that f (x) = 0

on [a, b]

||rf || =∫ ba |rf (x)|dx = |r |

∫ ba |f (x)|dx = |r |||f ||.

||f + g || =∫ ba |f (x) + g(x)|dx ≤

∫ ba |f (x)|dx +

∫ ba |g(x)|dx =

||f ||+ ||g ||.

Therefore, this function, defines a norm.Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

c) ||f || = maxa≤x≤b |f (x)| .

||f || = maxa≤x≤b |f (x)| > 0 provided f (x) 6= 0. Now, if||f || = maxa≤x≤b |f (x)| = 0 this implies that f (x) = 0 on [a, b]

||rf || = maxa≤x≤b |rf (x)| = maxa≤x≤b |r ||f (x)| =

|r |maxa≤x≤b |f (x)| = |r |||f ||.

||f + g || = maxa≤x≤b |f (x) + g(x)| ≤= maxa≤x≤b |f (x)|+ |g(x)| =

||f ||+ ||g ||.

Therefore, this function, defines a norm.



Examples

Orthogonal Polynomials

Let p0, p1, . . . be a sequence of orthogonal polynomials and let andenote the lead coefficient of pn. Prove that

||pn||2 = an < xn, pn >

Solution

If p0, p1, . . . is a sequence of orthogonal polynomials, then weknow that

i) deg(pn) = n

ii)< pn, xk >= 0, for 0 ≤ k ≤ n − 1



Examples

Thus, if pn(x) is given by

pn(x) = anxn + an−1x

n−1 + · · ·+ a1x + a0

then

||pn||2 =< pn(x), pn(x) >

||pn||2 =< anxn + an−1x

n−1 + · · ·+ a1x + a0, pn >=

an < xn, pn > +an−1 < xn−1, pn > + · · ·+ a1 < x , pn > +a0 < 1, pn >

but due to ii), we have that

||pn||2 = an < xn, pn > �



Examples

Orthogonal Polynomials

Let Tn(x) denote the Chebyshev polynomial of degree n and define

Un−1(x) =1

nT ′n(x); n = 1, 2, . . .

(a) Compute U0(x),U1(x), and U2(x).

(b) Show that if x = cos(θ) , then

Un−1 =sin(nθ)

sin(θ).

Solution

a) T1 = x ⇒ U0 = T ′1(x) = 1



Examples

T2 = 2x2 − 1⇒ U1 = 12T′2(x) = 2x

T3 = 4x3 − 3x ⇒ U2 = 13T′3(3) = 4x2 − 1

T4 = 8x4 − 8x2 + 1⇒ U3 = 14T′4(x) = 8x3 − 4x

b)

Un−1(x) =1

nT ′n(x)

but we know that Tn(cos θ) = cos(nθ), therefore

Un−1(x) =1

n

dTn(x)

dx=

1

n

dTn(x)

dθ

dθ

dx=

1

n

dTn(cos(θ))

dθ/dx

dθ⇒



Examples

Un−1(x) =1

n

d cos(nθ)

dθ/dx

dθ

Un−1(x) =1

n[−n sin(nθ)] /[− sin(θ)]

Un−1(x) =sin(nθ)

sin(θ)�



Examples

Markov Chain

Let A be an n × n stochastic matrix and let e be the vector in Rn

whose entries are all equal to 1. Show that e is is an eigenvector ofAT . Explain why a stochastic matrix must have λ = 1 as aneigenvalue.

Solution

Let A be the stochastic matrix with entries given by

A =

a11 a12 · · · a1na21 a22 · · · a2n

... · · ·an1 an2 · · · ann



Examples

By the definition of a stochastic matrix ( the entries of eachcolumn of A are nonnegative numbers that add up to 1 ), then Asatifies:

a1j + a2j + · · ·+ anj = 1; for all 1 ≤ j ≤ n

and At = (bij) satisfies

bi1 + bi2 + · · ·+ bin = 1; for all 1 ≤ i ≤ n

Therefore

ATe =

b11 b12 · · · b1nb21 b22 · · · b2n

... · · ·bn1 bn2 · · · bnn

11...1

=



Examples

b11 b12 · · · b1nb21 b22 · · · b2n

... · · ·bn1 bn2 · · · bnn

11...1

=

b11 + b12 + · · ·+ bnb21 + b22 + · · ·+ b2n

...bn1 + bn2 + · · ·+ bnn

⇒

b11 b12 · · · b1nb21 b22 · · · b2n

... · · ·bn1 bn2 · · · bnn

11...1

=

11...1

Thus, e is an eigenvector of AT with eigenvalue λ = 1. �



Examples

Markov Chain

In the Dark Ages, Harvard, Dartmouth, and Yale admitted onlymale students. Assume that, at that time, 80 percent of the sonsof Harvard men went to Harvard and the rest went to Yale, 40percent of the sons of Yale men went to Yale, and the rest splitevenly between Harvard and Dartmouth; and of the sons ofDartmouth men, 70 percent went to Dartmouth, 20 percent toHarvard, and 10 percent to Yale.

(i) Find the probability that the grandson of a man from Harvardwent to Harvard.

(ii) Modify the above by assuming that the son of a Harvard manalways went to Harvard. Again, find the probability that thegrandson of a man from Harvard went to Harvard.



Examples

Solution

i)

We first form a Markov chain with state space S = {H,D,Y } andthe following transition probability matrix :

P =

.8 0 .2.2 .7 .1.3 .3 .4

Recall: the ij th entry of the matrix Pn gives the probability that theMarkov chain starting in state i will be in state j after n steps.sd

Thus, the probability that the grandson of a man from Harvardwent to Harvard is the upper-left element of the matrix P2.



Examples

P2 =

.7 .06 .24.33 .52 .15.42 .33 .25

ii)

If all sons of men from Harvard went to Harvard, this would givethe following matrix for the new Markov chain with the same set ofstates:

P =

1 0 0.2 .7 .1.3 .3 .4

and The upper-left element of P2 is 1. Does that make sense ????



Examples

Markov Chain

Let consider a hypothetical market with Markov properties wherehistorical data has given us the following patterns: After a weekcharacterized of a bull market trend there is a 90% chance thatanother bullish week will follow. Additionally, there is a 7.5%chance that the bull week instead will be followed by a bearish one,or a 2.5% chance that it will be a stagnant one. After a bearishweek theres an 80% chance that the upcoming week also will bebearish, and so on.

Calculate the possibilities of a bull, bear or stagnant week for, 1and 5 weeks, into the future.

Solution

By compiling these probabilities into a table, we get the followingtransition matrix M:Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

M =

0.9 0.075 0.0250.15 0.8 0.050.25 0.25 0.05

Then create an initial vector x0 which contains information aboutwhich of the three different states any current week is in. In thisexample we will choose to set the current week as bearish,therefore x0 = (0, 1, 0)

Given the state of the current week, we can then calculate thepossibilities of a bull, bear or stagnant week for any number of nweeks into the future. In this way, after one and five weeks we willhave:



Examples

x1 = (0, 1, 0)

0.9 0.055 0.0250.15 0.8 0.050.25 0.25 0.05

1

= (0.15, 0.8, 0.05)

and

x5 = (0, 1, 0)

0.9 0.055 0.0250.15 0.8 0.050.25 0.25 0.05

5

= (0.48, 0.45, 0.07) �



Examples

Solve, in the traditional way ( using elementary operations -row reduction method ), each of the following systems

i)

x1 + x2 + x3 + x4 + x5 = 2x1 + x2 + x3 + 2x4 + 2x5 = 3x1 + x2 + x3 + 2x4 + 3x5 = 2

ii)

x1 + 2x2 + x3 = 12x1 + 4x2 + 2x3 = 3



Examples

Solution

i) Using row-reduction to the augmented matrix

1 1 1 1 1 21 1 1 2 2 31 1 1 2 3 2

⇒ 1 1 1 1 1 2

0 0 0 1 1 10 0 0 1 2 0

⇒ 1 1 1 1 1 2

0 0 0 1 1 10 0 0 0 1 −1

⇒ 1 1 1 0 0 1

0 0 0 1 1 10 0 0 0 1 −1

⇒ 1 1 1 0 0 1

0 0 0 1 0 20 0 0 0 1 −1



Examples

Thus, we have that the solution is

x5 = −1, x4 = 2, x3 = α, x2 = β, x1 = 1− α− β

x =

x1x2x3x4x5

=

1− α− β

βα2−1

=

1002−1

+α

−10100

+β

−11000

Thus, the solution is 2− dimensional subspace span by vectorsv1 = (1, 0, 1, 0, 0)T and v2 = (−1, 1, 0, 0, 0, )T



Examples

Determine whether the following are linear transformations fromP2 to P3.

a) L(p(x)) = xp(x)

(b) L(p(x)) = x2 + p(x)

(c) L(p(x)) = p(x) + xp(x) + x2p′(x)

Solution

a) For a and b real numbers, we have:

L(ap(x) + bq(x)) = x [ap(x) + bq(x)] = x [ap(x)] + x [bq(x)] =

a[xp(x)] + b[xq(x)] = aL(p(x)) + bL(q(x))

so, the Transformation is linear.



Examples

b) For a and b real numbers, we have:

L(ap(x) + bq(x)) = x2 + [ap(x) + bq(x)] 6=

x2 + [ap(x)] + x2 + [bq(x)] = aL(p(x)) + bL(q(x))

so, the Transformation is not linear.

c) For a and b real numbers, we have:

L(ap(x) + bq(x)) = (ap(x) + bq(x)) + x [ap(x) + bq(x)]+

x2[ap(x) + bq(x)]′ = a[p(x) + xp(x) + p′(x)]+

b[q(x) + xq(x) + q′(x)] = aL(p(x)) + bL(q(x))

so, the Transformation is Linear.



Examples

Let

b1 =

110

, b1 =

101

, b3 =

011

,

and let L be the linear transformation from R2 into R3 defined by

L(x) = x1b1 + x2b2 + (x1 + x2)b3

Find the matrix A representing L with respect to the ordered bases{e1, e2} and {b1,b2,b3}.



Examples

Solution

L(e1) = b1 + b3 = 1 · b1 + 0 · b2 + 1 · b3

L(e2) = b2 + b3 = 0 · b1 + 1 · b2 + 1 · b3

Thus, the matrix AL representing the transformation w.r.t. thegiven bases is

AL =

1 00 11 1

�



Examples

A linear transformation L : V→W is said to be one-to-one ifL(v1) = L(v2) implies that v1 = v2 (i.e., no two

distinct vectors v1, v2 in V get mapped into the same vectorw ∈W ). Show that L is one-to-one if and only if

ker(L) = {0V}.

Solution

if L is one to one, then

L(v1) = L(v2)⇒ v1 = v2

Now, if v ∈ ker(L) then L(v) = 0W, but we already know thatL(0V) = 0W ⇒ v = 0 for all v ∈ ker(L)⇒ ker(L) = {0V}



Examples

On the other hand, let’s assume that ker(L) = {0V}. Suppose that

L(v1) = L(v2)⇒

L(v1)− L(v2) = 0W ⇒

L(v1 − v2) = 0W ⇒

v1 − v2 ∈ ker(L)⇒

v1 − v2 = 0V ⇒

v1 = v2 ⇒

L is a one-to-one linear transformation. �



Examples

Let V be a subspace of F(R) spanned by functions ex , e−x . Let Lbe a linear operator on V such that

AL =

(2 −1−3 2

)is the matrix of L relative to the basis ex , e−x . Find the matrix ofL relative to the basis

cosh(x) =1

2(ex + e−x), sinh(x) =

1

2(ex − e−x)

Solution

From the above we know that

L(ex) = 2ex − 3e−x ; L(e−x) = −ex + 2e−x



Examples

and from the equations

cosh(x) = 12(ex + e−x), sinh(x) = 1

2(ex − e−x)⇒

we have

ex = cosh(x) + sinh(x) e−x = cosh(x)− sinh(x)⇒

L(ex) = 2ex−3e−x = 2(cosh(x)+sinh(x))−3(cosh(x)−sinh(x))⇒

L(ex) = 2ex − 3e−x = − cosh(x) + 5 sinh(x))

and

L(e−x) = −ex + 2e−x =−(cosh(x) + sinh(x)) + 2(cosh(x)− sinh(x))⇒

L(e−x) = −ex + 2e−x = cosh(x)− sinh(x)⇒



Examples

L(cosh(x)) = L(12(ex + e−x)) = 12L(ex) + 1

2L(e−x)⇒12 [− cosh(x) + 5 sinh(x)] + 1

2 [cosh(x)− sinh(x)]

L(cosh(x)) = 2 sinh(x)

Similarly

L(sinh(x)) = L(12(ex − e−x)) = 12L(ex)− 1

2L(e−x)⇒12 [− cosh(x) + 5 sinh(x)]− 1

2 [cosh(x)− sinh(x)]

L(sinh(x)) = − cosh(x) + 3 sinh(x)

and the matrix BL, representing L in this new basis is

BL =

(0 −12 3

)�



Examples

Let L be a linear operator on Rn. Suppose that L(x) = 0 for somex 6= 0. Let A be the matrix representing L with respect to thestandard basis {e1, e2, . . . , en}. Show that A is singular.

Solution

A be the matrix representing L w.r.t. to standard basis, then theequation

L(x) = 0⇒ Ax = 0

is valid for a vector x 6= 0, therefore system of the equations

Av = 0

has not a unique solution, namely, v = 0 and v = x , so thatimplies that A is a singular matrix, otherwise the system wouldhave a unique solution, which is not the case. �



Examples

let

A =

1 2 10 3 10 5 −1

(i) Find all eigenvalues of the matrix A.

(ii) For each eigenvalue of A, find an associated eigenvector.

(iii) Is the matrix A diagonalizable? Explain.

(iv) Find all eigenvalues of the matrix A2.

Solution

(i) Eigenvalues are given by the characteristic equation:



Examples

|A− λI| =

∣∣∣∣∣∣1− λ 2 1

0 3− λ 10 5 −1− λ

∣∣∣∣∣∣ = 0⇒

(1− λ)

∣∣∣∣ 3− λ 15 −1− λ

∣∣∣∣ = (1− λ) [(3− λ)(−1− λ)− 5] =

(λ− 1)(λ− 4)(λ+ 21) = 0

Thus the eigenvalues are: λ1 = 1, λ2 = 4, λ3 = −2 .

(ii) An eigenvector corresponding to the eigenvalue λ1 = 1 is givenby

(A− λ1I)x =

1− λ1 2 10 3− λ1 10 5 −1− λ1

x1x2x3

=



Examples

0 2 10 2 10 5 −2

x1x2x3

=

000

Thus, an eigenvector is

v1 =

100

=

An eigenvector corresponding to the eigenvalue λ2 = 4 is given by

(A− λ2I)x =

1− λ2 2 10 3− λ2 10 5 −1− λ2

x1x2x3

=



Examples

−3 2 10 −1 10 5 −5

x1x2x3

=

000


v2 =

111

=

An eigenvector corresponding to the eigenvalue λ3 = −2 is givenby

(A− λ2I)x =

1− λ3 2 10 3− λ3 10 5 −1− λ3

x1x2x3

=



Examples

3 2 10 5 10 5 1

x1x2x3

=

000


v3 =

11−5

Thus, the matrix A has three linearly independent vectors, namely:v1, v2, v3.



Examples

(iii) The matrix A is diagonalizable since

A = XDX−1 =

1 1 10 1 10 1 5

1 0 00 4 00 0 −2

1 1 10 1 10 1 5

−1

(iii) The matrix A2 is given by

A2 =(XDX−1

) (XDX−1

)⇒

A2 = XD2X−1

whereD2 =

1 0 00 16 00 0 4

so the eigenvalues of A2 are 1, 16, 4 �



Examples

Find the kernel and range of each of the following linear operatorson P3 :

(a) L(p(x)) = xp′(x)

(b) L(p(x)) = p(x)− p′(x)

(c) L(p(x)) = p(0)x + p(1)

Solution

ker(L) is the set of all polynomials such that L(p(x)) = 0

a)

L(p(x)) = xp′(x) = 0⇒ p′(x) = 0⇒ p(x) = a0 = cte.



Examples

b)

L(p(x)) = p(x)− p′(x) = 0⇒ p′(x) = p(x)⇒ p(x) = 0 is theonly polynomial satisfying that eqation.

c)

L(p(x)) = p(0)x + p(1) = 0⇒ p(0) = 0 and p(1) = 0

Thus, the kernel is the set of all quadratic polymomials satisfayingthe above condition, that is p(x) = α(x − x2). �



Examples

Find the singular value decomposition of each of the followingmatrices:

i)

A =

(1 12 2

)ii)

B =

(2 −21 2

)Solution

i)

The matrix ATA is given by :

ATA =

(5 55 5

)Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

whose eigenvalues and eigenvectors are: λ1 = 10, λ2 = 0

v1 =

(11

); v2 =

(−11

)

The matrix is AAT :

AAT =

(2 44 8

)whose eigenvalues and eigenvectiors are λ1 = 10 and λ2 = 0

u1 =

(12

); u2 =

(−21

)Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

Thus the SVD decomposition of the matrix A

A = UΣVT =

(1 −22 1

)( √10 00 0

)(1 −11 0

)T

ii)

The matrix BTB is given by:

BTB =

(5 −2−2 8

)whose eigenvalues and eigenvectors are: λ1 = 9, λ2 = 4

v1 =

(−12

); v2 =

(21

)



Examples

The matrix is BBT :

BBT =

(8 −2−2 5

)whose eigenvalues and eigenvectiors are λ1 = 9 and λ2 = 4

u1 =

(−21

); u2 =

(12

)Thus the SVD decomposition of the matrix B

B = UΣVT =

(−2 11 2

)(3 00 2

)(−1 22 1

)T



Examples

Given x1 = 12(1, 1, 1,−1) and x2 = 1

6(1, 1, 3, 5)verify that these vectors form an orthonormal set in R4. Extendthis set to an orthonormal basis for R4 by finding an orthonormalbasis for the null space of

A =

(1 1 1 −11 1 3 5

)[Hint: First find a basis for the null space and then use theGram-Schmidt process.]

Solution< x1, x2 >= 1

12 [1 + 1 + 3− 5] = 0

< x1, x1 >= 1 < x2, x2 >= 1

Therefore {x1, x1} form an orthonormal set.Dr. Marco A Roque Sol Linear Algebra. Session 11


Examples

Now, let’s find the Null Space of the matrix A, i. e. , solve thesystem

Av = 0⇒(

1 1 1 −11 1 3 5

)v1v2v3v3

=

(00

)Using row-reduction we have

Av = 0⇒(

1 1 1 −1 01 1 3 5 0

)⇒(

1 1 1 −1 00 0 2 6 0

)⇒

(1 1 1 −1 00 0 1 3 0

)⇒(

1 1 0 −4 00 0 1 3 0

)⇒



Examples

(1 1 0 −4 00 0 1 3 0

)So, the solution is given by

v1 = 4α− β, v2 = β, v3 = −3α, v4 = α

Thus, a basis for the Null space is v1 = (−1, 1, 0, 0) andv2 = (4, 0,−3,−1)

and using the Gram-schmidt process we come up with

u1 = v1||v1||

u2 = v2 − <v2,u1><u1,u1>

u1/||v2 − <v2,u1><u1,u1>

u1||

u1 = 1√2

(−1, 1, 0, 0) and u2 =√26 (2, 2,−3, 1) �



Examples


Linear Algebra. Session 11 - math.tamu.eduroquesol/Math_304...Dr. Marco A Roque Sol Linear Algebra....

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Transcript of Linear Algebra. Session 11 - math.tamu.eduroquesol/Math_304...Dr. Marco A Roque Sol Linear Algebra....